Measurable equidecompositions for group actions with an expansion property

Given an action of a group $\Gamma$ on a measure space $\Omega$, we provide a sufficient criterion under which two sets $A, B\subseteq \Omega$ are measurably equidecomposable, i.e., $A$ can be partitioned into finitely many measurable pieces which can be rearranged using the elements of $\Gamma$ to form a partition of $B$. In particular, we prove that every bounded measurable subset of $R^n$, $n\ge 3$, with non-empty interior is measurably equidecomposable to a ball via isometries. The analogous result also holds for some other spaces, such as the sphere or the hyperbolic space of dimension $n\ge 2$.


Introduction
In this paper, we present a general sufficient criterion for equidecomposing a given pair of sets using measurable pieces. In order to state quickly some concrete results obtainable by our method, we discuss first the Euclidean space R n , which is probably the most important special case.

Euclidean space R n
Let us call two subsets A and B of R n (set-theoretically) equidecomposable if it is possible to find a partition of A into finitely many pieces and rearrange these pieces using orientation-preserving isometries to form a partition of B. The most famous result about F1. Grabowski In particular, for n ⩾ 3, every two bounded measurable subsets of R n with non-empty interior and of the same Lebesgue measure are measurably equidecomposable.

Remarks 1.3.
(i) The assumption that n ⩾ 3 is needed in Theorem 1.2. For example, in R 2 (resp. R 1 ), Laczkovich in [23, Theorem 3] (resp. [21,Theorem 3.3]) constructed continuum many Jordan domains with boundaries differentiable everywhere (resp. bounded sets that are countable unions of intervals) that all have measure 1 but no two of these sets are set-theoretically equidecomposable.
(ii) Theorem 1.2 can be combined with the results of Dougherty and Foreman [9] (or recent generalisations by Marks and Unger [30]) to show that if the sets A and B in the theorem also have the property of Baire, then the obtained pieces can be additionally required to have the property of Baire, see Corollary 1.10(iii).
(iii) Various questions were posed early in the 20th century regarding the axiomatic theory of the Lebesgue measure, see e.g. [49,Chapter 11]. A remaining key open problem is whether, for n ⩾ 3, every mean (that is, a finitely additive and isometry invariant function) κ : A → [0, ∞) where A is the family of all bounded Borel subsets of R n , is a constant multiple of the Lebesgue measure. In a breakthrough, Margulis [29] showed that the answer is in the affirmative if we take for A the (larger) family of all bounded measurable sets. This resolved the famous Banach-Ruziewicz Problem whose origins can be traced to the 1904 monograph of Lebesgue [24] (see the discussion in [49,Page 267]).
The special case Ω = R n of our Theorem 1.14 reduces the family A in the result of Margulis to those sets that additionally have the property of Baire. In brief, the connection comes from the observation that an equidecomposition between A and B with all pieces in A gives a certificate that κ(A) = κ(B).
(iv) An upper bound on the number of pieces needed in Theorem 1.2 can be derived from our proof. For example, our estimates indicate that one can measurably equidecompose a ball into a cube in R 3 using at most 20 10 10 pieces, see Remark 7.5. On the other hand, we do not know any non-trivial lower bounds for the number of necessary pieces.

A general criterion for measurable equidecompositions
In this section, we present our general criterion for measurable equidecomposability. We start by noting the following assumption that applies to the whole paper. Assumption 1.5. Assume that µ is a σ-finite measure on (Ω, B) which is non-zero on all non-empty open sets and that the action a : Γ ↷ Ω preserves the measure µ.
We need a few definitions first. The result of the action of γ ∈ Γ on x ∈ Ω is denoted by γ.x := a(γ, x). Similarly, when U ⊆ Ω we put γ.U := {γ.u : u ∈ U }, and when T ⊆ Γ we put T.U := ∪ γ∈T γ.U . We say that a set A ⊆ Ω covers another set B ⊆ Ω if there is a finite set T ⊆ Γ such that B ⊆ T.A.
Let us also define various families of subsets of Ω that we will use. The completion of B with respect to µ is denoted by B µ . Elements of the σ-algebra B µ are called (Lebesgue) measurable. Let T denote the σ-algebra consisting of sets with the property of Baire. Let C consist of all subsets of Ω that have compact closure and cover a non-empty open set. While C is closed under finite unions, it is not an algebra. Clearly, the σ-algebras B and B µ are invariant (under the action a : Γ ↷ Ω). If, for example, Γ acts by homeomorphisms, then T and C are invariant too.
Given an a-invariant set family A ⊆ 2 Ω , two subsets A, B ⊆ Ω are called Aequidecomposable (with respect to the action a : Γ ↷ Ω) if for some m ∈ N there exist group elements γ 1 , ... , γ m ∈ Γ and a partition A = A 1 ⊔ ... ⊔ A m with each piece belonging to A such that γ 1 .A 1 , ... , γ m .A m partition B. If A is equal to B, B µ , T and B µ ∩ T , then an A-equidecomposition is called respectively Borel, measurable, Baire and Baire-Lebesgue. If A = 2 Ω , then we usually omit any reference to A; however, if we need to emphasize that no restriction is imposed on the pieces, then we will use the term set-theoretic equidecomposition. Given µ, we say that A essentially covers B if A covers B \ N for some null set N . If A, B ⊆ Ω are measurable sets and there exist null sets N and N ′ such that A \ N and B \ N ′ are Borel equidecomposable, then we say that A and B are essentially Borel equidecomposable. It is easy to show (see Proposition 3.4(i)) that the existence of an essential Borel equidecomposition together with a set-theoretic equidecomposition is equivalent to the existence of a measurable equidecomposition.
We say that C ∈ B µ is a domain of expansion if 0 < µ(C) < ∞ and for every real η > 0 there is a finite set R ⊆ Γ such that for all measurable sets Y ⊆ C we have Informally speaking, this states that all measurable subsets of C "uniformly expand" inside C under a suitable finite subset R of Γ , unless their R-images cover most of C. This property is crucial in the following general criterion, based on a result of Lyons and Nazarov [27].

Paradoxical actions
Equidecompositions have also been considered for spaces other than the Euclidean space R n , often with the aim of establishing "paradoxes" and concluding that certain kinds of measures do not exist. We refer the reader to the excellent monograph on the subject by Tomkowicz and Wagon [49]. Having a rich family of set-theoretic equidecompositions will be very useful when applying Theorem 1.6(ii). For the purposes of this paper, we make the following (non-standard) definition. (ii) For every γ ∈ Γ and every compact C ⊆ Ω, the closure of γ.C is compact.
(iii) Any two subsets of Ω with compact closure and non-empty interior are equidecomposable.
Of course, the crucial part of this definition is the last property. The other (mild technical) properties will be needed in some of our arguments. For example, Property (ii) implies that the family C (which consists of sets with compact closure that cover a non-empty open set) is invariant. As a small detour, let us note the following general proposition, whose second part relies on a powerful result of Marks and Unger [30]. Proposition 1.8. Let a : Γ ↷ Ω be paradoxical and let A ∈ C. Then the following holds.
(i) A subset B ⊆ Ω is equidecomposable to A if and only if B ∈ C.
(ii) Suppose additionally that A ∈ T and that the family of all meager subsets of Ω is a-invariant. Then a subset B ⊆ Ω is T -equidecomposable to A if and only if B ∈ C ∩ T .
affine bijections of R 2 generated by SL(2, Z) (the linear maps given by 2 × 2 matrices with determinant 1 and all entries in Z) and all translations (that is, maps of the form x → x + u for some vector u ∈ R 2 ). This group naturally acts on R 2 .
Hausdorff [15] showed that a "third" of S n , n ⩾ 2, is equidecomposable to a "half" of S n , which was enough to his intended application, namely, the non-existence of a mean defined on all subsets of the sphere. The paradoxicality of SO(n + 1) ↷ S n , n ⩾ 2, as stated in Definition 1.7(iii), was established by Banach and Tarski [2]. The paradoxicality of Iso(R n ) ↷ R n for n ⩾ 3 is the content of Theorem 1.1 (of Banach and Tarski [2]) while, as we mentioned already, this fails for n ⩽ 2 by the results of Banach [1]. One can ask what happens if we allow a richer group of transformations of R 2 . The paradoxicality of G 2 ↷ R 2 was established in the influential paper of von Neumann [36] that introduced the concept of a non-amenable group. In fact, some (explicit) smaller subgroups of G 2 suffice here, see Mycielski [34,Corollary 5] (compare also with Wagon [51,Theorem 2]).
Laczkovich [22] showed that the natural action SL(2, R) ↷ R 2 \ {0} is paradoxical. (See also Tomkowicz [47] for a strengthening of this result.) As noted by Mycielski [33,Page 143], the paradoxicality of Iso(H n ) ↷ H n , n ⩾ 3, can be established by observing that the subgroup of isometries that fix a point of H n acts on its every non-trivial orbit in the same way as SO(n) acts on S n−1 . Mycielski [33] showed that the action Iso(H 2 ) ↷ H 2 is also paradoxical. (Some small gaps in Mycielski's proof were fixed by Mycielski and Tomkowicz [35]; the proof, with some further modifications, can also be found in [49,Theorem 4.17].) We also refer the reader to Tomkowicz [48] for a general short proof of the paradoxicality of many of the above actions.
Let us collect these five (probably, best known) examples of paradoxical actions together. For later reference, we also state the standard invariant measure µ on Ω in each case. Example 1.9 (Some known paradoxical actions Γ ↷ Ω).

Expanding actions
Let us call an action Γ ↷ Ω satisfying Assumptions 1.4 and 1.5 expanding (with respect to the measure µ) if every set in B µ ∩ C is a domain of expansion. This notion is of interest because of the following corollary that can be derived with some work from Theorem 1.6 and Proposition 1.8. Corollary 1.10. In addition to Assumptions 1.4 and 1.5, assume that the action is paradoxical and expanding. Let A ∈ B µ ∩ C. Then the following statements hold. ( and µ(B) = µ(A). (iii) Suppose additionally that A has the property of Baire and that the action preserves the family of meager sets. Then a subset B ⊆ Ω is Baire-Lebesgue equidecomposable to A if and only if B ∈ B µ ∩ C ∩ T and µ(B) = µ(A).
On the other hand, many natural actions can be shown to be expanding: Theorem 1.11. Each of the actions in Example 1.9 is expanding.
Each these actions is paradoxical and acts by homeomorphisms (in particular, preserving meager sets), so it satisfies all conclusions of Proposition 1.8 and Corollary 1.10. It is remarkable that one can obtain exact characterisations for so many types of equidecompositions under rather general assumptions (that in particular include all actions in Example 1.9). We did not see such a characterisation anywhere in the previous literature, surprisingly not even for set-theoretic equidecompositions in any case of Example 1.9. In fact, Tomkowicz and Wagon [49,Page 176] write: "It is not completely clear which subsets E of S 2 are SO 3 (R)-equidecomposable with all of S 2 ." One answer to this question (exactly those sets that cover the whole of S 2 ) is given by Proposition 1.8(i).
As an illustration, here is one (easy to state) direct consequence of Corollary 1.10(ii) and Theorem 1.11. Corollary 1.12. For each action Γ ↷ Ω of Example 1.9, every two measurable sets A, B ∈ C of the same measure are equidecomposable with measurable pieces.

Connections to (local) spectral gap and finitely additive means
It is not hard to show (see Proposition 5.3) that if µ is a finite measure and the action Γ ↷ Ω has spectral gap, then the action is expanding. Thus the case of SO(n) ↷ S n−1 , n ⩾ 3, of Theorem 1.11 can be derived from the known spectral gap results established by Drinfel'd [10], Margulis [28], and Sullivan [43].
The case of the infinite measure space R n in Theorem 1.11 is not directly covered by the above approach. However, we were able to derive it from the spectral gap of SO(3) ↷ S 2 , using lengthy but rather elementary arguments. Later, we became aware of the general powerful results by Boutonnet, Ioana and Salehi Golsefidy [5] that can be used here. Let us discuss this connection in general.
Let X ⊆ Ω be a measurable set of finite positive measure. For f ∈ L 2 (Ω, µ), we define ∥f ∥ 2,X := ( X |f (x)| 2 dµ(x)) 1/2 . We say that an action Γ ↷ Ω satisfying Assumptions 1.4 and 1.5 has local spectral gap with respect to X if there exist a finite set Q ⊆ Γ and a constant c > 0 such that for any This notion is of interest to us because, as we will show in Lemma 5.2, the action has local spectral gap with respect to a set X ⊆ Ω if and only if X is a domain of expansion. Modulo Lemma 5.2, Boutonnet et al [5, Theorem A] presented a sufficient condition for the action to be expanding (stated here as Theorem 5.1). While it seems that Theorem 5.1 can be used to derive the full Theorem 1.11, we use it here for SL(2, R) ↷ R 2 \ {0} and the hyperbolic space only, presenting more direct proofs of the other cases of Theorem 1.11.
We decided to include also our initial proof of Theorem 1.11 for Iso(R n ) ↷ R n . As we have already mentioned, it is rather elementary, apart from using the spectral gap property of SO(3) ↷ S 2 . Also, it can be used to estimate the number of pieces in the obtained measurable equidecompositions; see Remark 7.5 for an example of a such calculation. Last but not least, our proof also gives the following result that does not seem to follow from [5] nor from other known spectral gap results. Theorem 1.13. For each n ⩾ 3, there is a closed nowhere dense bounded subset X of R n such that µ(X) > 0 and the action Iso(R n ) ↷ R n has local spectral gap with respect to X (i.e., by Lemma 5.2, X is a domain of expansion).
For an a-invariant family A ⊆ 2 Ω which is closed under finite unions, a mean on A is an a-invariant finitely additive function κ : A → [0, ∞). The analogue of the question discussed in Remark 1.3(iii), namely whether every mean on Borel sets with compact closure is a constant multiple of the measure µ, is also open for all actions listed in Example 1.9 to the best of the authors' knowledge. The following theorem provides some partial progress in this direction. Theorem 1.14. Let Γ ↷ Ω be any action from Example 1.9 and let A be the family of all measurable subsets of Ω that have the property of Baire and compact closure. Then every mean on A is a constant multiple of the measure µ.
The result of Margulis [29] (namely, the version of Theorem 1.14 when we take the action Iso(R n ) ↷ R n and enlarge the family A by dropping the requirement that the sets in A have the property of Baire) can be derived by a straightforward modification of our proof. Alternatively, it is a consequence of Theorem 1.11, Lemma 5.2 and the implication (4) =⇒ (1) of [5,Theorem 7.6].

Organisation of the paper
Section 2 contain some further notation and various auxiliary results. Theorem 1.6 is proved in Section 3 where we also give a full proof of the result of Lyons and Nazarov [27] on the existence of a.e.-perfect Borel matchings. Section 4 is dedicated to proving Proposition 1.8 and also contains the derivation of Corollary 1.10.
Section 5 discusses the relation to (local) spectral gap in detail, in particular showing that the action has local spectral gap with respect to X if and only if X is a domain of expansion.
Section 6 contains the proofs of all cases of Theorem 1.11. In particular, Section 6.5 contains a few different proofs that the action Iso(R n ) ↷ R n is expanding for n ⩾ 3, with our new (more elementary) proof appearing in Section 6.5.2. This proof is then used to derive Theorems 1.13 (at the end of Section 6.5.2) and to give upper bounds on the number of pieces in some of our equidecompositions (in Section 7). As we mentioned already, Corollary 1.12 clearly follows from Corollary 1.10 and Theorem 1.11. Theorem 1.14 is proved in Section 8, as a consequence of a more general Lemma 8.1. As described in Section 1.1, all new results stated there are direct consequences of the above results.
Some concluding remarks and remaining open questions can be found in Section 9.

Some further notation and auxiliary results
Let us collect some frequently used notation, also recalling some definitions that already appeared in the Introduction.
Let N := {0, 1, ... } consist of non-negative integers. For k ∈ N, we denote [k] := {1, ... , k}. When we write X = A ⊔ B, we mean that the sets A and B partition X (i.e., A ∪ B = X and A ∩ B = ∅). By π i we will denote the projection from a product to its i-th coordinate; formally, for sets X 1 , ... , X m and i ∈ [m], the projection π i maps (x 1 , ... , x m ) ∈ m j=1 X j to x i ∈ X i . Under Assumption 1.4, we will use the following shorthands for S, T ⊆ Γ , γ ∈ Γ , y ∈ Ω and X ⊆ Ω: We call γ.X a translate of X. The group action and the group multiplication take precedence over all other set operations; for example, S.X ∩ Y means (S.X) ∩ Y . The identity of the group Γ is denoted by e. Also, we write S −1 := {γ −1 : γ ∈ S} and call the set S symmetric if S −1 = S.
By a multiset Q ⊆ Γ we mean a function Q : Γ → N with Q(γ) encoding the multiplicity of γ in Q. It is finite if |Q| := γ∈Γ Q(γ) is finite. For f : Γ → R, we define γ∈Q f (γ) := γ∈Γ Q(γ) f (γ). The closure of X ⊆ Ω in the topological space (Ω, τ ) is denoted by X. The complement of X is X c := Ω \ X. The indicator function 1 X of X assumes value 1 on X and value 0 on Ω \ X.
Two sets A, B ⊆ Ω cover each other if there is a finite set S ⊆ Γ with S.A ⊇ B and S.B ⊇ A. When we talk about equidecompositions for R n , S n , or H n without specifying the group, we mean by default the group of orientation-preserving isometries.
Also, recall these families of subsets of Ω: B (Borel), B µ (measurable), C (having compact closure and covering a non-empty open set) and T (having the property of Baire).
Let G be a (bipartite) graph, by which we mean a triple (V 1 , V 2 , E), where V 1 and V 2 are sets (that are called the parts of G) and E (called the edge set of G) is a subset of A matching in G is a subset M of the edge set E such that for every x in V 1 (resp. V 2 ) there is at most one vertex y with (x, y) ∈ M (resp. (y, x) ∈ M ). For i = 1, 2, the projection π i (M ) consists of matched vertices in V i . It will be sometimes convenient to view a matching as a partial bijection; then π 1 (M ) and π 2 (M ) are just the domain and the range of M . The matching M is called perfect if π 1 (M ) = V 1 and π 2 (M ) = V 2 , that is, M as a function is a bijection from V 1 to V 2 .
For X ⊆ V 1 and Y ⊆ V 2 their neighbourhoods are respectively The degree of a vertex x is deg(x) := |N ({x})|. We call G locally finite if the degree of each vertex is finite.
There may be some ambiguity when V 1 ∩V 2 ̸ = ∅. There are two, essentially equivalent, ways to deal with this formally. The first one is to work with the actionã : Γ ⊕ C 2 ↷ Ω × C 2 instead, where C 2 := ({−1, 1}, ×) is the cyclic group with two elements, Γ acts on each copy of Ω the same way as before while the non-identity element of C 2 swaps these two copies. Then we can replace V i by V i × {(−1) i }, thus making the parts disjoint. However, then we have to make (routine) verifications that our claimed results, when proved forã, transfer to the original action a. Alternatively, we can operate with the unordered graph (2.1) which carries the same information as G. For example, instead of the degree of x ∈ V i we should have, strictly speaking, defined the degree of (x, (−1) i ), etc. Since the meaning will usually be clear from the context, we will be working mostly with G, switching toG occasionally.
We call a bipartite graph G = (V 1 , V 2 , E) Borel if V 1 , V 2 ⊆ Ω and E ⊆ Ω 2 are all Borel. One way to generate Borel subsets of Ω 2 that is relevant to this paper is as follows. A Borel arrow is a pair (U, γ), where U ⊆ Ω is Borel and γ ∈ Γ . Given a countable set of Borel arrows This set is Borel (as the countable union of the graphs of Borel partial functions). A special case is when A = {(Ω, γ) : γ ∈ S} for some countable S ⊆ Γ (that is, each U i is equal to Ω); then we denote which is just the union over γ ∈ S of the graphs of the functions a(γ, ·) : Ω → Ω.
We will be using the Lusin-Novikov Uniformisation Theorem (see e.g. [17, Theorem 18.10]) a number of times, often without explicitly mentioning it. We need only the weaker form of the theorem which states that, for Polish spaces, a continuous countablyto-one image of a Borel set is Borel. For example, one of its consequences is that for every Borel graph G = (V 1 , V 2 , E) with countable vertex degrees and every Borel X ⊆ V i its neighbourhood N (X) is Borel. Indeed, if e.g. i = 1 then N (X) = π 2 ((X × V 2 ) ∩ E) is the countable-to-one image of a Borel set under the (continuous) projection π 2 . Another useful consequence of the Uniformisation Theorem is as follows.
Lemma 2.1. Under Assumptions 1.4, let S ⊆ Γ be a countable subset and let E ′ ⊆ E S be Borel. Then there are Borel sets U γ ⊆ Ω for γ ∈ S such that A := {(U γ , γ) : γ ∈ S} bijectively generates E ′ (meaning that E ′ = E(A) and for every (x, y) ∈ E ′ there is exactly one γ ∈ S with x ∈ U γ and y = γ.x).
Proof. In brief, we have to pick in a Borel way, for every edge of E ′ , exactly one element of S that generates this edge. Such a selection can be obtained by taking a maximal Borel independent set, which exists by a result of Kechris, Solecki and Todorcevic [18,Proposition 4.2], in the (non-bipartite) Borel graph with vertex set {((x, y), γ) ∈ E ′ × S : γ.x = y} whose two vertices are connected if they correspond to the same edge of E ′ .
For completeness, we include a more direct proof.
Let ≼ be a linear ordering of S coming from some injection of S into N. We construct the required sets U γ one by one by taking the maximal possible set given the previous sets. Namely, we inductively define U γ for γ ∈ S as The obtained sets can be shown to be Borel by induction on γ ∈ S. For example, each auxiliary set is Borel by the Lusin-Novikov Uniformisation Theorem. By definition, E ′ ⊇ E(A), where A := {(U γ , γ) : γ ∈ S}. Also, for every (x, y) ∈ E ′ there is exactly one γ ∈ S with x ∈ U γ and y = γ.x, namely, the smallest element of {γ ∈ S : y = γ.x}. (Note that this set is non-empty since E ′ ⊆ E S .) It follows that A has all required properties.
The following result points out a well-known connection between equidecompositions and graph matchings. Proof. For every γ ∈ S, let A γ consist of those x ∈ A such that (x, γ.x) ∈ M and γ is the smallest element of S with this property (under a fixed ordering of S). Since M is a matching, the sets A γ partition A and their translates γ.A γ partition B. Thus the sets A and B are equidecomposable using elements from S only. If, moreover, the matching M is Borel, then all pieces A γ are Borel (which can argued similarly as in the proof of Lemma 2.1).
Let us call the action Γ ↷ Ω minimal if there is no closed subset X ⊆ Ω such that X is proper (i.e., X ̸ = ∅ and X ̸ = Ω) and Γ.X = X. Lemma 2.3. Under Assumption 1.4, suppose that the action Γ ↷ Ω is paradoxical. Then it is minimal. Also, if A ∈ C is equidecomposable to some set B, then B ∈ C. Also, any two sets A, B ∈ C cover each other.
Proof. Suppose that a closed proper subset X ⊆ Ω satisfies Γ.X = X. Pick some x ∈ X and y ∈ Ω \ X. By the local compactness of Ω, choose open sets U ∋ x and W ∋ y with compact closures where, additionally, we can assume that W ∩ X = ∅. Then Γ.W , as a subset of the invariant set Ω \ X, does not contain x ∈ U . Thus the sets U, W ∈ C are not equidecomposable, contradicting the paradoxicality of the action.
Next, suppose that B ⊆ Ω is equidecomposable to A ∈ C. The set B has to cover A and, by the transitivity of the covering relation, B also covers some non-empty open set. On the other hand, B is covered by finitely many copies of A. Since the closure of A is compact and the action of each element of Γ preserves this property by Definition 1.7(ii), the closure of B can be covered by finitely many compact sets and so it is compact itself. Thus B ∈ C as required.
Finally, let A, B ∈ C be arbitrary. By the definition of C, the sets A and B cover some non-empty open sets U and W respectively. By shrinking U and W , we can additionally assume that they have compact closures. By the paradoxicality of the action, A ∪ U is equidecomposable to B ∪ W , from which it easily follows that A and B cover each other.
Let Assumption 1.5 apply to the rest of this section. In particular, the action Γ ↷ Ω preserves the measure µ on (Ω, B). The following reformulation of the measure preservation property will be useful to us. Lemma 2.4. Under Assumptions 1.4 and 1.5, let Λ be a countable subgroup of Γ , let C ⊆ Ω be Borel, and let ψ : C → Ω be a Borel injective map such that, for every x ∈ C, we have ψ(x) ∈ Λ.{x}. Then ψ(C) is a Borel set and µ(ψ(C)) = µ(C).
Proof. Of course, the first claim (that ψ(C) is Borel) follows from the Lusin-Novikov Uniformisation Theorem; we also get it as a by-product of our proof of the second claim.
We say that some property holds almost everywhere (a.e. for short) if the set of x ∈ Ω where it fails is a null set with respect to the measure µ. For example, two sets A, B ⊆ Ω essentially cover each other if there is a finite S ⊆ Γ with S.A ⊇ B and S.B ⊇ A a.e.
Recall that a measurable set C ⊆ Ω is a domain of expansion if 0 < µ(C) < ∞ and for every real η > 0 there is a finite set R ⊆ Γ such that for all measurable sets Y ⊆ C we have Such a set R will be called η-expanding for C.  Proof. Let us assume, for example, that A is a domain of expansion. Let T ⊆ Γ be a finite set such that A ⊆ T.B and B ⊆ T −1 .A a.e. Let t := |T |. By the invariance of the measure, we have that 0 < µ(A)/t ⩽ µ(B) ⩽ t µ(A) < ∞. Let ε > 0 be arbitrary. Let η := ε/t 2 and let S be an η-expanding set for A. We claim that T −1 ST is an ε-expanding set for B. Indeed, let Y ⊆ B be a measurable set. We have that Y ⊆ T −1 .A a.e., so there exists γ ∈ T such that µ(γ. If µ(A \ ST.Y ) ⩽ η µ(A), then we have by the choice of T that Otherwise, we have µ(ST.Y ∩ A) ⩾ µ(Y )/(ηt). Using that A ⊆ T.B a.e., we deduce that for some γ ∈ T we have µ(ST.Y ∩ γ.B) ⩾ µ(Y )/(ηt 2 ), and so Remark 2.6. Recall that we defined the action Γ ↷ Ω to be expanding if every C ∈ B µ ∩ C is a domain of expansion. It follows from Lemmas 2.3 and 2.5 that, for paradoxical actions, it is enough to check just one arbitrary member of B µ ∩ C.
We call a bipartite Borel graph G = (A, B, E) a bipartite c-expander if 0 < µ(A) = µ(B) < ∞ and, for every measurable (equivalently, Borel) subset Y of A or B, it holds that 3 Proof of Theorem 1.6

Augmenting paths
First, we need to define some graph-theoretic concepts adopted to our setting. Let G = (A, B, E) be a bipartite graph and let M ⊆ E be a matching in G.
An alternating path (starting at A) is a non-empty sequence of points P = (x 0 , ... , i is even. Since M is a matching and x 0 ∈ A is unmatched, Item (ii) implies that the evenindexed vertices x 0 , x 2 , ... ∈ A are also pairwise distinct. This definition works also when A ∩ B ̸ = ∅, since the parity of the position in P determines the part a vertex is assigned to. If we are to work with the unordered graphG as defined in (2.1), then the corresponding definition is that is a path inG that starts with an unmatched vertex and whose edges alternate betweeñ E \M andM . The length of P is ℓ, the number of edges.
An augmenting path (starting at A) is an alternating path P = (x 0 , ... , x ℓ ) of odd length ℓ ⩾ 1 such that x ℓ ∈ B \ π 2 (M ). The augmentation of M along P is the matching M ′ := M △ E(P ) which is obtained by taking the symmetric difference between M and In order words, we modify M by including (x i−1 , x i ) for all odd i ∈ [ℓ] and removing (x i , x i−1 ) for all even i ∈ [ℓ]. Note that the augmented matching M ′ covers two new vertices: . Then a Borel augmenting family is a tuple (U, β 1 , ... , β ℓ ), where U is a Borel subset of A and β 1 , ... , β ℓ are elements of R (with repetitions allowed) so that (i) for every x ∈ U the sequence is an augmenting path for M ; (ii) for distinct x, y ∈ U the corresponding augmenting pathsP x,β 1 , ... ,β ℓ andP y,β 1 , ... ,β ℓ inG are vertex-disjoint. Informally speaking, a Borel augmenting family is a Borel collection of vertex-disjoint augmenting paths inG. The length of such a family is ℓ, the number of edges in each path. The augmentation M ′ of M along a Borel augmenting family (U, β 1 , ... , β ℓ ) is obtained from M by augmenting along the paths in (3.1) for all x ∈ U . (Since the paths are vertex-disjoint inG, all these augmentations can be done in parallel.) Note that M ′ is a matching, By swapping the roles of A and B, we can define in the obvious way an alternating path starting at B. Note that there is no need to define an augmenting path starting at B: by reversing the order of vertices we can consider augmenting paths starting at A only.

Proof of the result of Lyons and Nazarov on measurable matchings
In this section we provide all details of the result of Lyons and Nazarov [27, Remark 2.6] (stated as Theorem 3.3 here) which gives a sufficient condition for the existence of a Borel a.e.-perfect matching. We need the following result first, which does not use measure and, in fact, was proved in [13] for arbitrary (not necessarily bipartite) Borel graphs. . ⊆ E such that (i) for every i ⩾ 1, the matching M i admits no augmenting path of length at most 2i − 1; (ii) for every i ⩾ 0, there is a countable sequence of Borel matchings K 0 , K 1 , ... ⊆ E such that (a) K 0 = M i ; (b) for every j ⩾ 1, K j is the augmentation of K j−1 along some Borel augmenting family of length at most 2i + 1; (c) every (x, y) ∈ E eventually belongs to either all or none of the matchings Proof. The first item is [13, Proposition 1.1]. The second item follows directly from the proof of [13, Proposition 1.1]. In brief, the proof, when adopted to our setting, proceeds as follows.
Suppose that i ⩾ 0 and we have already constructed matchings M 0 , ... , M i . By a result of Kechris et al [18], there is a Borel vertex colouring ϕ : where each x j belongs to X := ∪ i+1 t=1 R 2t−1 × N so that each element of X appears as x j for infinitely many choices of j.
Each augmentation as above that flips any given edge (x, y) ∈ E strictly decreases the set of currently unmatched vertices inG at distance at most 2i from the edge. This set is finite, e.g. by König's Infinity Lemma. Thus (x, y) can be flipped only finitely many times, that is, it eventually belongs to all or none of the matchings K j . It follows that M i+1 := ∪ ∞ j=1 ∩ ∞ t=j K t is a Borel matching that does not admit any augmenting path of length at most 2i+1 as every potential path was considered for augmentation for infinitely many values of j. [16] states that if a matching K admits no augmenting path of length at most 2i − 1 and we flip an augmenting path of length 2i + 1 (that is, shortest possible), then no new augmenting paths of length at most 2i + 1 appear. Thus, when we construct M i+1 from M i in the proof of Lemma 3.1, it is enough in fact to take for X just some enumeration of R 2i+1 × N.

Remark 3.2. A result of Hopcroft and Karp
that is, M consists of edges that are eventually included in every M i . Clearly, M is a Borel matching. Thus it remains to show that the set of vertices not matched by M has measure 0. As we will see, this will be a consequence of Claims 3.3.1 and 3.3.6 via a Borel-Cantelli-type argument.
vertices not matched by M i . By Lemma 2.4, the sets π 1 (M ) and π 2 (M ) (and thus the sets X i and Y i ) have the same measure. Note that the set π 1 (M i △ M i+1 ) consists of those x ∈ A on which M i and M i+1 differ as partial functions (in particular, it includes those x ∈ A for which exactly one of these functions is defined).
Proof of Claim. In brief, when we construct M i+1 from M i , each individual augmentation that matches an extra point of A changes the current matching at at most i + 1 elements of A so the claim follows from the invariance of the measure µ.
Let us give a full proof. Given i, .. ⊆ E be the Borel matchings returned by the second part of Lemma 3.1.
Take j ⩾ 1 and let (U j , γ 1 , ... , γ 2ℓ+1 ) with 0 ⩽ ℓ ⩽ i be the Borel augmenting family for K j−1 that augments it into K j . Since each augmentation increases the set of matched vertices and thus x. Then ψ 0 , ... , ψ ℓ are Borel injections whose images partition π 1 (K j △ K j−1 ). By Lemma 2.4, we have Again, note that augmentations can only increase the set of matched vertices. Thus, for The claim is proved.
Fix i ⩾ 1. For integer j ⩾ 0, let X i j consist of the end-points of alternating paths for M i (that start at A) whose length is at most j and has the same parity as j. Since i is fixed (until (3.3)), we abbreviate X j := X i j . Since, for each x ∈ X i , the length-0 path (x) is alternating, we have in particular that X 0 = X i . As the graph is bipartite, it holds that , the matching M i gives a bijection between X ′ 2j and X ′ 2j−1 .
Proof of Claim. Let x be any element of X ′ 2j ⊆ A. By definition, there is an alternating path (x 0 , ... , x 2j ) with x 0 ∈ X 0 and x 2j = x. This path has even length, so (x, y) ∈ M i , where y := x 2j−1 . The truncated alternating path (x 0 , ... , x 2j−1 ) shows that y ∈ X 2j−1 . Suppose, on the contrary to the claim, that y ̸ ∈ X ′ 2j−1 . Then y ∈ X 2j−3 . Let this be witnessed by some alternating path Q of length at most 2j − 3. The odd-length path Q ends with an edge in E \ M i . Either Q already contains x or we can extend Q by adding x, in each case obtaining a contradiction to x ̸ ∈ X 2j−2 .
Conversely, take any y in : otherwise an even-length alternating path Q witnessing this has to end with the pair (x, y) ∈ M i (as y is the unique M i -match of x) and a truncation of Q shows that y ∈ X 2t−1 ⊆ X 2j−3 , a contradiction. In particular, x cannot occur in P as an even-indexed vertex. Thus we can add x to P , obtaining an alternating path which shows that x ∈ X 2j . We have already argued that x ̸ ∈ X 2j−2 . Thus we conclude that x ∈ X ′ 2j , as desired.
Recall that X j is a subset of A (resp. B) if j is even (resp. odd).
Proof of Claim. The inclusion X 2j+1 ⊆ N (X 2j ) is clear. For the other direction, take arbitrary x ∈ X 2j and y ∈ N ({x}) ⊆ B. Pick an alternating path P of length at most 2j starting at X 0 and ending in x. Suppose that y does not belong to P , as otherwise a truncation of P shows that y is in X 2j−1 ⊆ X 2j+1 , giving the required. In particular, we have that (x, y) ̸ ∈ M i as otherwise y precedes x on P . Thus we can extend P by adding y. This shows that y ∈ X 2j+1 , as desired.
Claims 3.3.2 and 3.3.3 show by induction on j ∈ [i] that the set X ′ j is Borel. (In fact, this is true for every j since X ′ j is defined by a local rule; however the stated range of j will suffice for our purposes.) .

Proof of Claim. Claim 3.3.2 implies by induction on
The stated lower bound on µ(X 2j−1 ) is a direct consequence of Claim 3.3.3 and the bipartite expansion property of G assumed by the theorem.
For j ∈ N, define Y j to consist of the end-points of alternating paths which start in Y 0 := Y i whose length is at most j and has the same parity as j. (This becomes the same definition as that of X j when we swap the roles of the sets A and B.) By symmetry, Claim 3.3.4 also holds when we swap A and B, and replace each X t by Y t .
Proof of Claim. Assuming the contrary, pick alternating paths P = (x 0 , ... , x ℓ ) and Q = (y 0 , ... , y k ) starting at A and B respectively (thus The minimality of ℓ+k implies that the pathsP andQ inG intersect only in their endpoints. Also, if ℓ is odd (resp. even), then exactly one of the edges ( . Therefore the concatenation of P with the reversal of Q at the common end-point, that is, (x 0 , ... , x ℓ , y k−1 , ... , y 0 ), is an augmenting path of length at most 2i − 1, contradicting the choice of M i .
Proof of Claim. Suppose first that i is odd. By Claim 3.3.5, the sets X i−1 and Y i are disjoint. As both are subsets of A, at least one of these two sets, suppose X i−1 , has measure at most µ(A)/2. By Claim 3.3.4, the measures of the sets X 0 ⊆ X 2 ⊆ ... ⊆ X i−1 increase by factor at least 1 + c each time. Thus, by X 0 = X i , we have giving the required. The obvious modifications of this proof (swapping the roles of A and B if needed) also apply to the remaining cases (when µ(Y i ) ⩽ µ(A)/2 or when i is even), giving the stated bound. Now we are ready to finish the proof of the theorem (essentially by applying the Borel-Cantelli Lemma). Indeed, π 1 (M ) contains every vertex of A which matched by M i and on which all matchings M j with j ⩾ i agree. Thus, for every i ∈ N, we have The last series is summable by Claims 3.3.1 and 3.3.6. Furthermore, Claim 3.3.6 also gives that µ(X i ) tends to 0. Since we can pick an arbitrarily large i, we have that µ(A \ π 1 (M )) = 0, which finishes the proof of Theorem 3.3.

Finishing the proof of Theorem 1.6
Proof of Theorem 1.6(i). The only non-trivial part is to show the converse direction, namely that if A is a domain of expansion (in particular, 0 < µ(A) < ∞), A and B essentially cover each other, and µ(B) = µ(A), then A is essentially Borel equidecomposable to B. By removing null sets, we can assume that A and B are Borel. Choose a finite symmetric set T ⊆ Γ such that T.A ⊇ B and T.B ⊇ A a.e. By Lemma 2.5, B is also a domain of expansion. Let t := |T |. Fix η > 0 with 2tη < 1. Take a finite subset R ⊆ Γ which is η-expanding for both A and B.
Let us we verify that the Borel graph G := (A, B, E T R∪(T R) −1 ∩ (A × B)) satisfies the bipartite expansion condition (2.5) with c := 1. Let Y be a measurable subset of, say, A. Since R is η-expanding for A, at least one of the following two alternatives hold. If Thus, G is indeed a bipartite 1-expander. Now, by Theorem 3.3, G contains a Borel matching M such that the unmatched sets A \ π 1 (M ) and B \ π 2 (M ) have measure 0. By Lemma 2.2(ii), the matched sets π 1 (M ) ⊆ A and π 2 (M ) ⊆ B are Borel equidecomposable, as desired.
Let us state one step needed in the proof of Theorem 1.6(ii) as Part (i) of the following auxiliary proposition. Proof. We will prove both items at the same time. Pick null sets N A and N B such that Let Λ be the subgroup of Γ generated by {γ 1 , ... , γ m , δ 1 , ... , δ n } and let N := Λ.N ′ . Since Λ is countable and N ′ is a Borel null set, we have that N also is a Borel null set. Furthermore, by the Λ-invariance of the set N , we have that It follows that A and B are equidecomposable using the partition and the group elements γ 1 , ... , γ m , δ 1 , ... , δ n . Informally speaking, we use the pieces V i on N and the pieces U i on the complement of N . It is clear that the sets U i \ N are Borel (in particular they are measurable and have the property of Baire). Furthermore, the sets V i ∩ N are contained in the null set N , so they are measurable. This finishes the proof of the first part.
Additionally, if the sets V i have the property of Baire then the sets V i ∩ N have the property of Baire as well since N is Borel. This observation finishes the proof of the second part.
Proof of Theorem 1.6(ii). Let us show the converse direction (as the forward direction is trivial).
Since A and B are equidecomposable, they also cover each other. By the converse direction in Theorem 1.6(i) that we have already proved, these sets are essentially Borel equidecomposable. Thus Proposition 3.4(i) gives the required measurable equidecomposition of A and B.

Proof of Proposition 1.8
There was some freedom as to how to define when an action is paradoxical. Our Definition 1.7(iii) is of the same form as the known paradoxes are usually stated. We had to add some further properties, namely, the local compactness (to ensure that the family of sets covered by the above statement is rich enough) and the preservation of compact closures (a rather weak restriction which holds, for example, if the whole space Ω is compact or the group acts by homeomorphisms). Note that the latter property implies that the family C is invariant. (Recall that C consists of those subsets of Ω that cover a non-empty open set and have compact closure.) The proof of Proposition 1.8 will occupy this section. We will need the Banach-Schröder-Bernstein Lemma (see e.g. [49,Theorem 3.6]), stated in terms of equidecompositions.
and B is equidecomposable to some A ′ ⊆ A, then A and B are equidecomposable.
Also, it will be convenient to use the semigroup of equidecomposability types S that was introduced by Tarski [45]. For more details, see [49, Chapter 10] whose presentation we follow. Informally speaking, we consider all multi-subsets of Ω of bounded multiplicity, identified under the appropriately defined equidecomposability relation. Formally, let P * (Ω) be the family of all subsets of Ω * := Ω × N whose projection on N is finite. Let Γ * be the direct product of Γ and the group S N of all permutations of N. Define its action a * on Ω * by a * ((γ, σ), (x, n)) := (a(γ, x), σ(n)) (that is, Γ * acts component-wise). For . In brief, its proof proceeds by considering the bipartite graph G whose parts are the sets Y, Z ⊆ Ω × N and whose edge set corresponds to a fixed witness of n[Y ] ≼ n[Z]. Thus every vertex in Y (resp. Z) has G-degree exactly n (resp. at most n). A simple double-counting of edges shows that every finite subset X of Y has at least |X| neighbours in Z. Now, Rado's theorem [38] (that uses the Axiom of Choice) gives that G has a matching covering all elements of Y and thus Y ≼ Z, as claimed.
Thus Finally, it remains to show that arbitrary A, B ∈ C are equidecomposable. By Lemma 2.3 we know that A and B cover each other. As we proved in the previous paragraph, each element U ∈ C can be doubled. By Lemma 2.3, each of the two obtained copies of U necessarily belongs to C. Thus by induction, for any integer n ⩾ 2, there is a partition A = A 1 ⊔ ... ⊔ A n such that each A i is equidecomposable to A (and belongs to C). By choosing n such that n copies of A cover B, one can conclude that B is equidecomposable to a subset of A. The same also holds when we swap the roles of A and B. Now the desired conclusion follows from Lemma 4.1.
In order to establish the second part of Proposition 1.   [30]). Suppose that G = (V 1 , V 2 , E) is a locally finite bipartite Borel graph on a Polish space such that Then there is a Borel matching M in G such that the set of unmatched vertices in each part is meager.
Proof of Proposition 1.8(ii). Again, the forward implication is trivial so we prove the converse direction only. Let A, B ∈ C ∩ T be arbitrary. By Part (i) of the proposition, that we have already proved, these sets are equidecomposable. Let S ⊆ Γ be a finite set that suffices for this equidecomposition. By enlarging S, assume that S −1 = S ∋ e. Also, as we have showed in the proof of Part (i), every set in C can be set-theoretically doubled. By repeating, we can find a finite symmetric set T ∋ e such that the relations 2|S| This graph satisfies (4.3). Indeed, take some finite set X, say X ⊆ A. By the choice of T , the set Y := T.X ∩ A contains 2 |S| disjoint copies of X and thus |Y | ⩾ 2 |S| |X|.
has at least 2 |X| elements, as claimed.
We cannot apply Theorem 4.3 yet, as the sets A and B need not be Borel. So we proceed as follows. Find Borel sets A ′ ⊆ A and B ′ ⊆ B such that A \ A ′ and B \ B ′ are meager and choose a Borel meager set N containing (A \ A ′ ) ∪ (B \ B ′ ). (In fact, we can require N to be an F σ -set, see e.g. [17,Proposition 8.23].) Let Λ ⊆ Γ be the subgroup generated by R. Of course, Λ is countable. Since the action preserves meager and Borel sets, the set Λ.N is meager and Borel. When we remove the set Λ.N from the graph G, we remove whole components. Thus the new graph G ′ still satisfies (4.3). Also, G ′ is a Borel graph since its parts, A \ Λ.N = A ′ \ Λ.N and B \ Λ.N = B ′ \ Λ.N , are Borel. Thus Theorem 4.3 applies to G ′ and gives a Borel matching M such that all unmatched vertices are inside some Borel meager set N ′ . By enlarging the set N ′ , we can also assume that it contains N as a subset. The Borel meager set Λ.N ′ is again a union of some components of G. Thus the set-theoretic equidecomposition between A and B gives an equidecomposition between the meager sets A ∩ Λ.N ′ and B ∩ Λ.N ′ while M gives a Borel equidecomposition between A \ Λ.N ′ and B \ Λ.N ′ . Putting these together, we get the required Baire equidecomposition between A and B (using the finite set R ⊆ Γ ).
Proof of Corollary 1.10. All forward implications in Corollary 1.10 are trivial so let us show the converse direction. In all cases, the sets A and B are in C (resp. essentially are in C) so by Lemma 2.3 they cover (resp. essentially cover) each other. Now, Item (i) of the corollary is a direct consequence of Theorem 1.6(i). In order to derive Item (ii) from Item (i), it suffices by Proposition 3.4(i) to show that A and B are set-theoretically equidecomposable, and this follows from Proposition 1.8(i).
Finally, let us derive Item (iii) from Item (ii). By the latter, we know that A and B are measurably equidecomposable. Proposition 1.8(ii) gives that these sets are also Baire equidecomposable. These two equidecompositions can be combined into a Baire-Lebesgue equidecomposition by Proposition 3.4(ii).

Relation to (local) spectral gap
Recall that the local spectral gap property was defined in Section 1.5. Note that we do not assume in the definition that Γ is countable. The authors of [5] consider only countable Γ , in which case our definition coincides with the one in [5]. However, the definition of local spectral gap, as stated in Section 1.5, makes sense also when Γ is not countable.
For a finite multiset Q ⊆ Γ , let T Q : L 2 (Ω, µ) → L 2 (Ω, µ) be the averaging operator defined by We say that the operator T Q has spectral gap if there exists a constant c > 0 such that for Also, we say that the action a : Γ ↷ Ω has spectral gap if there is a finite multiset (equivalently, a finite set) Q such that the operator T Q has spectral gap. If µ(Ω) < ∞, then the latter property is easily seen to be equivalent to the local spectral gap of the action a with respect to the whole space Ω. Boutonnet et al [5, Theorem A] proved the following sufficient condition for local spectral gap.
Theorem 5.1 (Boutonnet et al [5]). Let Γ be a connected Lie group with a fixed left Haar measure m Γ . Suppose that the Lie algebra g of Γ is simple. Let Ad : Γ → GL(g) denote its adjoint representation. Let Λ be a dense countable subgroup of Γ and let B be a basis of g such that for every g ∈ Λ the matrix of Ad(g) in the basis B has all entries algebraic. Then the left translation action Λ ↷ (Γ, m Γ ) has local spectral gap with respect to every measurable set X with compact closure and non-empty interior.
The above notions and results are of interest to us because of the following equivalence, mentioned in the Introduction. Proof. Boutonnet et al [5,Theorem 7.6] proved that, under the additional assumptions that Γ is countable and the action Γ ↷ Ω is ergodic, Property (B) (i.e., the local spectral gap with respect to X) is equivalent to the following: (C) If a sequence A n , n = 0, 1, ... , of measurable subsets of Ω satisfies µ(A n ∩ X) > 0 for all n, and for all γ ∈ Γ , then lim n→∞ µ(A n ∩ X) = µ(X). In fact, the part of the proof in [5] that shows the equivalence between (B) and (C) does not use the ergodicity of the action Γ ↷ Ω. Clearly, (A) looks closer in spirit to (C) than to (B) and, indeed, it is fairly easy to derive the equivalence of (A) and (C). Since we need only the implication (B) =⇒ (A) for our equidecomposition results, we present a direct proof of this for reader's convenience. In the other direction, we show only that (A) implies (C), leaving to the reader to check that the proof of the implication (C) =⇒ (B) from [5, Theorem 7.6] does not use ergodicity.
In order to prove (B) =⇒ (A), we need some preliminaries. Let us say that the action Γ ↷ Ω has a highly local spectral gap with respect to a measurable set B ⊆ Ω with 0 < µ(B) < ∞ if there are a finite set S ⊆ Γ and a real κ such that One motivation behind this definition (besides that it is useful for our proof) is that each side of (5.3) depends only on the restriction of f to B (but not on any values of f outside B as is the case in (1.1)).
Clearly, the highly local spectral gap for B implies the local spectral gap for B (with the same choice of S and κ). Let us show that the converse implication also holds.
Claim 5.2.1. If the action has local spectral gap with respect to a measurable set B ⊆ Ω with 0 < µ(B) < ∞, then it has the highly local spectral gap property with respect to B.
Proof of Claim. Let a finite set S and a real κ satisfy the local spectral gap property with respect to B ⊆ Ω. By enlarging S, we can assume that e ∈ S. Define κ ′ := κ |S| and where we view S ′ as a set, not as a multiset. (Alternatively, if one views S ′ as a multiset so e.g. |S ′ | = |S| 2 , then κ ′ = κ suffices in the argument below.) Let us show that S ′ and κ ′ satisfy the highly local spectral gap condition. Take any f ∈ L 2 (Ω, µ) with B f dµ = 0.
We define f ′ ∈ L 2 (Ω, µ) as follows. Fix some total order on S with the identity e being the smallest element. For x ∈ Ω, if there is γ ∈ S such that γ.x ∈ B then let γ x be the smallest such γ and define f ′ (x) := f (γ x .x); otherwise (i.e., if S.{x} ∩ B = ∅), we let f ′ (x) := 0 (while γ x is undefined). Since e comes before any other element of S, we have that f ′ (x) = f (x) for all x ∈ B. For g, h ∈ S, define It follows from the definition that, for every g ∈ S, we have B = ⊔ h∈S B g,h , that is, the sets B g,h , h ∈ S, are disjoint and partition B. Also, trivially, B g,h ⊆ B ∩ gh −1 .B.
As f and f ′ coincide on B, we have B f ′ dµ = B f dµ = 0. Also, the square of the L 2 (Ω, µ)-norm of f ′ is finite, as it is at most |S| times the square of ∥f ∥ 2,B . By the properties stated above and the inequality ( h∈S x h ) 1/2 ⩽ h∈S x 1/2 h valid for any nonnegative reals x h , we have that that is, the real κ ′ and the set S ′ establish the highly local spectral gap for B. This proves Claim 5.2.1. Now, we can give a direct proof that (B) implies (A). Since the properties in question are invariant under scaling the measure by a constant factor, assume that µ(X) = 1. By Claim 5.2.1, we can find a real κ and a symmetric finite set S ⊆ Γ with S ∋ e that satisfy the highly local spectral gap condition for X. Now, given η > 0, let ℓ ∈ N satisfy (1 + η/(2κ 2 |S| 2 )) ℓ ⩾ 1/η and consider the set Q := S ℓ = {γ 1 ... γ ℓ : γ 1 , ... , γ ℓ ∈ S} of all possible ℓ-wise products of elements of S. We will show that this set Q is ηexpanding for X. We need an auxiliary claim first.
Now we are ready to show that Q is η-expanding for X. Take an arbitrary measurable subset Y ⊆ X. Let Y 0 := Y and, inductively for i ∈ Y ∩ X has measure at least 1 − η, as required. Otherwise, we obtain from Claim 5.2.2 by induction on i = 0, ... , ℓ that µ(Y i ) ⩾ (1+η/(2κ 2 |S| 2 )) i µ(Y ). Taking i = ℓ, we get the the required lower bound µ(Q.Y ∩X) ⩾ µ(Y ℓ ) ⩾ µ(Y )/η. Since η > 0 was arbitrary, we conclude that X is a domain of expansion. We have shown that (B) implies (A). Now, let us show that (A) implies (B). Let X be a domain of expansion. For each positive integer n, fix some finite set S n ⊆ Γ which is (1/n)-expanding for X. Let Λ be the subgroup of Γ generated by ∪ ∞ n=1 S n . First, let us show that the countable group Λ satisfies (C). Suppose on the contrary that some sequence (A n ) n∈N violates this property. Let A ′ n := A n ∩ X. By passing to a subsequence, we can assume that there is an integer m ⩾ 2 such that µ(A ′ n ) < µ(X)(1 − 2/m) for every n. Let S := S m , that is, S is a (1/m)-expanding set for X. Thus, for every n ∈ N, By passing to a subsequence again, we can assume that (i) for all n we have µ(S.A ′ n ∩ X) ⩾ 1 − 1 m µ(X), or (ii) for all n we have µ(S.A ′ n ∩ X) ⩾ m µ(A ′ n ). If (i) holds then clearly µ((S.A ′ n ∩ X) \ A ′ n ) ⩾ 1 − 1 m µ(X) − (1 − 2 m )µ(X) = µ(X)/m and thus, for some γ ∈ S and infinitely many n, we have µ((γ.A ′ n ∩ X) \ A ′ n ) ⩾ µ(X)/(m|S|), and so also for infinitely many n. This is in contradiction with the assumption (5.2) of Property (C). Suppose now that (ii) holds. Since m ⩾ 2, we have µ((S.A ′ n ∩ X) \ A ′ n ) ⩾ µ(A ′ n ) for all n. Therefore for some γ ∈ S and infinitely many n we have µ( which again is a contradiction to (5.2). Thus (C) holds for X with respect to the action Λ ↷ Ω. By [5, Theorem 7.6], the action of the countable group Λ on Ω has local spectral gap with respect to X, that is, (B) holds for the group Λ. Of course, when we enlarge the group to Γ , then (B) still holds.
The following proposition will be needed later, for estimating the number of pieces in some equidecompositions given by our proofs. Proposition 5.3. In addition to Assumptions 1.4 and 1.5, assume that µ is a finite measure. Let S ⊆ Γ be a finite symmetric multiset, and let c ∈ (0, 1) be such that for every f ∈ Then the following statements hold.
Proof. By scaling the measure, we can assume that µ(Ω) = 1. Take any measurable Y ⊆ Ω with y := µ(Y ) > 0. Analogously to (5.5), define f := (1 − y)1 Y − y1 Ω\Y . We have ∥f ∥ 2 2 = y(1 − y) and Ω f (x) dµ(x) = 0. Let Z := S.Y and z := µ(Z). Clearly, for every x ∈ Ω \ Z we have that (T S f )(x) = −y. By the invariance of the measure, it also holds that Ω T S f dµ = |S| −1 γ∈S Ω γ.f dµ = 0. Under these constraints on T S f , its L 2 -norm is minimised when the function is constant on Z, that is, assumes the value (1 − z)y/z there. Thus By the spectral gap property, the left-hand side is at most (1−c) 2 ∥f ∥ 2 2 = (1−c) 2 y(1−y). Solving the obtained linear inequality in z, we obtain the first inequality of Part (i). The second inequality is obtained by observing that by c, y ∈ [0, 1], We prove Part (ii) similarly as we did after Claim 5.2.2. Define Y 0 := Y . Inductively the set S is non-empty and µ is invariant). Otherwise, the measure of each new set Y i , i ⩽ ℓ, increases by factor at least 1 + c ′ η by Part (i) and thus µ(S ℓ .Y ) ⩾ µ(Y )/η by the choice of ℓ.
6 Proof of Theorem 1.11 6.1 Proof of Theorem 1.11 for SO(n) ↷ S n−1 , n ⩾ 3 Here, Ω is the sphere S n−1 with the uniform probability measure µ. It was shown independently by Margulis [28] and Sullivan [43] for n ⩾ 5, and by Drinfel'd [10] for n = 3, 4, Recall that G 2 is the subgroup of affine bijections of R 2 generated by the special linear group SL(2, Z) and all translations. In order to prove that the action a : G 2 ↷ R 2 is expanding, we consider another action b defined as follows.
Identify the torus R 2 /Z 2 with X := [0, 1) 2 and let ν denote the uniform probability measure on Borel subsets of X. The group SL(2, Z) acts naturally on R 2 , commuting with the reduction of vectors modulo Z 2 . Thus we obtain the standard measure-preserving action b : SL(2, Z) ↷ (X, ν). A classical result of Rosenblatt [40] states that b has spectral gap. By Proposition 5.3, the set X is a domain of expansion for the action b.
By Remark 2.6, it is enough to show that X is a domain of expansion for the paradoxical action a. Let π : R 2 → X be the natural projection that reduces each coordinate modulo 1. For γ ∈ SL(2, Z), let γ ′ denote the corresponding element of G 2 under the natural inclusion of SL(2, Z) into G 2 . For example, the restriction of the composition π • a(γ ′ , ·) to X is equal to b(γ, ·), the b-action of γ.
Take any η > 0. Since X is a domain of expansion for the action b, there exists an ηexpanding finite set S ⊆ SL(2, Z) for X under the action b. We construct an η-expanding set T for X ⊆ R 2 under the action a : G 2 ↷ R 2 as follows. For every γ ∈ S and (m, n) ∈ Z 2 such that γ ′ .X intersects the square [m, m + 1) × [n, n + 1), add the product t −1 m,n γ ′ ∈ G 2 into T , where t m,n ∈ G 2 is the translation by vector (m, n). Since |S| < ∞, the constructed set T is finite too. Furthermore, for every γ ∈ S and U ⊆ X, we have that γ.U = π(γ ′ .U ) ⊆ T.U ∩ X: indeed, for every integer square intersecting γ ′ .U , the set T contains the composition of γ ′ with the integer translation moving this square back to X = [0, 1) 2 . Thus, S.U ⊆ T.U ∩ X and, if U is measurable, then where the last inequality follows from the facts that S is η-expanding for (X, ν) under the action b and the measures ν and µ coincide on X ⊇ S.U . Thus T ⊆ G 2 is an η-expanding set for X. As η > 0 was arbitrary, X is a domain of expansion for the action a : G 2 ↷ R 2 , as desired.

Proof of Theorem 1.11 for the hyperbolic space H n
For an introduction to hyperbolic spaces see e.g. Bridson and Haefliger [6, Section 2] or Ratcliffe [39]. One representation of H n (that we will use here) is to take the bilinear form  ↷ (H n , µ).
The Lie algebra of SO(n, 1) 0 (or SO(n, 1)) is so(n, 1) which consists of (n + 1) × (n + 1)-matrices M such that M T I n,1 + I n,1 M = 0, (6.3) where I n,1 is the diagonal matrix having n entries equal to 1 and the last entry equal to −1.
It is well-know that so(n, 1) is simple; in fact, a complete characterisation of simple real Lie algebras is available, see e.g. Knapp [19,Theorem 6.105].
The linear system of equations (6.3) has integer coefficients, so we can choose a basis B for so(n, 1) that consists of matrices with all entries rational.
The adjoint Ad(γ) for γ ∈ SO(n, 1) 0 maps M ∈ so(n, 1) to the matrix product γM γ −1 . If a matrix γ has algebraic entries, then so does its inverse; it follows that the matrix of the map Ad(γ) when expressed in the basis B has all entries algebraic. Thus the only non-trivial remaining assumption of Theorem 5.1 is the existence of a countable dense set X ⊆ SO(n, 1) 0 such that each γ ∈ X as a matrix has algebraic entries (as then we can take Λ to be the subgroup generated by X). We can identify SO(n, 1) with the variety in V ⊆ R (n+1)×(n+1) defined by the system of polynomials with integer coefficients, stating that the determinant is 1 and the bilinear form in (6.1) is preserved. Since SO(n, 1) 0 is a connectivity component of SO(n, 1), the existence of X follows from the following general lemma. Proof. Although this lemma has surely been proved before, we could not find a suitable statement (of the real case) anywhere in print. So we present our proof.
One of the consequences of the Tarski-Seidenberg Theorem [46,41] is that the ordered field of reals R and the ordered field A of real algebraic numbers satisfy the same set of first-order sentences, where the language includes the constants 0 and 1, the multiplication and addition functions, the equality relation, and the binary order relation. For an r-vector x, we can express the statement that f i (x) = 0 for each i ∈ [m] in first-order logic. Thus, for every Euclidean ball B ⊆ R n whose centre and radius are rational, Now, suppose on the contrary to the claim that we have some x ∈ V and rational r > 0 such that the radius-r ball B ⊆ R r around x has no algebraic points from V . Pick a rational vector x ′ ∈ R n within distance r/3 from x and let B ′ be the ball around x ′ of radius r/2. Then B ′ ⊆ B, so B ′ ∩ V has no algebraic points. However, B ′ ∩ V is non-empty as it contains x, a contradiction.
Thus, by Theorem 5.1 the action b : Λ ↷ (Γ, m Γ ) of the countable subgroup Λ of Γ generated by X has local spectral gap with respect to every measurable subset of Γ with compact closure and non-empty interior. We will show that the restriction a ′ of the measure-preserving action a : Γ ↷ (H n , µ) to Λ is expanding, thus finishing the proof.
Take any Borel B ⊆ H n which has non-empty interior and compact closure. Let us show that B is a domain of expansion.
One can derive from the formula for the hyperbolic distance that the topology on H n is the induced topology from R n+1 . Thus B is a bounded subset of R n+1 . Since the density Let us argue that the closure B ′ of B ′ is compact. Take an arbitrary infinite sequence γ 1 , γ 2 , ... ∈ B ′ . Viewing H n as a subset of R n+1 , fix some n elements x 1 , ... , x n ∈ H n so that the n + 1 vectors x 0 , x 1 , ... , x n ∈ R n+1 are linearly independent. Each isometry γ i ∈ Iso(H n ) can be represented by a linear H n -preserving transformation R n+1 → R n+1 given by some matrix M i ∈ SO(n, 1) 0 . The images of the special point x 0 by M i are in the compact set B, so by passing to a subsequence we can assume that they converge to some z 0 ∈ B. Since each M i is an isometry of H n , the images of x 1 , ... , x n all lie in some large ball in H n around z 0 (and thus in some large ball in R n+1 ). Again by passing to a subsequence, we can assume that, for every j ∈ [n], M i x j converges to some z j as i → ∞. Let X, Z, and X i for i ∈ N be the (n + 1) × (n + 1)-matrices with columns respectively x 0 , ... , x n , z 0 , ... , z n , and M i x 0 , ... , M i x n . By the finite-dimensionality, we have that X i → Z as i → ∞. Also, by the choice of the vectors x j , the matrix X is invertible. Thus M i = X i X −1 converges to M := ZX −1 as i → ∞. This limiting matrix M ∈ SO(n, 1) 0 has to belong to B ′ , so this set is indeed compact. Thus, Theorem 5.1 applies to the set B ′ , giving that the action of Λ on (Γ, m Γ ) has local spectral gap with respect to this set. This means by Lemma 5.2 that B ′ is a domain of expansion. Let η > 0 be arbitrary. Thus there are γ 1 , ... , γ m ∈ Λ such that the η-expansion property, as defined in (2.4), is satisfied for every Borel X ⊆ B ′ . Let us argue that the same isometries are η-expanding for B. Take any Borel It remains to show is that f is measure-preserving. This is exactly the statement of, for example, [39, Lemma 4 of Section 11.6] (which follows with some work from the uniqueness of the Haar measure). Thus B is a domain of expansion.
By Remark 2.6, this finishes the proof of the case Iso(H n ) ↷ H n of Theorem 1.11.
6.4 Proof of Theorem 1.11 for SL(2, R) ↷ R 2 \ {0} Here, we follow the same strategy as in Section 6.3. One new caveat is that, unlike for the hyperbolic space, the stabiliser H of a point is not a compact subgroup. So we cannot define B ′ by the direct analogue of (6.4) as its closure would not be compact. We get around this by, essentially, taking a compact subset of positive measure in each coset of H in a continuous way. Also, we could not find a version of [39, Lemma 4 of Section 11.6] that we could just cite here, so we prove it (as well as a few other claims) via direct explicit calculations.
Proof of Theorem 1.11 for SL(2, R) ↷ R 2 \ {0}. Recall that m Γ is a left Haar measure on Γ = SL(2, R). We view each element γ ∈ Γ as a 2 × 2-matrix (γ ij ) 2 i,j=1 . It is easy to to find a countable dense subgroup Λ of SL(2, R): just take the subgroup of matrices with all entries rational. The Lie algebra sl 2 (R) of SL(2, R) consists of 2 × 2 matrices with trace zero and is well known to be simple. For the basis B, one can take, for example, For γ ∈ SL(2, R), the adjoint Ad(γ) maps a 2 × 2-matrix M = (M ij ) 2 i,j=1 ∈ sl 2 (R) to γM γ −1 . Clearly, we have M = M 1,2 X + M 2,1 Y + M 1,1 Z. If we write the linear map Ad(γ) in the basis B, then each entry of the corresponding 3 × 3-matrix is, in fact, a quadratic polynomial with integer coefficients in the entries of the matrix γ. Therefore, if γ ∈ Λ then γ (and thus the matrix of Ad(γ)) has all entries rational. Clearly, the map ı : Γ → Ω, where ı(γ) := γe 1 for γ ∈ Γ , is continuous and the pre-images under ı of the points in Ω are exactly the left cosets of H. Consider the map σ : Ω → Γ defined by The map σ is continuous on Ω = R 2 \ {0} and σ is the right inverse of ı: ı • σ = Id Ω . Let us show that the push-forward of µ ′ by ϕ is a constant multiple of the Haar measure m Γ on Γ . By the uniqueness of the Haar measure, it is enough to argue that, for every γ ∈ Γ , viewed as the map Γ → Γ where β → γ.β, the composition γ ′ := ϕ −1 • γ • ϕ from Ω × H to itself preserves the measure µ × ρ. In this concrete case, it is easy to show this by writing explicit formulas. Namely, the inverse of ϕ is It routinely follows that γ ′ ((x, y), M (u)) = (γ.(x, y), M (u + F )), where F = F (γ, x, y) does not depend on u. We see that the bijection γ ′ is the measure preserving map a(γ, ·) in the first coordinate. Also, if the first coordinate is fixed, then the second coordinate of γ ′ corresponds to some translation of R under the parametrisation M from (6.5). It follows by an application of Tonelli's theorem that γ ′ is measure-preserving, as claimed. Thus, by scaling the measure m Γ , we can assume that ϕ is a measure-preserving homeomorphism between (Ω ′ , µ ′ ) and (Γ, m Γ ). For the rest of the proof, it will be more convenient to replace the action b with its conjugate by ϕ. Namely, let the measurepreserving action b ′ : Λ ↷ (Ω ′ , µ ′ ) be defined by b ′ (γ, (x, h)) := ϕ −1 (b(γ, ϕ(x, h))) for (x, h) ∈ Ω × H and γ ∈ Λ. By the calculations in the previous paragraph, for every x ∈ Ω, there is the (unique) map γ x : H → H so that b ′ (γ, (x, h)) = (a(γ, x), γ x (h)); moreover, each γ x under the isomorphism H ∼ = R corresponds to some translation of R and thus preserves the Haar measure ρ. Now, we are ready to show that the action a : Γ ↷ Ω is expanding. Take any Borel B ⊆ Ω with non-empty interior and compact closure. Let B ′ := B × I. Clearly, B ′ ⊆ Ω ′ has compact closure and non-empty interior and µ ′ (B ′ ) = µ(B). Since ϕ is a measurepreserving homeomorphism, Theorem 5.1 (when applied to b) gives that the conjugated action b ′ has local spectral gap with respect to every measurable subset of Ω ′ with compact closure and non-empty interior. Thus there are a real κ and a finite set S ⊆ Λ such that Let us show that the same choice of κ and S also witnesses the local spectral gap of the action a with respect to B. Take any f ∈ L 2 (Ω, µ) with B f dµ = 0. The idea is to apply the inequality in (6.6) to the function (x, h) → f (x) except, in order to have an L 2function, we set it to 0 at the points which do not matter when we consider local spectral gap for B ′ under S (namely, those points that do not appear in (6.6)). So, let f ′ (x, h) := f (x) for (x, h) ∈ B ′ ∪ S −1 .B ′ and let f ′ be zero otherwise. The obtained function f ′ is in L 2 (Ω ′ , µ ′ ) because is it obtained by patching together compositions of g ∈ L 2 (Ω ′ , µ ′ ) with finitely many measure-preserving maps and γ x preserves the measure ρ for each x ∈ Ω, we have for each s ∈ S that It follows by (6.6) that Thus the action a : Γ ↷ Ω is expanding by Lemma 5.2 (and Remark 2.6), as required.
6.5 Proof of Theorem 1.11 for Iso(R n ) ↷ R n , n ⩾ 3 We give a few proofs of this case: by deriving it from known results in Section 6.5.1 and then giving a more direct proof in Section 6.5.2.

Derivation from known results
Theorem 1.11(i) can be rather straightforwardly derived from known (deep) results in a few different ways.
First proof of Theorem 1.11 for Iso(R n ) ↷ R n , n ⩾ 3. Margulis [29] showed that, for this action, every isometry-invariant positive mean defined on compactly supported bounded measurable functions is a scalar multiple of the integral with respect to the Lebesgue measure. Also, it can be shown via Lebesgue's density theorem that every dense subgroup of Iso(R n ) acts ergodically on R n . Thus the implication (1) =⇒ (4) of [5, Theorem 7.6] applies here and gives that the action Iso(R n ) ↷ R n has local spectral gap with respect to every measurable set in C. We conclude by Lemma 5.2 that the action Iso(R n ) ↷ R n is expanding.
Alternatively, the desired local spectral gap property of Iso(R n ) ↷ R n can be derived from [5, Theorem A] (which is Theorem 5.1 here) in a similar way as it was done for the action Iso(H n ) ↷ H n in Section 6.3. Yet another proof is to use the following more recent result of Boutonnet and Ioana [4, Theorem A].
Theorem 6.2 (Boutonnet and Ioana [4]). Let Λ be a countable dense subgroup of Iso(R n ), n ⩾ 3, such that the left-translation action Θ(Λ) ↷ SO(d) has spectral gap, where Θ : Iso(R n ) → SO(n) denotes the natural quotient. Then the natural action Λ ↷ R n has local spectral gap with respect to every measurable set with compact closure and non-empty interior.
In order to apply Theorem 6.2, one can let Λ be the subgroup generated by a finite subset of SO(n) ⊆ Iso(R n ) having the spectral gap property and some countable dense subset of Iso(R n ).

Approach via more direct computations
In this subsection we will reprove the case of Iso(R n ) ↷ R n , n ⩾ 3, of Theorem 1.11 in a more direct way. We prove the case n = 3 first and derive the general case n ⩾ 4 as a consequence. We will not use the results of Margulis [29] nor Boutonnet et al [5,4]. However, when proving the base case n = 3, we still need as an input the spectral gap property of the action of SO(3) on the 2-dimensional sphere. While the original proof of this by Drinfel'd [10] requires a fair amount of background, there are more elementary proofs now: see, for example, Benoist and de Saxcé [3].
The rest of this section is devoted to domains of expansion in R n . However, we start by proving a sufficient criterion for being a domain of expansion for a general action Γ ↷ Ω.
Let us informally motivate the upcoming definitions. Suppose that we fix some ρ > 1 and would like to show that the annulus Y := {y ∈ R 3 : 1 ⩽ ∥y∥ 2 ⩽ ρ} is a domain of expansion. For U ⊆ Y and z ∈ [1, ρ] let U z := {y ∈ U : ∥y∥ 2 = z} be the z-leaf of U .
For i ∈ [3], let µ i denote the i-dimensional Hausdorff measure on R 3 .
By the spectral gap property of SO(3) ↷ (S 2 , µ 2 ) and Proposition 5.3, the sphere S 2 is a domain of expansion under this action. So, for every δ > 0, there is a finite δ-expanding set S δ ⊆ SO(3). For each z ∈ [1, ρ], the group SO(3) also acts on (Y z , µ 2 ) (which is just the sphere of radius z) and the set S δ is also δ-expanding for Y z . Thus, for every Borel U ⊆ Y , the leaf U z under the action of S δ occupies at least (1 − δ)-fraction of Y z or expands by factor at least 1/δ in the measure µ z . It follows that, if U does not expand in measure under S δ , then this means that the set U is "close" to a ring set (a union of some spheres Y z ).
In order to deal with such sets, we add a finite set T of isometries of R 3 with the property that there is c > 0 such that if U is any Borel ring set then µ 2 ((T.U ) z ) ⩾ cµ(U ) for every z ∈ [1, ρ]. We will call such a set T a diffuser. Informally speaking, a diffuser spreads any ring set across all radii of interest fairly uniformly. If we apply a spherical expanding set S β with β ≪ δ to any such "uniformly spread" set T.U , then we ensure that, for each z ∈ [1, ρ], its z-leaf (T.U ) z expands or occupies most of Y z . It follows from the above properties that S β .T has a good expansion property when applied to any (almost) ring set. Thus, our proof strategy is as follows: if U expands under S δ then we are done; otherwise S δ .U is close to a ring set and consequently expands under S β T . The exact value of ρ will be chosen to simplify finding a diffuser set; in fact, we will choose ρ so that the diffuser can be taken to be a one-element set (see Figure 1 and Lemma 6.8).
Let us give all formal general definitions (that are motivated by the above discussion). Assumptions 1.4 and 1.5 apply everywhere in this section.
for every Borel X ⊆ Z, the set Y X := ∪ z∈X Y z is a Borel subset of Y , (iv) each µ z is a finite measure on (Ω, B) supported on Y z such that for every Borel set U ⊆ Y the function z → µ z (U ) is Borel and integrable, and it holds that Note that if ν happens to be the push-forward of µ under the map Y → Z that sends each Y z to z (when necessarily µ z is a probability measure for ν-a.e. z ∈ Z), then (6.7) states that the map z → µ z gives a disintegration of the measure µ, see e.g. [50,Section 5.1.2].
Under Definition 6.3, the support of a Borel set U ⊆ Ω is defined as Of course, Supp(U ) = Supp(U ∩Y ). Also, for z ∈ Z and X ⊆ Z we denote U z := U ∩Y z and and U X := x∈X U x ⊆ Y . Note that U z = U ∩ Y z is Borel by Item (iii) of the above definition. Also, since each µ z is supported on Y z , we have µ z (U ) = µ z (U z ).
Definition 6.4. Let Y be a subset of Ω with a foliation ((Z, B Z , ν), (Y z , µ z ) z∈Z ). For ε > 0, we say that a finite set S ⊆ Γ is leaf-wise ε-expanding if, for every Borel U ⊆ Y , it holds that We say that Y is a domain of leaf-wise expansion if for every ε > 0 there is a leaf-wise ε-expanding finite set S ε ⊆ Γ . For c > 0, we say a finite set T ⊆ Γ is a c-diffuser for Y if for every Borel R ⊆ Z and every z ∈ Z we have The following lemma states, informally speaking, that a diffuser spreads well not only ring sets Y R but also those sets that are "close" to them. Lemma 6.5. Let (Y z , µ z ) z∈Z be a foliation of Y ⊆ Ω. Let T be a diffuser for Y and let c > 0 satisfy (6.8). Given ε ∈ (0, 1), let δ := εc/(ε + |T |). Then, for any Borel set V ⊆ Y of positive measure satisfying µ Proof. Note that if we divide ν by some α > 0 and multiply each µ z by the same constant α, then all statements of Definitions 6.3 and 6.4 remain valid. (This will be used later in Remark 7.5 and this is why we divide by ν(Z) in (6.8).) Thus, we can assume without loss of generality that ν(Z) = 1.
Let R := Supp(V ). Note that the set R is Borel by Item (iv) of Definition 6.3. The measure of V Z\R is 0 by (6.7), so by removing this set from V we may assume that Since T is a c-diffuser, we obtain that, for all z ∈ Z \ R ′ , Thus, in order to finish the proof, it is enough to show that ν(R ′ ) ⩽ ε. By (6.7), we have that Since µ z (W ) = µ z (Y \ V ) ⩽ δ µ z (Y ) for each z ∈ R by the assumption on V and since W ⊆ Y R , we have, again by (6.7), that µ(W ) ⩽ δ µ(Y R ). Hence giving the required bound ν(R ′ ) ⩽ ε by the choice of δ.
We are ready to state our criterion for being a domain of expansion. Recall that S η for η > 0 is a finite leaf-wise η-expanding subset of Γ . Let us additionally assume that S η ∋ e.
If Y is a domain of leaf-wise expansion which admits a diffuser T , then Y is a domain of expansion.
Proof. By scaling the measures ν and µ by the same constant, we can assume for convenience that ν(Z) = 1. By taking R := Z in (6.8) and integrating it over all z ∈ Z, we see that c ⩽ 1. Thus δ ⩽ cε ⩽ ε ⩽ 1/4 and β ⩽ 1/32.
We proceed to apply Proposition 6.6. Here, we have that Ω = R 3 with the L 2 -norm ∥ · ∥ and Γ = Iso(R 3 ). Define is the interval [1, ρ] with the Lebesgue measure. Here, µ z is the restriction of µ 2 to Y z := {y ∈ R 3 : ∥y∥ = z}, the sphere of radius z. By our scaling, µ z (Y z ) = 4πz 2 for z ∈ Z. The map Y → (Z, ν) × (S 2 , µ 2 ), which sends x to (∥x∥, x/∥x∥), is a measurepreserving Borel isomorphism. Since a product of two standard finite measure spaces gives a foliation by Tonelli's theorem, we conclude that we indeed have a foliation of Y .
The following lemma is an exercise in elementary geometry.
Lemma 6.7. There is a solid 3-dimensional cube K ⊆ R 3 with the following properties.
(i) The side length of K is 1. (ii) There are two opposing faces A and B of K such that A ∩ Y 1 consists of the four corners of A while B is tangent to Y ρ . (iii) For every z ∈ Z and every point x ∈ K ∩ Y z the angle between the plane tangent to Y z at x and the plane extending the face A is at most π/4.

Sketch of Proof.
The cube K is pictured in Figure 1. We construct it as follows. We start by inscribing into Y 1 a 2-dimensional square A of side length 1 whose third coordinate is a positive constant. Then we extend A to a cube of side length 1 in such a way that A is its lower face (lower with respect to the third coordinate in R 3 ). The first property of the lemma is clear by construction. Let h be the height of a triangle whose vertices are the centre of Y 1 together with two opposite vertices of A. One can easily compute that h = √ 2/2. Thus h + 1 = ρ and the second property follows. Let us show the third property. By elementary considerations the angle in question is maximized at points x which are corners of the face A (when z = 1). At such x the angle in question is equal to the angle between the diagonal of A and the tangent plane to Y 1 , and that angle is easily computed to be equal to π/4. The following lemma, which gives f ∈ Iso(R 3 ) such that {f } is a diffuser, is a routine calculus exercise. Lemma 6.8. Let f : K → K be an isometry which maps the face A to one of the side faces of K. Let R ⊆ [1, ρ] be a Borel subset. Then for all z ∈ [1, ρ] we have Proof. Let g : K → A be the orthogonal projection onto A. Let L be the set of straight line segments contained in the face A that connect f (B) to f (A) and are orthogonal to both f (B) and f (A).
Let us fix z ∈ [1, ρ] and let (g • f ) z be the restriction of g • f to f −1 (K z ). Note that Let U ⊆ ℓ be a Borel subset. Note that f maps ℓ isometrically into the sphere K z . It follows that we have Since the map x → ∥x∥ does not increase distances and maps K R ∩ ℓ onto R, we have µ 1 (K R ∩ ℓ) ⩾ ν(R). Consequently we have Let us fix ℓ ∈ L. We clearly have A = (A ∩ f (A)) × ℓ. Furthermore, the elements of L are exactly the sets {x} × ℓ in this decomposition. Therefore, by Tonelli's theorem, and since the side length of K is equal to 1, Note that for ℓ ∈ L we have (g • f ) z (K R ) ∩ ℓ = (g • f ) z (K R ∩ ℓ), and so by (6.12), Note that (g • f ) z (K R ) = g(f (K R ) z ). Hence, since g is an orthogonal projection, we have µ z (f (K R )) = µ 2 (f (K R ) z ) ⩾ µ 2 ((g • f ) z (K R )). On the other hand, by (6.7) we have µ(K R ) ⩽ ν(R) sup{µ z (K) : z ∈ R}. Note that for z ∈ R the set K z projects onto the face A. Hence, by the third item of Lemma 6.7, we have that Putting these inequalities together, we obtain that which is exactly our claim.
Corollary 6.9. Let ρ := 1 + √ 2/2 and let X ⊆ [1, ρ] be a Borel set of positive measure. Then the set Y := {y ∈ R 3 : ∥y∥ ∈ X} is a domain of expansion for the action Proof. Recall that the set Y comes with the spherical foliation by Z = [1, ρ]. By the spectral gap property of SO(3) ↷ S 3 and Proposition 5.3, the set Y is a domain of leafwise expansion. By Lemma 6.8, Y admits a (single-element) diffuser. Now, the corollary follows from Proposition 6.6.
We are ready to give another proof for the Euclidean case of Theorem 1.11.
Second proof of Theorem 1.11 for Iso(R n ) ↷ R n , n ⩾ 3. We use induction on n. If n = 3, then Y := {y ∈ R 3 : 1 ⩽ ∥y∥ ⩽ ρ} is a domain of expansion by Corollary 6.9. Lemma 2.5 gives that every measurable subset X ⊆ R 3 that belongs to C is a domain of expansion (since X and Y cover each other), as desired.
Suppose that n ⩾ 4. By Lemma 2.5 again, we only need to show that Y := [0, 1] n is a domain of expansion. Note that Y becomes foliated when we take Z := [0, 1] and Y z := [0, 1] n−1 × {z} for z ∈ Z (with the uniform probability measures).
Let γ be the isometry of R n which is the identity on the first n − 2 coordinates and which is the rotation by π/2 around the centre of [0, 1] 2 on the last two coordinates. It is easy to see that for any Borel set R ⊆ Z and any z ∈ Z we have µ z (γ.Y R ) = µ(Y R ). In particular, the singleton set {γ} is a diffuser (with the constant c = 1). Now, Proposition 6.6 shows that Y is a domain of expansion, finishing the proof.
Also, it is easy now to derive Theorem 1.13 from the Introduction, namely the existence of a domain of expansion Y in R n , n ⩾ 3, which is closed and nowhere dense.
(Note that such a set Y is meager and cannot essentially cover a non-empty open set.) Proof of Theorem 1.13. Let X be a closed nowhere dense subset of [0, 1 + √ 2/2] of positive measure, for example, a "fat" Cantor set.
If n = 3, let Y := {y ∈ R 3 : ∥y∥ ∈ X}; otherwise let Y := [0, 1] n−1 × X. The only non-trivial property that we have to check is that Y is a domain of expansion. If n = 3, this is the result of Corollary 6.9.
Suppose that n ⩾ 4. Here, we take the product-space foliation of Y = [0, 1] n−1 ×X by the last factor Z := X. Since the action Iso(R n−1 ) ↷ R n−1 is expanding, Y is a domain of leaf-wise expansion. Similarly to the previous proof, one can show that Y admits a diffuser consisting of a single isometry (which is trivial on the first n − 2 coordinates and is a rotation by angle π/2 in the last two coordinates). Thus Y is a domain of expansion by Proposition 6.6.
In this section we give sample estimates of the number of pieces in some of our constructed equidecompositions. We will need the following result of Lubotzky, Phillips and Sarnak [25,26].
Theorem 7.1. For every integer k ≡ 2 (mod 4) with k − 1 ⩾ 3 being a prime, there is a symmetric k-set Q ⊆ SO(3) such that, for the action SO(3) ↷ S 2 , the averaging operator k δ∈Q δ.f on L 2 (S 2 , µ) has the second largest eigenvalue (in absolute value) at most 2 √ k − 1/k and thus satisfies Proof. Let us outline just the construction of such a set. Take integer quaternions of the form q = a 0 + a 1 i + a 2 j + a 3 k with 3 i=0 a 2 i = k − 1, a 1 , a 2 , a 3 even and a 0 > 0 odd, and then take the standard map that sends a unit quaternion q/ √ k − 1 to an element of SO (3), see e.g. Note that the Alon-Boppana bound [37] implies that no larger value of c can work in (7.1) for a symmetric k-set Q.
First, we do one example where calculations are easy.
Next, we give a rough estimate of the number of pieces needed for measurable equidecomposition between a cube and a Euclidean ball in R 3 . Our purpose is just to demonstrate how an explicit bound can be derived from our results.
First, we need to understand how large the constructed η-expanding sets are. For a domain of expansion Y ⊆ Ω and η ∈ (0, 1), let s(η; Y ) be the smallest s ∈ N such that there is a real δ ∈ (0, η) and a δ-expanding set S ⊆ Γ for Y with |S| ⩽ s.
Proof. Let Q ⊆ SO(3) be the symmetric k-set returned by Theorem 7.1 that satisfies (7.1).
By Proposition 5.3(ii), the set Q ℓ is η-expanding for the whole space S 2 . The size of Q ℓ can be upper bounded by the number of words over Q of length at most ℓ that do not contain an element next to its inverse, giving the stated bound.
Lemma 7.4. Let A and B be, respectively, a cube and a ball in R 3 of the same volume. Let Y be the annulus, as defined in (6.11). Then A and B are measurably equidecomposable with at most 1400 · s(1/1109; Y ) + 2000 pieces.
Proof. First we rescale A and B by the same factor so that each fits inside the annulus. Namely, we scale the ball to have radius r := (ρ − 1)/2. Then the cube has side length λ := ( 4π 3 ) 1/3 r and elementary calculations show that it fits into Y (e.g. touching the unit sphere Y 1 in the centre of its face). We can cover Y by at most m := ⌈2ρ/λ⌉ 3 scaled cubes, arranged as a grid. Also, if we take a maximal packing of balls of radius r/2 with centres in Y , then balls with the same centres but of the twice larger radius r cover Y . Since the smaller balls are disjoint and lie entirely between the spheres of radii 1 − r/2 and ρ + r/2, we have at most n := ⌊((ρ + r means that these isometries can be used to construct a bipartite (multi-)graph H with parts A and B such that every vertex of A (resp. B) has degree at most 8 (resp. exactly 8). By Rado's theorem [38], H has a matching covering every vertex of B. Thus we can show [B] ≼ [A] with at most 1000 isometries. In total, we need to add at most 2000 further elements to R to satisfy the lemma.
8 Proof of Theorem 1.14 Theorem 1.14 follows from the following result where we have collected all the properties that are used in our proof: it is trivial to check that each action listed in Example 1.9 (which is expanding by Theorem 1.11) satisfies all assumptions of Lemma 8.1. Proof. If |Ω| = 1, then there is nothing to do. Otherwise, by e.g. the paradoxicality of the action, Ω has infinitely many points. Fix pairwise disjoint non-empty open sets U 1 , U 2 , U 3 ⊆ Ω with compact closures. Since the open set W with µ(W ) < ∞ covers every other set in C by Lemma 2.3 and µ is invariant, we have for every Y ∈ C that µ(Y ) < ∞ (and that µ(Y ) > 0 since Y covers a non-empty open set). In particular, 0 < µ(A ′ ) = µ(B ′ ) < ∞. Since the action a is expanding, the sets A ′ and B ′ are domains of expansion. By Proposition 1.8(ii) (resp. Theorem 1.6(ii)), the sets A and B are Baire (resp. Lebesgue) equidecomposable. Thus, by Proposition 3.4(ii), these sets are Baire-Lebesgue equidecomposable, say with pieces A ′ = A 1 ⊔ ... ⊔ A m and B ′ = B 1 ⊔ ... ⊔ B m . Each piece belongs to A: it is in B µ ∩ T by definition and has compact closure as a subset of A ′ or B ′ . By the finite additivity and a-invariance of κ, we have First, assume that r = p/q with p, q ∈ N. We use a consequence of Sierpiński's theorem [42] (for a modern proof see e.g. [12,Proposition A.1]) that if (Ω ′ , A ′ , µ ′ ) is an arbitrary non-atomic measure space with µ ′ (Ω ′ ) < ∞, then for every ρ ∈ [0, µ ′ (Ω ′ )] there is Y ∈ A ′ with µ ′ (Y ) = ρ. By applying this result to the σ-algebra B µ ∩ T ⊇ A of all Baire-Lebesgue subsets of Ω, we can find (B µ ∩ T )-measurable partitions A = A 0 ⊔ A 1 ⊔ ... ⊔ A p and B = B 1 ⊔ ... ⊔ B q such that each of the p + q sets A 1 , ... , A p , B 1 , ... , B q has measure exactly µ(B)/q. Again, each piece is in A. We conclude by the first part of the claim that all these sets, except A 0 , have the same κ-value which has to be κ(B)/q. Thus κ(A) ⩾ p i=1 κ(A i ) ⩾ p κ(B)/q, as required.
Now, take an arbitrary positive r ∈ R. Suppose on the contrary that κ(A) < r κ(B). Then, trivially, κ(B) > 0 and there are p, q ∈ N with κ(A) < p q κ(B) and p/q < r. However, this contradicts the conclusion of the previous paragraph.
Observe that, for every µ-null set A ∈ A, we have that κ(A) = 0. Indeed, since A = (A \ U 1 ) ⊔ (A ∩ U 1 ), it is enough by the additivity of κ to prove this when A ⊆ Ω \ U i for some i ∈

Concluding Remarks
Note that there are paradoxical actions which are not expanding. For example, take two isometries in SO(3) such that they generate a rank-2 free group F and the action b : F ↷ S 2 has spectral gap, let c : Z ↷ S 1 where the generator of Z is a rotation by some angle θ ̸ ∈ πQ, and consider the action a : (F × Z) ↷ Ω, where Ω := S 2 × S 1 and a((γ, n), (x, y)) := (b(γ, x), c(n, y)). Since b is paradoxical (e.g. by [49,Theorem 5.5]), we can double the whole space Ω also under the action a (by using only elements of the form (γ, 0) ∈ F × Z). Since every orbit of b or c is dense (in the former case by, for example, the proof of [25,Theorem 2.7]), this also holds for the new action a. Thus, with respect to the action a, every set with non-empty interior covers Ω and is equidecomposable to Ω by the proof of Proposition 1.8(i). It follows that the action a is paradoxical.
On the other hand, the products S 2 × Y , where we take for Y almost b-invariant sets of measure 1/2, are almost a-invariant sets. This shows that the action a is not expanding.
An interesting open question that remains is whether any two bounded Borel subsets of R n , n ⩾ 3, that have non-empty interior and the same Lebesgue measure are equidecomposable with Borel pieces. Our Theorem 1.2 implies that the answer is in the affirmative, provided we can first remove a Borel null set from each set. Also, it would be interesting to find some alternative characterization of bounded measurable subsets of R n , n ⩾ 3, that are domains of expansion. By Lemma 2.5, essentially covering a non-empty open set is a sufficient condition. However, it is not necessary by Theorem 1.13. Of course, the above questions can also be asked for other group actions. In particular, they are open for all actions listed in Example 1.9, as far as the authors know.
Another interesting question is to get reasonable upper bounds on the sizes of expanding sets (namely to estimate the function s(η; X) from Section 7) when, for example, X = [0, 1] 3 is the unit cube under the action Iso(R 3 ) ↷ R 3 .