Multiplier properties for the AP -Henstock integral

: In this paper, we investigate some properties of the AP -Henstock integral on a compact set and prove that the product of an AP -Henstock integrable function and a function of bounded variation is AP -Henstock integrable. Furthermore, we prove that the product of an AP -Henstock integrable function and a regulated function is also AP -Henstock integrable. We also define the AP -Henstock integral on an unbounded interval, investigate some properties, and show similar multiplier properties.


Introduction and Main Results
T he Henstock integral of real-valued functions was first defined by Henstock [1,2]. It is a direct generalization of the Riemann integral because it uses the concept of the tagged partition and the Riemann sum. In Henstock integral, the concept of the norm of a tagged partition in the Riemann integral is replaced by the positive gauge function. Therefore, the definition of the Henstock integral is as simple as the definition of the Riemann integral. On the other hand, to introduce the Lebesgue integral, a good amount of measure theory is required. It is one of the reasons why the Henstock integral is simpler than the Lebesgue integral. However, the Henstock integral is also a generalization of the Lebesgue integral. Every Lebesgue integrable function is Henstock integrable, and both integrals are the same. One of the Lebesgue integral deficits is that not every continuous function that is differentiable everywhere, possible except for a countable number of points, is recovered from its derivative by the Lebesgue integrable. In this sense, we say that the Lebesgue integral does not recover a function from its derivative. On the other hand, the Henstock integral overcomes this drawback: every continuous function F : [a, b] → R that is differentiable everywhere except for countable number of points on [a, b] can be recovered from its derivative by the Henstock integral, and x a F ′ = F(x) − F(a).
The approximate Henstock integral (AP-Henstock integral) [3] further generalizes the Henstock integral by using the concept of the approximate derivative [4], and the gauge function in the Henstock integral is generalized to the choice in the AP-Henstock integral. Every Henstock integrable function is AP-Henstock integrable, and the integrals are the same. Furthermore, the AP-Henstock integral recovers an approximate continuous function from its approximate derivative: Theorem 1. [4] For the detailed introduction of the AP-Henstock integral, the reader is referred to [5][6][7][8][9].
Although the space of integrable functions is closed under the addition and the scalar multiplication, the product of two integrable functions is not necessarily integrable. Therefore, it is an important question of what kind of properties of an integrable function guarantees the integrability of the product of two integrable functions. We call those properties the multiplier properties and the related theorems the multiplier theorems. For example, in the case of the Lebesgue, Denjoy, and Henstock integral, the product of an integrable function f : [a, b] → R and a function of bounded variation(which is integrable in any sense mentioned above) G : where F(x) = x a f , and the last integral is the Riemann-Stieltjes integral. For the Perron integral, a restricted condition on the function of bounded variation is required, see [6]. In this paper, we develop the same kind of multiplier properties that ensure the AP-Henstock integrability of the product of two AP-Henstock integrable functions.
On the other hand, the Henstock integral can be defined on unbounded intervals [10]. It is now known as Hake's theorem that there is no such thing as an "improper integral" for the Henstock integral. By modifying the definition of the Henstock integral on unbounded intervals, we define the AP-Henstock integrals on unbounded intervals. In this setting, we investigate some properties for the AP-Henstock integral and prove that, under some additional conditions, the product of an AP-integrable function and a function of bounded variation. Furthermore, the product of an AP-integrable function and a regulated function is AP-Henstock integrable.
We state the mean value theorem for the Riemann-Stieltjes integral which will be used in the main body of our work.

Definition and basic properties
be a finite collection of non-overlapping tagged intervals. If ([x i−1 , x i ], t i ) is fine to the choice S for each i = 1, · · · , n, then we say that P is S-fine. Let E ⊂ [a, b]. If P is S-fine and t i ∈ E for each i = 1, · · · , n, then P is said to be S-fine on E.

Proof.
Since g − f = 0 almost everywhere on I, by the theorem, Since ϵ > 0 is arbitrary, we conclude that f r ∈ AH(I r ) and I r f r = I f .
is a S ϵ -fine partition of I (r) . If we let x i := y i /r and t i := s i /r, then Since ϵ > 0 is arbitrary, we conclude that f (r) ∈ AH(I (r) ) and I (r) f (r) = r I f .

Multiplier Properties on bounded intervals
It is well known that the product of two AP-Henstock integrable functions is not necessarily AP-Henstock integrable, even when one of them is bounded or continuous. In this section, we provide some conditions under which the product of two AP-Henstock integrable functions is AP-Henstock integrable. We also establish the Mean Value Theorems. We start by showing the squeeze theorem for the AP-Henstock integral.
for all x ∈ I, and such that Proof. Let f ∈ AH(I) and let ϵ > 0. We can take φ ϵ := ψ ϵ := f . Conversely, assume that for every ϵ > 0 there exist functions φ ϵ and ψ ϵ in Then for each ϵ > 0, it follows that S(φ ϵ ; P ) ≤ S( f ; P ) ≤ S(ψ ϵ ; P ) for any tagged partition P of I. Since φ ϵ ∈ AH(I), there exists a choice S 1 on I such that if P is a S 1 -fine partition of I, then |S(φ ϵ ; P ) − I φ ϵ | ≤ ϵ, whence it follows that I φ ϵ − ϵ ≤ S(φ ϵ ; P ). Similarly there exists a choice S 2 on I such that if P is a S 2 -fine partition of I, then S(ψ ϵ ; P ) ≤ I ψ ϵ + ϵ. Now let S := {S 1 ∩ S 2 : S 1 ∈ S 1 , S 2 ∈ S 2 } and let P be a S-fine tagged partition of I. Then we have Adding these inequalities, we obtain Since ϵ > 0 is arbitrary, f satisfies the Cauchy Criterion( [4], Theorem 16.6) for the AP-Henstock integral. Therefore f ∈ AH(I).
It easy to see that a function f is regulated on I if and only if there is a sequence {s n } ∞ n=1 of step functions on I that converges uniformly to f on I. Proof. Let f : I → R be regulated on I and let ϵ > 0. Then there exists a step function s ϵ : for all x ∈ I, then the functions φ ϵ and ψ ϵ are AP-Henstock integrable on I and it follows from Theorem 7 that f is AP-Henstock integrable on I.
Theorem 9. Let f ∈ AH(I := [a, b]) be bounded below and g be regulated on I. Then the product f g belongs to AH(I).
Proof. Assume that f (x) ≥ 0 for x ∈ I. It is clear that if s is a step function, then s f belongs to AH(I).
Let A > I f ≥ 0 and let ϵ > 0. Since g is regulated on I, there exists a step function s ϵ on I such that for all x ∈ I, then φ ϵ , ψ ϵ ∈ AH(I) and it follows that φ ϵ (x) ≤ f (x)g(x) ≤ ψ ϵ (x) for all x ∈ I, and that Therefore, it follows from Theorem 7 that f g ∈ AH(I). Now, let f (x) ≥ M on I. Then by writing f g = ( f − M)g + Mg, we see that f g ∈ AH(I).
Theorem 10. Let f and | f | be AP-Henstock integrable on I := [a, b] and let g be a bounded, measurable function on I.
Then the product f g is AP-Henstock integrable on I.

Proof.
Let h := f g. Since h is measurable, there exists a sequence {s n } ∞ n=1 of step functions such that {s n } ∞ n=1 converges to h almost everywhere on I. Let s n be the middle function of −M| f |, s n , and M| f |. Then −M| f | ≤ s n ≤ M| f | and {s n } converges to h almost everywhere on I. Therefore, it follows from the Dominated Convergence Theorem for the AP-Henstock integral [9] that h ∈ AH(I).
Theorem 11. Let I := [a, b], f ∈ AH(I), φ is of bounded variation on I, and F(x) := x a f on I. If F is Riemann-Stieltjes integrable with respect to φ on I, then the product f φ belongs to AH(I) and where the second and third integrals are the Riemann-Stieltjes integrals.
Proof. Since F is Riemann-Stieltjes integrable with respect to φ, the third integral exists, and the existence of the second integral and the validity of the second equality follows from the well-known integration by part formula for the Riemann-Stieljes integral ( [4], Theorem 12.14). Therefore, we only need to show the first equality. To this end, let ϵ > 0. Since φ is Riemann-Stieltjes integrable with respect to F, there exist δ > 0 such is any tagged partition of I with norm less than 2δ, then Let |φ(x)| ≤ M for all x ∈ I. Since f ∈ AH(I), there exist a choice S on I such that if P and it follows from the Saks-Henstock Lemma for the AP-Henstock integral that Since ϵ > 0 is arbitrary, f φ ∈ AH(I) and I f φ = I φdF.
Note that the indefinite AP-Henstock integral is only approximately continuous, not necessarily continuous on I. It is shown in [6] (Exercise 12.10) that if two bounded functions F and φ on I share a common point of discontinuity in I, then F is not Rieman-Stieltjes integrable with respect to φ on I. Therefore, the condition in the above theorem that F is Reimann-Stiltjes integrable with respect to φ cannot be removed. On the other hand, it is well-known fact that if F is continuous and φ is of bounded variation on I, then F is Riemann-Stieltjes integrable with respect to φ [6]. Therefore, the following corollary follows.

Corollary 2. Let f be AP-Henstock integrable on I := [a, b], φ be bounded variation on I, and F(x) :=
x a f on I. If F is continuous on I, then f φ ∈ AH(I) and where the second and third integrals are the Riemann-Stieltjes integrals.
In addition to the multiplier theorem above, we provide a version of integration by parts theorem.
Theorem 12. Let I := [a, b] and let F, G : I → R be approximately continuous on I. If f , g ∈ AH(I), and F ′ ap = f , G ′ ap = g except for countably many points in I, then Fg + f G ∈ AH(I) and Moreover, Fg ∈ AH(I) if and only if f G ∈ AH(I), in which case Proof. By the hypothesis, there exist countable sets C f and C g such that Also, by the hypothesis, FG is approximately continous on I. Therefore, by Theorem 1, (FG) ′ ap ∈ AH(I) and I (FG) ′ ap = F(g)G(g) − F(a)G(a). Since C is a countable set, f G + Fg ∈ AH(I) and We now establish the Mean Value Theorems for the AP-Henstock integral. Proof. Since f is continuous, f is bounded on I. We invoke the fact that a nonnegative AP-Henstock integrable function is Lebesgue integrable and that the product of a Lebesgue integrable function and a bounded, Lebesgue integrable function is Lebesgue integrable. Therefore, p is Lebesgue integrable and f p is Lebesgue integrable on I. Proof. Since g is of bounded variation, it follows from Theorem 11 that f g ∈ AH(I) and where the third integral is the Riemann-Stieltjes integral. Then, by Theorem 1.2, there exists ξ ∈ I such that the last two terms equal

Multiplier Properties on unbounded intervals.
In this section, we define the AP-Henstock integral on unbounded intervals and investigate some properties of the integral including some multiplier properties.
We extend any function f : [a, ∞) → R to a function defined on [a, ∞] in the extended real numbers R * := R ∪ {∞, −∞} by defining f (∞) = 0. We then take a tagged partition of the interval [a, ∞] : that contains x as a point of density. We require that S x is bounded for each x ∈ R and S ∞ = [d, ∞] for some d > a. We say that the tagged partition P is S-fine if x i−1 , x i ∈ S t i for i = 1, · · · , n + 1, and d ≤ x n . Because S x is bounded for x ∈ R, t n+1 = ∞. Define 0 · ∞ = 0 so that the contribution of the final term in P to the Riemann sum is f (∞) · ∞ = 0. Now, we give the definition of the AP-integral of a function f : [a, ∞] → R.
Now, suppose that for any ϵ > 0 there exists a choice S ϵ on I such that if P and Q are any partitions of I that are S ϵ -fine of I, then |S( f ; P ) − S( f ; Q)| ≤ ϵ. For each n ∈ N, let S n = {S n,x : x ∈ I} be a choice on I such that if P and Q are S n -fine, then |S( f ; P ) − S( f ; Q)| ≤ 1/n. We may assume that S n+1,x ⊂ S n,x for all x ∈ I, n ∈ N. For each n ∈ N, let P n be a S n -fine partition of I. If m > n, then P m and P n are S n -fine. Therefore, for m > n, |S( f ; P n ) − S( f ; P m )| ≤ 1/n, and it follows that {S( f ; P n )} ∞ n=1 is a Cauchy sequence. Let A := lim n→∞ S( f ; P n ). By taking m → ∞, we have |S( f ; P n ) − A| ≤ 1/n. Now, for any given ϵ > 0, let K ∈ N be such that 1/K ≤ ϵ/2. If Q be a S K -fine partition of I, then The following theorem is the additive property of the AP-Henstock integral of a function on [a, ∞]. . Suppose that f ∈ AH(I 1 ) and f ∈ AH(I 2 ). Let f 1 be the restriction of f to I 1 and let f 2 be the restriction of f to I 2 . Let A 1 := I 1 f 1 and let A 2 := I 2 f . Given ϵ > 0, let S ′ ϵ := {S ′ ϵ,x : x ∈ I 1 } be a choice on I 1 and let S ′′ ϵ := {S ′′ ϵ,x : x ∈ I 2 } be a choice on I 2 such that if P 1 is a S ′ ϵ -fine partition of I 1 and P 2 is a S ′′ ϵ -fine partition of I 2 , then |S( f 1 ; P 1 ) − A 1 | ≤ 1 2 ϵ and |S( f 2 ; P 2 ) − A 2 | ≤ 1 2 ϵ. We define a choice S ϵ := {S ϵ,x : x ∈ I} on I by Let P be a S ϵ -fine partition of I and suppose that each tag occurs only once. Then the point c must be a tag of an subinterval in P. Let ([u, v], c) be the tagged interval in P of which the tag is c. Then P is of the form P a ∪ ([u, v], c) ∪ P b where the tags of P a are less than c and the tags of P b are greater than c. Let P 1 := P a ∪ ([u, c], c) and let P 2 := P b ∪ ([c, v], c). Then P 1 is a S ′ ϵ -fine partition of I 1 and P 2 is a S ′′ ϵ -fine partition of I 2 . Therefore, Since ϵ > 0 is arbitrary, f is integrable on I to I 1 f + I 2 f . Now, suppose that f ∈ AH([a, ∞]). For each ϵ > 0, letS ϵ := {S ϵ,x : x ∈ I} be a choice on I that satisfies the Cauchy Criterion (Theorem 15). Let f 1 denote the restriction of f to I 1 and letS ′ ϵ := {S ϵ,x ∩ I 1 : x ∈ I 1 } be the restriction ofS ϵ to I 1 . Let P 1 , Q 1 beS ′ ϵ -fine partitions of I 1 . By adjoining the same tagged partition of I 2 , extend P 1 , Q 1 to partitions P, Q of I that areS ϵ -fine . Then, |S( f 1 ; P 1 ) − S( f 1 ; Q 1 )| = |S( f ; P ) − S( f ; Q)| ≤ ϵ.
Therefore, by Theorem 15, f 1 is integrable on I 1 . In the same way, the restriction of f to I 2 is integrable on I 2 Since ϵ > 0 is arbitrary, lim c→∞ x m x n f ≤ ϵ, the sequence x n a f ∞ n=1 is a Cauchy sequence. Let lim n→∞ x n a f := A and N be an integer such that x N ≥ K(ϵ) and x n a f − A < ϵ whenever n ≥ N. If c > x N , then Since ϵ > 0 is arbitrary, lim c→∞