When is the composition of functions measurable?

When is the composition of functions measurable? F. Javier Fernández e-mail: fjavier[email protected] F. Adrián F. Tojo e-mail: [email protected] CITMAga, 15782, Santiago de Compostela, Spain, Departamento de Estatística, Análise Matemática e Optimización, Universidade de Santiago de Compostela, 15782, Facultade de Matemáticas, Santiago, Spain. Abstract In this article we explore under which conditions on the interior function the composition of functions is measurable. We also study the sharpness of the result by providing a counterexample for weaker hypotheses. Keywords: measurable functions, composition, Lebesgue measure MSC 2020: 28-02 1 Introduction It is a well known fact among young analysts that the composition of measurable functions is not necessarily measurable, although this fact, to be true, has to be stated precisely, as it depends on what we mean when we say that a function is measurable. In this case, we are referring to the fact that the inverse image of an open set is a Lebesgue-measurable set of the real line. It is therefore convenient to start with the definition of measurable function in a broader sense. Definition 1.1. Let (X,M),(X,N)be measurable spaces. We say f:(X,M)→(X,N)is measurable if f−1(A)∈ M for every A∈ N . It is clear then, from this definition, that whether a function is measurable or not depends on the measurable spaces chosen. Let Land Bbe the Lebesgue and Borel σ-algebras on R respectively. In general, if f,g:(R,L)→(R,B)are measurable g◦f:(R,L)→(R,B)does not have to be so, which is the case we first talked about. In order for g◦f:(R,L)→(R,B)to be measurable, it would be enough if f:(R,L)→(R,L)or g:(R,B)→(R,B)were measurable (this last condition holds when gis continuous), but in general this is not the case. Here we are interested in sufficient conditions on f:(R,L)→(R,B)that can guarantee that, if g:(R,L)→(R,B)is measurable, so is g◦f:(R,L)→(R,B). We will consider then µ∗and µ, the Lebesgue exterior measure and measure respectively. For brevity, we will speak of L/Band L/Lmeasurable functions. 1 Rather counterintuitively, great regularity or monotonicity of fdoes not guarantee that the composition g◦fwill be L/Bmeasurable when gis. We illustrate this point with the following example. Example 1.2. We will first proceed to construct a strictly increasing C∞function fwhich is not L/Lmeasurable, that is, such that there exists D∈ L satisfying f−1(D)∈ L . We start by recalling the construction of the Smith–Volterra–Cantor set (or fat Cantor set) [3, p. 39]. Let λ∈(0, 1)and define r:=1−λ 3−2λ. Observe that r∈(0,1/3). The Smith–Volterra–Cantor set C of measure λis created iteratively in the following way. Let C0:= [0, 1]and, given Cn−1 define Cnby deleting the central open interval of length rnfrom each the connected components of Cn−1. It is clear by construction that Cn⊂Cn−1and Cnis compact for every n⩾1. Thus, the set C:=T∞ n=0Cn, the Smith–Volterra–Cantor set, is well defined, non-empty and compact. Since, by definition, Cnhas 2n−1connected components, µ(Cn) = µ(Cn−1)−2n−1rn=1−r(1−2nrn) 1−2r for every n∈N, we conclude that µ(C) = lim n→∞ µ(Cn) = 1−r 1−2r=1−3r 1−2r=λ. Let C⊂I:= [0,1]be a Smith–Volterra–Cantor set such that µ(C)∈(0,1). The positive measure of this set will be crucial for this example –cf. Theorem 1.11. Consider the function ψ(x):=e−(1−x2)−1 for x∈(−1,1).ψ∈ C ∞((−1, 1),R). Since Cis closed, I\Cis open, so it has a countable number of connected components each of which is an open interval. Let us define h(x) = 0 if x∈Cand h(x) = 2−(b−a)−1ψ2x−b−a b−a if x∈(a,b)where (a,b)is a connected component of I\C. Let Mn:=max|ψ(n)|. We will show now that h∈ C ∞(I,R). It is clear that his C∞in I\C. If x∈C, let us check that limy→x−f(x) = 0 (in case the limit can be taken). If x=bfor some (a,b)⊂I\C, this is obvious. Otherwise, there is a sequence of points in Cconverging to xfrom the left. Thus, given ϵ∈R+, there exists y∈C, x−ϵ < y<x, so any z∈(y,x)\Cbelongs to an interval (a,b)with b−a< ϵ and, therefore, h(z) = 2−(b−a)−1ψ((2x−b−a)/(b−a)) <2−(b−a)−1<b−a< ϵ. Hence, limy→x−f(x) = 0. Repeating the argument for limits from the right and observing that h|C=0, we conclude that his continuous. Assuming his n−1 times differentiable and taking into account that |h(n)(x)|=2−(b−a)−12n(b−a)−nψ(n)2x−b−a b−a⩽2−(b−a)−12n(b−a)−nMn, for any x∈(a,b)⊂I\C, we can reason as before to conclude that h∈ C ∞(I,R). Define f(x) = Ry 0h(x)dy.f∈ C ∞(I,R)and fis strictly increasing. Indeed, since Cis totally disconnected, given x,y∈I,x<y, there exist t,s∈(x,y),t<ssuch that [t,s]⊂I\C, so h(x)>0 in [t,s]and hence f(y)−f(x) = Ry xh(z)dz>0. 2 Given a connected component (a,b)of I\C, µ(f(a,b)) = f(b)−f(a) = Zb a h(z)dz. Thus, given that fis strictly increasing, I\Chas a countable number of connected components and that the Lebesgue measure is σ-additive, µ(f(I\C)) = RI\Ch(z)dz.f(I)is also measurable because it is an interval. Since fis strictly increasing, f(C)∩f(I\C) = ;and f(C) = f(I)\f(I\C) and, therefore, f(C)is measurable. Hence, µ(f(C)) =µ(f(I)) −µ(f(I\C)) = f(1)−f(0)−ZI\C h(z)dz =Z1 0 h(z)dz−ZI\C h(z)dz=ZC h(z)dz=0. Now, since µ(C)>0, there exists D⊂Csuch that D∈ L [3, Exercise 29, pg. 39]. We have that f(D)⊂f(C), so µ∗(f(D)) ⩽µ∗(f(C)) = 0 and, therefore, µ∗(f(D)) = 0, so by the completeness of µ,f(D)∈ L . Finally, f−1(f(D)) = D∈ L , so fis not L/Lmeasurable. We now define a L/Bmeasurable function gin such a way that g◦fis not L/Bmeasurable. Let gbe the characteristic function of the set f(D).gis measurable since f(D)is. Furthermore, {1}∈B, but (g◦f)−1({1}) = f−1(f(D)) = D∈ L , so g◦fis not L/Bmeasurable. Example 1.2 has shown that, if we are to provide sufficient conditions for g◦fto be L/L measurable, we will have to look further away than the regularity of f. In fact, it is clear from Example 1.2 that the behavior of fwhen it takes inverse images is determinant on the behavior of the composition, so we will try first to impose conditions on f−1in the case fis invertible. For the next results we consider Ω,Λ∈ L ,f:(Ω,L)→(R,B),g:(Λ,L)→(R,B), f(Ω)⊂Λ. Lemma 1.3. If g is L/Bmeasurable, f invertible and f −1absolutely continuous, then g ◦f is L/Bmeasurable. Proof. Let B∈ B. Then g−1(B)∈ L . Since f−1is absolutely continuous it takes Lsets to L sets [7, p. 250], so f−1(g−1(B)) ∈ L .■ Observe that Lemma 1.3 crucially avoids the circumstances of Example 1.2, since, in that case, the function f−1was not absolutely continuous (since it did not map Lsets to Lsets). This illustrates that, in general, even if fis absolutely continuous, f−1needs not to be –see [1,9]. A necessary sufficient condition for f−1to be absolutely continuous (in the case the domain is an interval) can be found in the following result –cf. [2, Lemma 2.2],[9]. Lemma 1.4. Let J = [c,d]be a compact real interval and f a strictly increasing and absolutely continuous function on J. f −1is absolutely continuous if and only if f ′(x)=0for a.e. x ∈J. Lemma 1.4 raises the question as to whether the condition f′(x)=0 a.e. is enough to show that g◦fis measurable without asking for fto be invertible, which is a very stringent condition. We provide an answer to this question in Corollary 1.14, but before stating it we will need the following results. Lemma 1.5. Assume f and g are L/Bmeasurable. If for every N ⊂Rsuch that µ∗(N) = 0we have that f −1(N)∈ L , then g ◦f is L/Bmeasurable. 3 Proof. Let B∈ B. Then g−1(B)∈ L , so, by the regularity of µ,g−1(B) = C⊔Nwhere µ∗(N) = 0 and C∈ B. By hypothesis, f−1(N)∈ L . On the other hand, since C∈ B and fis L/B measurable, f−1(C)∈ L . Therefore, f−1(g−1(B)) = f−1(C⊔N) = f−1(C)⊔f−1(N)∈ L . We conclude that g◦f:(Ω,L)→(R,B)is measurable. ■ Remember that the Lebesgue measure is complete, that is, if µ∗(A) = 0 then A∈ L and µ(A) = 0, so we obtain the following corollary. Corollary 1.6. Assume f and g are L/Bmeasurable. If for every N ∈ L such that µ(N) = 0we have that µ∗(f−1(N)) = 0then g ◦f is L/Bmeasurable. Remark 1.7. The hypotheses in Lemma 1.5 are sharp in that, if there exists N⊂Rsuch that µ∗(N) = 0 and f−1(N)∈ L , then there exists a L/Bmeasurable gsuch that g◦fis not L/B measurable. It is enough to take g=χN. Remark 1.8. The condition occurring in Corollary 1.6 is reminiscent of the Lusin N-property: µ(A) = 0 implies µ(f(A)) = 0 for every A∈ L , (N) which is intimately related to absolute continuity, as the following theorem shows. Theorem 1.9 ([8, Theorem 7.7]). Let I be a bounded interval and f :I→Rcontinuous. Then f is absolutely continuous if and only if f satisfies satisfies (N) on I and RAf′<∞where A ⊂I is the set of points x where f is differentiable and f ′(x)∈(0, ∞). As stated before, the goal now is to move away from the invertibility of f. Furthermore, we would like to work in a context where fneed not be differentiable, as this conditions is definitely much stronger than mere measurability. With this aim we try to generalize the concept of the derivative not being zero to non differentiable functions, which leads to the following definition of the sets Df(x)and Sf. Definition 1.10. Let f:Ω⊂R→R. For every x∈Ωwe define the set D f (x):=\ δ∈R+ conv§f(y)−f(x) y−x:y∈Ω, 0 <|y−x|< δªR where XRdenotes the closure in the extended real line with the compact interval topology and conv the convex hull. Observe that, if x∈Ω∩Ω′, then D f (x)is a nonmepty closed convex subset of R. Let Sf:={x∈Ω∩Ω′: 0 ∈D f (x)}. These definitions lead to the main result we want to introduce. Theorem 1.11. If f and g are L/Bmeasurable and µ∗(Sf) = 0, then g ◦f is L/Bmeasurable. Proof. By Corollary 1.6, it is enough to check that µ∗(Sf) = 0 implies that, for every A∈ L , if µ∗(A) = 0 then µ∗(f−1(A)) = 0. Equivalently, let µ∗(f−1(A)) =0 and let us prove that µ∗(A)=0. Let B⊂f−1(A)such that Bis bounded and µ∗(B)>0. We can assume, without loss of generality, that Bcontains no isolated points of Ω, since Ω\Ω′has to be a countable set because the usual topology is second countable, and thus µ∗(Ω\Ω′) = 0. This way we guarantee that Df(x)=;for x∈B. Let C:=f(B)⊂A. It will be enough to show that µ∗(C)=0. 4 Since µ∗(Sf) = 0 and µ∗(B)>0, we have that either µ∗({x∈B:D f (x)⊂R+})>0 or µ∗({x∈B:D f (x)⊂R−})>0. Let us assume the first case and the second will be analogous. Let E={x∈B:D f (x)⊂R+}and, for n∈N, En:=§x∈B:f(y)−f(x) y−x>1 nfor all y∈Ω, 0 <|y−x|<1 nª. Observe that E=Sn∈NEn. Indeed, let x∈E. Since 0 ∈ D f (x)and D f (x)is a closed set, inf D f (x)>0. Let ϵ∈(0, inf D f (x)). By definition of D f (x), there exists δ∈(0, ϵ)such that f(y)−f(x) y−x> ϵ for every y∈Ω, 0 <|y−x|< δ. Take n∈Nsuch that δ > 1 n. Then, f(y)−f(x) y−x>1 nfor every y∈Ω, 0 <|y−x|<1 n, that is x∈En. On the other hand, if x∈Enfor some n∈N, f(y)−f(x) y−x>1 nfor every y∈Ω, 0 <|y−x|<1 n, and thus, z⩾1 nfor every z∈D f (x), so x∈E. We conclude that E=Sn∈NEn. Hence, there exists k∈Nsuch that µ∗(Ek)>0 for, otherwise –see [5, Corollary 12.1.1], µ∗(E) = µ∗[ n∈N En=lim n→∞ µ∗(En) = 0, which would be a contradiction. Let δ∈(0,µ∗(Ek)) be fixed arbitrarily. Using a characterization of the exterior measure – see [6, Lemma 2.2], consider a countable infinite cover of Ekconsisting pairwise disjoint intervals {[an,bn)}n∈Nsuch that µ∗(Ek) + δ > X n∈N (bn−an)⩾µ∗(Ek) and bn−an<1 kfor every n∈N. Since Ek⊂Bis bounded, µ∗(Ek)<∞, so let m∈Nbe such that ∞ X n=m+1 (bn−an)< δ. Let X=Sm n=1[an,bn). Then, µ∗(Ek\X)< δ since {[an,bn)}∞ n=m+1is a cover of Ek\X. Thus, µ∗(Ek)⩽µ∗(Ek∩X) + µ∗(Ek\X)< µ∗(Ek∩X) + δ. Therefore, µ∗(Ek∩X)> µ∗(Ek)−δ. Now, there exists j∈ {1, . . ., m}such that µ∗(Ek∩[aj,bj)) > 1 m(µ∗(Ek)−δ)for, if we assume otherwise, then, for every n=1, . . ., m, µ∗(Ek∩[an,bn)) <1 m(µ∗(Ek)−δ), 5 and, therefore, µ∗(Ek∩X)⩽ m X n=1 µ∗(Ek∩[an,bn)) ⩽µ∗(Ek)−δ, which is a contradiction. Hence, for every x,y∈Ek∩[aj,bj), we have that |x−y|<1 k, so 1 k<f(y)−f(x) y−x. This implies fis strictly increasing on Ek∩[aj,bj). Consider f−1:f(Ek∩[aj,bj)) →Ek∩[aj,bj). We have that, for every x,y∈f(Ek∩[aj,bj)), |f−1(y)−f−1(x)|<k|y−x|, so f−1is a Lipschitz function and, therefore, 0<1 m(µ∗(Ek)−δ)< µ∗(Ek∩[aj,bj)) = µ∗(f−1(f(Ek∩[aj,bj)))) ⩽kµ∗(f(Ek∩[aj,bj))) ⩽kµ∗(C). Thus, µ∗(C)>0 and this ends the proof. ■ Remark 1.12. Returning to Example 1.2, observe that, f′=hand h(y) = 0 for y∈C, so C⊂Sf and, since µ(C)>0, we have that µ(Sf)>0, so we cannot apply Theorem 1.11. Remark 1.13. If Ω⊂Ris an interval and fis continuous (in fact, it is enough for fto be a Darboux function) then D f (x):=\ δ∈R+§f(y)−f(x) y−x:y∈Ω, 0 <|y−x|< δªR , which simplifies the calculations. If fis differentiable at x, then, by definition of derivative, D f (x) = {f′(x)}, so we have the following corollary. Corollary 1.14. If f and g are L/Bmeasurable, f is differentiable a.e. and µ∗((f′)−1(0)) = 0, then g ◦f:(Ω,L)→(R,B)is measurable. Even if Corollary 1.14 presents a simple condition to check it is clear that it leaves out common and interesting cases. For instance, the theorem cannot be applied in the case where fis a constant function and µ(Ω)>0. In order to palliate this fact, we will present a corollary of Corollary 1.14 where local constancy is permitted but, in order to do so, we will need to introduce the concept of essentially open set which will allow us to control those places where the function is locally constant. Definition 1.15. Let X⊂R. We say that Xis essentially open if there exists an open set U⊂R such that µ∗(X∆U) = 0 where X∆U:= (U\X)⊔(X\U). The notion of a set being essentially open can be rephrased in several ways, as the following Lemma shows. Lemma 1.16. Let X ⊂R. The following statements are equivalent: 1. X is essentially open. 6 2. X ∈ L and there exist an open set U ⊂Rand V,W∈ L , such that V ⊂U, W ⊂X\U, µ(V) = µ(W) = 0and X = (U\V)⊔W. 3. X ∈ L and there exist a set Y ⊂Rwith a countable number of connected components and V,W∈ L such that V ⊂Y , W ⊂X\Y , µ(V) = µ(W) = 0and X = (Y\V)⊔W. Proof. (I)⇒(II) Assume Xis essentially open and let Ube an open set such that µ∗(X∆U) = 0. Since U\X,X\U⊂X∆U, we have that µ∗(U\X) = µ∗(U\X) = 0, so V:=U\X,W:=X\U∈ L and V∩W=;. Finally, since U∈ L , X= (X∩U)⊔(X\U) = [U\(U\X)] ⊔(X\U) = (U\V)⊔W∈ L . (II)⇒(I) If V,W,U∈ L are such that Uis open, µ(V) = µ(W) = 0 and X= (U\V)⊔W, then µ∗(X∆U) =µ∗((U\X)⊔(X\U)) ⩽µ∗(U\X) + µ∗(X\U) =µ∗(U\((U\V)⊔W)) + µ∗([(U\V)⊔W]\U) =µ∗(U∩V∩(U\W)) + µ∗(W\U)⩽µ∗(V) + µ∗(W) = 0. We conclude that Uis essentially open. (II)⇒(III) Since every open subset of Rhas a countable number of connected components, this is evident. (III)⇒(II) Let Y⊂Rwith a countable number of connected components and V,f W∈ L such that V∩f W=;,µ(V) = µ(f W) = 0 and X= (Y\V)⊔f W. Let {Yn}n∈Λbe the family of connected components of Y. The Ynare intervals. Let Λ1be the set of those indices k∈Λsuch that Ikis just a point, Λ2=Λ\Λ1. Define U:=Sn∈Λ2  Ynand W:=f W∪[ n∈Λ1 Yn∪[ n∈Λ2 Yn\ Yn\V. Uis open and, since Wis the union of a zero measure set and a countable set, W∈ L and µ(W) = 0. Finally, V∩W=;and X= (Y\V)⊔f W= (U\V)⊔W.■ The next result allows us to work with functions fwhich are not continuous by separating the continuous and discontinuous parts of f. Theorem 1.17. Assume f and g are L/Bmeasurable and f is a.e. differentiable and can be expressed as f1+f2where f1is absolutely continuous and f2(Ω)is countable. If (f′)−1(0)is essentially open, then g ◦f is L/Bmeasurable. Proof. Consider an open set U⊂Rand V,W∈ L such that µ(V) = µ(W) = 0 and (f′)−1(0) = (U\V)⊔W. Let B∈ B. Then g−1(B)∈ L . Now, f−1(g−1(B)) = [f−1(g−1(B)) ∩(U\V)] ⊔[f−1(g−1(B)) ∩W]⊔[f−1(g−1(B))\(f′)−1(0)]. Given that f′=0 in Ω\(f′)−1(0), in particular µ∗({x∈Ω\(f′)−1(0):f′(x) = 0}) = 0, so we have, by Corollary 1.14, that f−1(g−1(B))\(f′)−1(0)∈ L . Also, µ∗(f−1(g−1(B)) ∩W) = 0, so f−1(g−1(B)) ∩W∈ L . Finally, since Uis open, U=Sn∈ΛUnwhere the {Un}n∈Λis a countable 7 family of pairwise disjoint open intervals. Assume x∈(Un\V)∩f−1(g−1(B)). Let y∈Un,x<y. Since f′=0 on Un\Vand µ(V) = 0, we have that f(y)−f(x) =f1(y)−f1(x) + f2(y)−f2(x) = Z[x,y] f′ 1(z)dz+f2(y)−f2(x) =Z[x,y]\V f′ 1(z)dz+f2(y)−f2(x) = f2(y)−f2(x). Therefore, ftakes on Una countable number of values. fis also L/Bmeasurable, so f−1(y)∈ L for every y∈f(Ω). Thus, f−1(g−1(B)) ∩(U\V) =[ y∈g−1(B) f−1(y)∩[ n∈Λ Un\V=[ y∈g−1(B) n∈Λ f−1(y)∩(Un\V). The number of sets in this last union that is nonempty is countable and, since Un,V,f−1(y)∈ L , we have that f−1(g−1(B)) ∩(U\V)∈ L . We conclude that f−1(g−1(B)) ∈ L , so g◦fis L/Bmeasurable. ■ From now on we will assume that Ωis an interval. This is no restriction as we can always extend fto an interval by a constant. Every function of bounded variation fcan be decomposed as fa+fs+fjwhere fais absolutely continuous, fsis continuous and singular (that is f′ s=0 a.e.) and fjis the jump part [4, Theorem 3.16]. The jump part is a countable sum of step functions and contains all of the discontinuities of f. It can be expressed as f(x) = X t<x ∆+f(t) + X t⩽x ∆−f(t) and clearly it has a countable image, so we have the following corollary of Theorem 1.17. Corollary 1.18. Assume Ωis an interval. If g is measurable, f is of bounded variation, fs=0and (f′)−1(0)is essentially open, then g ◦f is L/Bmeasurable. Remark 1.19. Again, Corollary 1.18 cannot be applied to Example 1.2. Since f′is continuous, (f′)−1(0)is closed, but in this case it is not essentially open. To see this, first observe that (f′)−1(0) = C∪Ywhere Yis a countable set formed by the middle points of the connected components of I\C. Since µ(Y) = 0, (f′)−1(0)is essentially open if and only if Cis. Assume, that C is essentially open and we will arrive to a contradiction. Let U⊂Rbe an open set and V,W∈ L , V⊂U, such that V∩W=;,µ(V) = µ(W) = 0 and C= (U\V)⊔W. Take x∈U\V. Then there exist r∈R+such that (x−r,x+r)⊂U. Since x∈C,x∈ ∂(I\C), so there exists y∈(x−r,x+r)\C. Since I\Cis open, there exists s∈R+such that (y−s,y+s)⊂(x−r,x+r)\C. Thus, µ(C∩(x−r,x+r)) = µ(((x−r,x+r)\V)⊔W) = 2r. Therefore, µ((x−r,x+r)\C) = 0 and, since (y−s,y+s)⊂(x−r,x+r)\C, we have that µ((y−s,y+s)) = 0, which is a contradiction. Acknowledgments The authors were partially supported by Xunta de Galicia, project ED431C 2019/02, and by the Agencia Estatal de Investigación (AEI) of Spain under grant MTM2016-75140-P, co-financed by the European Community fund ERDF. 8 References [1]de Amo, E., Diaz Carrillo, M., Fernández-Sánchez, J.: Functions with unusual differentiability properties. An. Stiint. Univ.‘Al. I. Cuza’Iasi, Mat 62(1), 121–140 (2016) [2]Cabada, A., Pouso, R.L.: Extremal solutions of strongly nonlinear discontinuous second-order equations with nonlinear functional boundary conditions. Nonlinear Analysis 42(8), 1377– 1396 (2000) [3]Folland, G.B.: Real analysis: modern techniques and their applications, 2 edn. PAM. Wiley (1999) [4]Monteiro, G.A., Slavík, A., Tvrdý, M.: Kurzweil-Stieltjes Integral: theory and applications. World Scientific, Singapore (2018) [5]Munroe, M.E.: Introduction to measure and integration. Addison-Wesley Cambridge, Mass. (1953) [6]Márquez Albés, I., Tojo, F.A.F.: Existence and Uniqueness of Solution for Stieltjes Differential Equations with Several Derivators. Mediterranean Journal of Mathematics 18(5), 181 (2021) [7]Natanson, I.: Theory of functions of a real variable, Vol.I, rev.ed. 5 pr. edn. Ungar (1983) [8]Saks, S.: Theory of the Integral, 2 edn. Dover Books on Advanced Mathematics. Dover, New York (1964) [9]Spataru, S.: An absolutely continuous function whose inverse function is not absolutely continuous. Note di Matematica 1(2004) 9