A NOTE ON THE POLAR DECOMPOSITION IN METRIC SPACES

Abstract

The analogue of polar coordinates in the Euclidean space, a polar decomposition in a metric space, if well-defined, can be very useful in dealing with integrals with respect to a sufficiently regular measure. In this note we handle the technical details associated with such polar decompositions.

Supplementary Information

The online version contains the Armenian language version of the article available at https://doi.org/10.1007/s10958-023-06674-w.

Introduction

From a bird’s perspective, in dealing with integrals of measurable functions with respect to infinite measures, it is often technically very convenient to view the full integral as a double integral, where the “infiniteness” of the measure falls completely onto the outer integral, e.g.,

$$\begin{aligned} \int \limits _\mathcal {M}f(x)d\mu (x)=\int \limits _{[0,R)}\int \limits _{\mathcal {S}_r}f(x)d\omega _r(x)d\nu (r),\quad R\in (0,+\infty ]. \end{aligned}$$

This assumes a measure space bundle $(\mathcal {M},\mu )\xrightarrow []{\rho }([0,R),\nu )$, $\mathcal {S}_r=\rho ^{-1}(\{r\})$, and a disintegration

$$\begin{aligned} \mu (\cdot )=\int \limits _{[0,R)}\omega _r(\cdot )d\nu (r), \end{aligned}$$

with $\omega _r$ bounded and concentrated on $\mathcal {S}_r$ for $\nu$-a.e. $r\in [0,R)$. If $\mathcal {M}$ can be seen as a product $[0,R)\times \mathcal {S}$, i.e., there is a Borel almost isomorphism $([0,R)\times \mathcal {S},\tilde{\mu })\xrightarrow []{\Psi }\mathcal {M}$ such that $\rho \circ \Psi =\pi _1$ (here $\pi _1(r,\varphi )=r$), then new variables $\mathcal {M}\ni x\mapsto (r,\varphi )\in [0,R)\times \mathcal {S}$ can be introduced, so that

$$\begin{aligned} \int \limits _\mathcal {M}f(x)d\mu (x)=\int \limits _{[0,R)}\int \limits _{\mathcal {S}}\tilde{f}(r,\varphi )d\tilde{\omega }_r(\varphi )d\nu (r),\quad \tilde{f}(r,\varphi )=f\circ \Psi (r,\varphi ),\quad d\tilde{\omega }_r(\varphi )=d\omega _r\circ \Psi (r,\varphi ). \end{aligned}$$

The choice of the function $\rho$ is pretty arbitrary at this level, but in metric spaces there is a natural candidate - the distance from a fixed point $x_0\in \mathcal {M}$ of origin. If $\mathcal {M}=\mathbb {R}^n$ and $d\mu (x)=\dot{\mu }(x)dx$, $\dot{\mu }\in C^\infty (\mathbb {R}^n)$ a smooth positive density, then this simply amounts to switching to polar coordinates,

$$\begin{aligned} \int \limits _{\mathbb {R}^n}f(x)d\mu (x)=\int \limits _0^\infty \int \limits _{\mathcal {S}_r}f(r,\varphi )\omega (r,\varphi )d\varphi dr,\quad \forall f\in L^1_\textrm{loc}(\mathbb {R}^n), \end{aligned}$$

$$\begin{aligned} \mathcal {S}_r\doteq \left\{ x\in \mathbb {R}^n\,\vline \quad |x|=r\right\} ,\quad \omega (r,\varphi )=\frac{d\mu (x(r,\varphi ))}{d\varphi dr}. \end{aligned}$$

More generally, let $\mathcal {M}=(\mathcal {M},d)$ be a metric space and $\mu$ a measure on the Borel $\sigma$-algebra $\Sigma$ of $\mathcal {M}$. All measures in this paper will be assumed positive, $\mu (\cdot )\ge 0$, and $\sigma$-additive, i.e.,

$$\begin{aligned} \left( \forall \{A_k\}_{k=1}^\infty \subset \Sigma \right) \quad (\forall k,l\in \mathbb {N})\,k\ne l\,\Rightarrow \,A_k\cap A_l=\emptyset \quad \Rightarrow \quad \mu \left( \bigcup _{k=1}^\infty A_k\right) =\sum _{k=1}^\infty \mu (A_k). \end{aligned}$$

We will call every such measure a Borel measure, although some sources require that Borel measures have additional neat properties. Fix a point $x_0\in \mathcal {M}$ and denote by $\rho _{x_0}\in C(\mathcal {M},[0,R))$ the function given by

$$\begin{aligned} \rho _{x_0}(x)=d(x,x_0),\quad \forall x\in \mathcal {M}. \end{aligned}$$

(1)

We will speak of spheres $\mathcal {S}_r(x_0)$ and balls $\mathcal {B}_r(x_0)$ of radius $r\in [0,R)$ centred at $x_0$,

$$\begin{aligned} \mathcal {S}_r(x_0)\doteq \rho _{x_0}^{-1}(\{r\}),\quad \mathcal {B}_r(x_0)\doteq \rho _{x_0}^{-1}([0,r))=\bigcup _{0\le s<r}\mathcal {S}_s(x_0),\quad \forall r\in [0,R). \end{aligned}$$

The Borel measure $\nu _{x_0}$ on [0, R) is defined by

$$\begin{aligned} \nu _{x_0}\doteq \mu \circ \rho _{x_0}^{-1}, \end{aligned}$$

(2)

so that

$$\begin{aligned} \nu _{x_0}([0,r))=\mu \left( \mathcal {B}_r(x_0)\right) ,\quad \forall r\in [0,R). \end{aligned}$$

Definition 1

We will say that the metric measure space $(\mathcal {M},d,\mu )$ admits a polar decomposition at the point $x_0\in \mathcal {M}$ if there exists a field $r\mapsto \omega _r$ of bounded (i.e., finite) Borel measures $\omega _r$ on $\mathcal {M}$ for $\nu _{x_0}$-almost all $r\in [0,R)$ such that for every measurable function $f:\mathcal {M}\rightarrow \mathbb {C}$, for which the integral

$$\begin{aligned} \int \limits _\mathcal {M}f(x)d\mu (x) \end{aligned}$$

makes sense (finite or infinite), the following conditions hold:

• 1. The map

  $$\begin{aligned}{}[0,R)\ni r\mapsto \int \limits _\mathcal {M}f(x)d\omega _r(x) \end{aligned}$$

  is Borel measurable.

• 2. The measure $\omega _r$ is concentrated on $\mathcal {S}_r(x_0)$ for $\nu _{x_0}$-a.e. $r\in [0,R)$, i.e.,

  $$\begin{aligned} \int \limits _\mathcal {M}f(x)d\omega _r(x)=\int \limits _{\mathcal {S}_r(x_0)}f(x)d\omega _r(x). \end{aligned}$$

• 3. We have

  $$\begin{aligned} \int \limits _\mathcal {M}f(x)d\mu (x)=\int \limits _{[0,R)}\int \limits _{\mathcal {S}_r(x_0)}f(x)d\omega _r(x)d\nu _{x_0}(r). \end{aligned}$$

Remark 1

If the measure $\nu _{x_0}$ is absolutely continuous with respect to the Lebesgue measure on [0, R) then there exists a Borel measurable, non-negative, locally integrable function $\dot{\nu }_{x_0}$ such that $d\nu _{x_0}(r)=\dot{\nu }_{x_0}(r)dr$. In that case, the factor $\dot{\nu }_{x_0}$ can be incorporated into the measures $\omega _r$, so that part 3. of Definition 1 becomes

$$\int \limits _\mathcal {M}f(x)d\mu (x)=\int \limits _0^R\int \limits _{\mathcal {S}_r(x_0)}f(x)d\omega _r(x)dr.$$

This corresponds to the definition (1.2) of polar decomposition in [10], so that in this case the further integration-by-parts techniques used in that work are applicable.

From a measure theory perspective, a polar decomposition is merely an instance of measure disintegration, and our job in this paper is to study sufficient conditions for a measure disintegration of this kind in a metric space. Whether or not $\nu _{x_0}$ is absolutely continuous with respect to the Lebesgue measure is unrelated to the disintegration and is a reflection of the geometric shape of $\mathcal {M}$ and the distribution of $\mu$.

A discussion of assumptions

Finite measure of balls

A polar decomposition as in Definition 1 is useful when the condition 3. is satisfied non-trivially. Let us write it in the measure form,

$$\begin{aligned} \mu (\cdot )=\int \limits _{[0,R)}\int \limits _{\mathcal {S}_r(x_0)}\omega _r(\cdot )d\nu _{x_0}(r). \end{aligned}$$

If the measure $\mu$ is bounded, then everything works well, but the point of a polar decomposition is often introduced to deal with an infinite measure. If the measure $\mu$ is not locally finite, then certainly $C(\mathcal {M})\not \subset L^1_\textrm{loc}(\mathcal {M},\mu )$, and one is led to consider very thin spaces of integrable functions. Arguably, there is little use of such measures in functional analysis, and assuming the measure $\mu$ is locally finite is not a serious loss of generality.

If $\mu$ is locally finite at $x_0$, then let

$$\begin{aligned} R_*\doteq \inf \left\{ r\in [0,R)\,\vline \quad \mu (\mathcal {B}_r(x_0))=+\infty \right\} \in (0,+\infty ], \end{aligned}$$

(3)

with the convention that $\inf \emptyset =R$. If $R_*<R$, then for every measurable set A with $\rho _{x_0}(A)\subset [R_*,R)$ we have $\mu (A)=+\infty$, which makes the formula 3. in Definition 1 useless for $r\ge R_*$. As far as a polar decomposition around the fixed point $x_0$ is concerned, one can consider the subspace $\mathcal {M}'\doteq \mathcal {B}_{R^*}(x_0)$ instead of $\mathcal {M}$ and set $R\doteq R_*$. This way all balls $\mathcal {B}_r(x_0)$ for $r\in [0,R)$ will have finite volume. In particular, $\mu$ will be locally finite on $\mathcal {M}$.

Separability and completeness of $(\mathcal {M},d)$

The separability of the metric space $(\mathcal {M},d)$, and thus of its Borel $\sigma$-algebra and all Lebesgue functions on it, is of utmost importance for most functional analytical considerations. It is therefore not a serious restriction to assume $(\mathcal {M},d)$ is separable. However, we will see in Theorem 2 below that polar decompositions can be obtained even without separability, at the cost of completeness of $\mu$ and further restrictions.

If the metric space $(\mathcal {M},d)$ is not complete, then closed balls may fail to be compact and may have infinite measure even if $\mu$ is locally finite and $\mathcal {M}$ locally compact. Indeed, take $\mathcal {M}=(0,1)$ with the Euclidean metric and the measure $d\mu (x)=dx/x$. Assuming the completeness of $\mathcal {M}$ wards against pathological behaviour of locally finite Borel measures. However, our main results, Theorem 1 and Theorem 2, will not assume the completeness of $(\mathcal {M},d)$.

On the other hand, if we agree to embrace separability and completeness, then we are in the realm of Polish metric spaces $(\mathcal {M},d)$. Borel measures on Polish spaces are very tame and provide convenient technical ease.

$\sigma$-compactness of $\mathcal {M}$

Under this rubric we will present two technical statements which can be seen as partial converses of each other. Together they will show that $\sigma$-compactness of $\mathcal {M}$ is not too far from being the right context for polar decompositions, although it is not strictly necessary. Recall that a topological space is called $\sigma$-compact if it is the union of countably many compact subspaces.

First, we will show that if we are willing to assume that the metric space $(\mathcal {M},d)$ is Polish, then we could instead take $\mathcal {M}$ as $\sigma$-compact.

Proposition 1

Every locally finite Borel measure on a Polish space is concentrated on a $\sigma$-compact subspace, which is the union of a countable locally finite family of countable disjoint unions of compact sets.

Proof

Let $\mathcal {M}$ be a Polish space and $\mu$ a locally finite Borel measure. Let $\{U_x\}_{x\in \mathcal {M}}$ be a family of open subsets such that $\mu (U_x)<\infty$ for $\forall x\in \mathcal {M}$, by local finiteness of $\mu$. As a metrizable space, $\mathcal {M}$ is paracompact [12], and thus there is a locally finite open refinement $\{V_\alpha \}_{\alpha \in A}$ of $\{U_x\}_{x\in \mathcal {M}}$. But $\mathcal {M}$ is also second countable and hence Lindelöf, so that $\{V_\alpha \}_{\alpha \in A}$ has a countable locally finite subcover $\{V_k\}_{k=1}^\infty$.

Fix $k\in \mathbb {N}$. Clearly, $\exists x\in \mathcal {M}$ s.t. $V_k\subset U_x$, and therefore $\mu (V_k)\le \mu (U_x)<\infty$. As an open subset of the Souslin space $\mathcal {M}$, $V_k$ is itself a Souslin space (Lemma 6.6.5 in [3]), and the restriction of $\mu$ to $V_k$ is a bounded Borel measure. By Theorem 7.4.3 in [3], $\mu$ is tight on $V_k$, i.e.,

$$(\forall \epsilon >0)(\exists K_\epsilon \Subset V_k\,\text{ compact})\,\mu (V_k\setminus K_\epsilon )<\epsilon .$$

Let $X_1\doteq V_k$ and $K_1\Subset X_1$ compact be chosen as above with $\mu (X_1\setminus K_1)<1$. Then $X_2\doteq X_1\setminus K_1$ is an open subset of a Souslin space, and thus it is a Souslin subspace, and $\mu$ restricted to $X_2$ gives a bounded Borel measure, which is again tight by the same theorem. Continuing this process inductively, on the step $n\in \mathbb {N}$ we choose $K_n\Subset X_n$ compact such that $\mu (X_n\setminus K_n)<\frac{1}{n}$. Denote

$$Y_k\doteq \bigcup _{n=1}^\infty K_n\subset X.$$

Then

$$\mu (V_k\setminus Y_k)\le \mu \left( V_k\setminus \bigcup _{n=1}^NK_n\right) =\mu (X_N\setminus K_N)<\frac{1}{N},\quad \forall N\in \mathbb {N},$$

showing that $\mu (V_k\setminus Y_k)=0$.

Now, denote

$$Y\doteq \bigcup _{k=1}^\infty Y_k\subset \mathcal {M}.$$

Then

$$\mu \left( Y^\complement \right) =\mu \left( \bigcap _{\ell =1}^\infty Y_\ell ^\complement \right) =\mu \left( \bigcup _{k=1}^\infty \left[ V_k\cap \bigcap _{\ell =1}^\infty Y_\ell ^\complement \right] \right) \le$$

$$\mu \left( \bigcup _{k=1}^\infty V_k\cap Y_k^\complement \right) =\mu \left( \bigcup _{\ell =1}^\infty (V_k\setminus Y_k)\right) \le \sum _{k=1}^\infty \mu (V_k\setminus Y_k)=0,$$

showing that $\mu$ is concentrated on Y. Since the cover $\{V_k\}_{k=1}^\infty$ is locally finite, the assertion of the proposition holds. $\square$

Remark 2

Proposition 1 above barely falls short of showing that a locally finite Borel measure on a Polish space is concentrated on a locally compact subspace; if countable disjoint unions of compact sets were always subsets of locally compact sets, their locally finite union would also be a subset of a locally compact set. But this is indeed not the case, which we will demonstrate on a simple example.

Example 1

Let $\mathcal {M}=l^2(\mathbb {R})$ and let $\gamma :l^2_c(\mathbb {Q})\rightarrow \mathbb {N}$ be any enumeration of the countable dense subset $l^2_c(\mathbb {Q})\subset l^2(\mathbb {R})$ of rational sequences with finite support. For any Borel subset $U\subset l^2(\mathbb {R})$ set

$$\begin{aligned} \mu (U)\doteq \sum _{x\in U\cap l^2_c(\mathbb {Q})}\frac{1}{\gamma (x)^2}. \end{aligned}$$

This defines a bounded Borel measure $\mu$ on the Polish space $l^2(\mathbb {R})$, which is indeed concentrated on the countable union $l^2_c(\mathbb {Q})$ of compact subsets (singletons). But there does not exist a locally compact subspace Y such that $l^2_c(\mathbb {Q})\subset Y\subset l^2(\mathbb {R})$. Indeed, if such a space existed, then the point 0 would have a compact neighbourhood $K\Subset Y$. Then also $K\Subset l^2(\mathbb {R})$ and thus $\overline{K}^{l^2}=K$. But $\exists \epsilon >0$ such that $\mathcal {B}_\epsilon (0)\cap l^2_c(\mathbb {Q})\subset K$, and hence,

$$\begin{aligned} \overline{\mathcal {B}_\epsilon (0)\cap l^2_c(\mathbb {Q})}^{l^2}=\overline{\mathcal {B}_\epsilon (0)}\subset \overline{K}^{l^2}=K, \end{aligned}$$

which is a contradiction, because the closed balls are not compact in $l^2(\mathbb {R})$.

Below is a slightly more sophisticated example of the same nature.

Example 2

Let $\mathcal {M}=l^2(\mathbb {R})$ and let

$$\begin{aligned} \mathcal {M}=\bigsqcup _{k=1}^\infty \mathcal {M}_k,\quad \mathcal {M}_k\doteq \left\{ x\in l^2(\mathbb {R})\,\vline \quad k-1\le \Vert x\Vert _2<k\right\} ,\quad \forall k\in \mathbb {N}. \end{aligned}$$

For every $k\in \mathbb {N}$ let $\gamma _k:l^2_c(\mathbb {Q})\cap \mathcal {M}_k\rightarrow \mathbb {N}$ be any enumeration of points. Fix any non-negative sequence $\{a_k\}_{k=1}^\infty \in [0,+\infty )^\mathbb {N}$. For every Borel subset $U\subset l^2(\mathbb {R})$ set

$$\begin{aligned} \mu (U)\doteq \sum _{x\in U\cap l^2(\mathbb {Q})}m(x),\quad m(x)\doteq \frac{a_k}{\gamma _k(x)^2},\quad \forall x\in l^2(\mathbb {Q})\cap \mathcal {M}_k,\quad \forall k\in \mathbb {N}. \end{aligned}$$

Then $\mu$ is a Borel measure on $l^2(\mathbb {R})$, which is bounded or infinite depending on whether $\sum _ka_k$ converges or not. Weather $\mu$ is bounded or not, the measures of all balls satisfy $0<\mu (\mathcal {B}_\epsilon (x))<\infty$, $\forall x\in l^2(\mathbb {R})$, $\forall \epsilon >0$. As in the example before, $\mu$ is concentrated on $l^2_c(\mathbb {Q})$ which is not a subset of any locally compact subspace.

On the other hand, the main assumptions of Theorem 1 - separability of $\mathcal {M}$ and inner regularity of $\mu$ - can be achieved at once if $\mathcal {M}$ is assumed $\sigma$-compact.

Lemma 1

If $\mathcal {M}$ is $\sigma$-compact, then it is separable, and every semifinite (see Definition 211F in [7]) Borel measure on it is inner regular.

Proof

Let $\mathcal {M}=\bigcup \limits _{n=1}^\infty K_n$ with $K_n\Subset \mathcal {M}$ compact for $n\in \mathbb {N}$. Each $K_n$ is a compact metric space and hence Polish. Thus, the separability of $\mathcal {M}$ is obvious. Let

$$\begin{aligned} f\doteq \bigoplus _{n=1}^\infty {\text {id}}_{K_n},\quad f:\bigoplus _{n=1}^\infty K_n\rightarrow \mathcal {M}, \end{aligned}$$

then f is continuous. The topological sum of countably many Polish spaces is Polish, and therefore $\mathcal {M}$ is a Souslin space. By Theorem 423E in [9], all open subsets of $\mathcal {M}$ are K-analytic, and by Proposition 432C in [9], $\mu$ is inner regular. $\square$

Remark 3

Suppose that the finite measure of balls (and hence local finiteness of $\mu$) is taken for granted. Then the $\sigma$-compactness of the space $\mathcal {M}$ guarantees the existence of a polar decomposition by Lemma 1 and Theorem 1. On the other hand, if $\mathcal {M}$ is separable and complete, then by Proposition 1 we can restrict $\mathcal {M}$ to a $\sigma$-compact subspace of full $\mu$-measure.

The existence of polar decomposition

As noted before, a polar decomposition is a particular case of a measure disintegration. There are at least two sufficiently detailed modern expositions of measure disintegration that we know of: [3] by Bogachev and [9] by Fremlin. Theorem 10.4.8 together with Corollary 10.4.10 in [3] have the merit of delivering a slightly stronger result, a regular conditional measure as opposed to a mere disintegration, which is valid also for signed measures. The deficiency of the treatment in [3] is in the author’s preoccupation with bounded measures only, which they merrily admit in the introduction to the first volume [2]. Instead, Fremlin’s section 452 in [9] delivers a disintegration for $\sigma$-finite positive measures, which is just fine for our purposes. Otherwise the two expositions of the subject are essentially comparable.

The main result of this note is the following existence theorem.

Theorem 1

Let $(\mathcal {M},d)$ be a separable metric space and $x_0\in \mathcal {M}$. Let $\mu$ be a Borel measure on $(\mathcal {M},\Sigma )$ such that:

• 1. $\mu$ is inner regular¹ (i.e., tight), that is,

  $$\begin{aligned} (\forall A\in \Sigma )\quad \mu (A)=\sup \left\{ \mu (K)\,\vline \quad K\in \Sigma ,\quad K\subset A,\quad K\,\text{ compact }\right\} . \end{aligned}$$

• 2. Balls $\mathcal {B}_r(x_0)\subset \mathcal {M}$ of any radius²$r\in [0,R)$ have finite measure $\mu (\mathcal {B}_r(x_0))<\infty$.

Then $(\mathcal {M},d,\mu )$ admits a polar decomposition at $x_0$.

Proof

Recall the function $\rho _{x_0}\in C(\mathcal {M},[0,R))$ defined as in Eq. 1 and the Borel measure $\nu _{x_0}=\mu \circ \rho _{x_0}^{-1}$ on [0, R), so that $\rho _{x_0}:(\mathcal {M},\mu )\rightarrow ([0,R),\nu _{x_0})$ is an inverse-measure-preserving map as in Definition 235G in [9]. Since [0, R) is Lindelöf, $\nu _{x_0}$ is $\sigma$-finite if and only if it is locally finite, and that is guaranteed by $\nu _{x_0}([0,r))=\mu (\mathcal {B}_r(x_0))<\infty$ for $\forall r>0$. The countable generation of the Borel $\sigma$-algebra of $\mathcal {M}$ is given by the separability of $\mathcal {M}$. Now by Exercise 452X(l) in [9], there exists a disintegration $\{\omega _r\}_{r\in [0,R)}$ of $\mu$ over $\nu _{x_0}$ consistent with $\rho _{x_0}$, which is essentially unique. For $\forall r\in [0,R)$, the object $\omega _r$ is a Borel probability measure on $\mathcal {M}$. Since [0, R) is countably separated (Lemma 343E in [8]), by Proposition 452G(c) in [9] the disintegration $\{\omega _r\}_{r\in [0,R)}$ is strongly consistent with $\rho _{x_0}$ (see Definition 452E in [9]). Thus, every $\omega _r$ is concentrated on $\rho _{x_0}^{-1}(\{r\})=\mathcal {S}_r(x_0)$. By Proposition 452F in [9], for every measurable $f:\mathcal {M}\rightarrow \mathbb {C}$ such that $\int \limits _\mathcal {M}f(x)d\mu (x)$ makes sense, we have

$$\begin{aligned} \int \limits _\mathcal {M}f(x)d\mu (x)=\int \limits _{[0,R)}\int \limits _{\mathcal {S}_r(x_0)}f(x)d\omega _r(x)d\nu _{x_0}(r). \end{aligned}$$

The existence of a polar decomposition is established. $\square$

Corollary 1

Let $(\mathcal {M},d)$ be a $\sigma$-compact metric space and $x_0\in \mathcal {M}$. Let $\mu$ be a Borel measure on $\mathcal {M}$ such that all open balls centred at $x_0$ have finite $\mu$-measure. Then $(\mathcal {M},d,\mu )$ admits a polar decomposition at $x_0$.

Proof

Since all open balls $\mathcal {B}_r(x_0)$ have finite measure, $\mu$ is locally finite and hence semifinite. By Lemma 1, $\mathcal {M}$ is separable and $\mu$ is inner regular. It remains to apply Theorem 1. $\square$

For a metric space, separability, second countability and the Lindelöf property are mutually equivalent. Below we give a version of the above theorem which does not require the separability of $\mathcal {M}$, but instead puts harsher conditions on the measure $\mu$. Note that there is a discrepancy between the definition of a Radon measure space in [9] and that of a Radon measure in most of the standard literature. Commonly, a Radon measure is a locally finite inner regular Borel measure. But Fremlin defines a Radon measure space what would normally be called a locally determined Borel measure space with a complete Radon measure (see Definition 411H(b) in [9]).

Theorem 2

Let $(\mathcal {M},d)$ be a metric space and $x_0\in \mathcal {M}$. Let $\mu$ be a Borel measure on $(\mathcal {M},\Sigma )$ such that:

• 1. $(\mathcal {M},\Sigma ,\mu )$ is a Radon measure space in the sense of Definition 411H(b) in [9].

• 2. Balls $\mathcal {B}_r(x_0)\subset \mathcal {M}$ of any radius $r\in [0,R)$ have finite measure $\mu (\mathcal {B}_r(x_0))<\infty$.

Then $(\mathcal {M},d,\mu )$ admits a polar decomposition at $x_0$.

Proof

Since the measure $\nu _{x_0}$ is still $\sigma$-finite, by Theorem 211L(c) in [7] it is strictly localizable. By Proposition 452O in [9], there exists a disintegration $\{\omega _r\}_{r\in [0,R)}$ of $\mu$ over $\nu _{x_0}$ consistent with $\rho _{x_0}$, and every $(\mathcal {M},\omega _r)$ is a bounded Radon measure space. The rest proceeds as before. $\square$

Remark 4

Note that Theorem 2 produces a polar decomposition where $(\mathcal {S}_r(x_0),\omega _r)$ is a bounded Radon measure space as per Definition 411H(b) in [9].

The absolute continuity of the measure $\nu _{x_0}$

To understand the nature of the question of absolute continuity of the measure $\nu _{x_0}$ in Eq. 2 with respect to the Lebesgue measure, we will demonstrate very simple examples. The measure $\nu _{x_0}$ reflects the measure $\mu$ as distributed among spheres $\mathcal {S}_r(x_0)$, and its behaviour depends on the homogeneity of the measure $\mu$ as well as the commensurability of different spheres $\mathcal {S}_r(x_0)$. If $\mu$ is sufficiently homogeneous (whatever that means in a given context), then a non-absolutely continuous measure $\nu _{x_0}$ may arise due to an irregular foliation by non-commensurate spheres (again, whatever that means). On the plane, take $\mathcal {M}$ to be the union of an interval $[0,a]\times \{0\}$ on the horizontal axis and any arc on the unit circle centred at (0, 0), and let $x_0=(0,0)$. Let $\mu$ be the Lebesgue measure on $\mathcal {M}$ measuring the length of curves in the usual way. Then the measure $\nu _{x_0}$ will have a singular part concentrated at $r=1$. On the other hand, for any other $x_0\in \mathcal {M}$ the measure $\nu _{x_0}$ would be absolutely continuous. But we can add arcs of various circles to produce more, infinitely many points $x_0\in [0,a]$ with singular measures $\nu _{x_0}$.

Meanwhile, if the measure $\mu$ is not homogeneous enough, then singular $\nu _{x_0}$ can arise even on ideally shaped metric spaces $\mathcal {M}$. Take $\mathcal {M}$ to be any standard smooth geometric object with Euclidean distance metric, e.g., the unit interval [0, 1], and let $\mu$ contain a single point mass at any point. Then no matter where we take the point $x_0\in [0,1]$, the measure $\nu _{x_0}$ will not be absolutely continuous. The extreme example is the Cantor measure $\mu$, the Borel measure on [0, 1] given by the Cantor function, which is singular everywhere on [0, 1] [6]. Here, too, the measure $\nu _{x_0}$ is purely singular for all $x_0\in [0,1]$.

Conditions on $(\mathcal {M},d,\mu )$ and $x_0\in \mathcal {M}$ under which $\nu _{x_0}$ is absolutely continuous are a delicate subject which should be studied separately. One thing that can be noticed is that the problems arising from the non-homogeneity of $\mu$ are far harder than those due to the uneven shape of $\mathcal {M}$.

Riemannian manifolds

Let us first consider the problem of finite measure of open balls in a Riemannian manifold.

Remark 5

Let $(\mathcal {M},g)$ be a smooth connected Riemannian manifold with its geodesic distance $d_g$ and Riemannian volume $\upsilon _g$. For every ball $\mathcal {B}_r(x_0)$, if the Ricci curvature is bounded from below on $\mathcal {B}_r(x_0)$ then its Riemannian volume is finite, $\upsilon _g(\mathcal {B}_r(x_0))<\infty$.

This statement follows immediately from the Gromov-Bishop-Günther comparison theorem, see Theorem 8.45 in [5].

Note that the $\sigma$-compactness of a connected manifold is obvious, therefore, Corollary 1 is applicable to Riemannian manifolds with Ricci curvature bounded from below on balls, and Borel measures $\mu$ having bounded Radon-Nikodym derivative with respect to the Riemannian volume. The latter condition guarantees finite $\mu$-measure of open balls, while their $\upsilon _g$-measures are already finite by the above remark.

Let us now turn to the absolute continuity of the measure $\nu _{x_0}$.

Proposition 2

Let $(\mathcal {M},g)$ be a smooth connected Riemannian manifold with its geodesic distance $d_g$ and Riemannian volume $\upsilon _g$. Suppose that the Ricci curvature is bounded from below on every ball $\mathcal {B}_r(x_0)$. For every Borel measure $\mu$ on $\mathcal {M}$ that is absolutely continuous with respect to $\upsilon _g$, and for every point $x_0$, the measure $\nu _{x_0}$ is absolutely continuous.

Proof

It suffices to prove the statement for $\mu =\upsilon _g$. Fix $x_0\in \mathcal {M}$ and take $a\in (0,R)$, with R as in Eq. 3. Since the Ricci curvature is bounded from below on $\mathcal {B}_a(x_0)$, by Gromov-Bishop-Günther comparison theorem (Theorem 8.45 in [5]) there exists $c_a>0$ such that

$$\begin{aligned} 1\ge \frac{\mu \left( \overline{\mathcal {B}_r(x_0)}\right) }{c_ar^n}\ge \frac{\mu \left( \overline{\mathcal {B}_s(x_0)}\right) }{c_as^n},\quad n\doteq \dim \mathcal {M}, \end{aligned}$$

for all $0\le r\le s\le a$. Thus,

$$\begin{aligned} 0\le \mu \left( \overline{\mathcal {B}_s(x_0)}\right) -\mu \left( \overline{\mathcal {B}_r(x_0)}\right) \le \frac{\mu \left( \overline{\mathcal {B}_r(x_0)}\right) }{r^n}(s^n-r^n)\le nc_aa^{n-1}(s-r), \end{aligned}$$

which shows that the function

$$\begin{aligned} r\mapsto \mu \left( \overline{\mathcal {B}_r(x_0)}\right) =\nu _{x_0}([0,r]) \end{aligned}$$

(4)

is Lipschitz and hence absolutely continuous on any compact interval $[0,b]\subset [0,a)$. Because this is true for every $a\in (0,R)$, we find that the function Eq. 4 is absolutely continuous on every compact interval $[0,b]\subset [0,R)$. Therefore, by the fundamental theorem of Lebesgue integral calculus, the measure $\nu _{x_0}$ is absolutely continuous. $\square$

A very similar argument can be found in Proposition 2.1 in [11]. We are grateful to Karen Hovhannisyan for pointing out this paper to us.

Corollary 2

Let $(\mathcal {M},g)$ be a smooth connected Riemannian manifold with its geodesic distance $d_g$ and Riemannian volume $\upsilon _g$. Suppose that the Ricci curvature is bounded from below on every ball $\mathcal {B}_r(x_0)$. Let $\mu$ be a Borel measure on $\mathcal {M}$ such that $d\mu (x)=\dot{\mu }(x)d\upsilon _g(x)$ for $\dot{\mu }\in L^\infty (\mathcal {M},\upsilon _g)$. Then at every $x_0\in \mathcal {M}$, the metric measure space $(\mathcal {M},d,\mu )$ admits a polar decomposition in the form given in Remark 1.

Remark 6

The Ricci curvature is bounded globally on Riemannian manifolds with bounded geometry, including compact manifolds and homogeneous spaces. More generally, by Hopf-Rinow theorem, in a complete Riemannian manifold all balls are relatively compact, and thus the Ricci curvature is automatically bounded on each $\mathcal {B}_r(x_0)$. This makes the application of Corollary 2 completely straightforward in complete Riemannian manifolds.

Note that for complete Riemannian manifolds taken with their natural volume measure, there are more geometric and explicit polar decompositions available. See, e.g., Proposition III.3.1 in [4].

Invariant sub-Riemannian structures on Lie groups

Let G be a connected real Lie group, equipped with a left-invariant sub-Riemannian structure $(G,\mathcal {D},\langle ,\rangle )$ and the corresponding Carnot-Carathéodory metric d [1]. It is clear that (G, d) is complete as a metric space (Corollary 7.51 in [1]), so that by Proposition 3.47 in [1], all balls $\mathcal {B}_r(x_0)$ are relatively compact, and hence have finite measure $\mu (\mathcal {B}_r(x_0))<\infty$ for any locally finite Borel measure $\mu$. On the other hand, the $\sigma$-compactness of G is also clear, and by Corollary 1 we establish the existence of a polar decomposition at every point $x_0\in G$.

However, the question whether the measure $\nu _{x_0}$ is absolutely continuous appears to be open today.