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This is a equation from Stein-Sharkarchi Real Analysis.

Let $F$ be absolutely continuous and increasing on $[a,b]$ with $F(a)=A$ and $F(b)=B$. Suppose $f$ is any measurable function on $[A,B]$.

Show that $f(F(x))F'(x)$ is measurable on $[a,b]$.

I am really having a hard time starting this problem. I know that $F'(x)$ is definitely measurable, but $f(F(x))$ need not be.

user7090
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  • The composition of measurable functions is measurable. – Math1000 Nov 29 '14 at 19:27
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    The book specifically states that $f(F(x))$ need not be measurable. – user7090 Nov 29 '14 at 19:31
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    This depends on whether you are talking about Borel- or Lebesgue-measurability. If $f$ is Borel-measurable, then the composition $f \circ F$ is, too. If $f$ is Lebesgue-measurable, the composition need not be Lebesgue-measurable. – Lukas Geyer Nov 29 '14 at 20:06

2 Answers2

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Let us first consider a closed interval $I = [c,d]\subset [A,B] =F([a,b])$ (for the last equation use monotonocity and continuity of $F$ (intermediate value theorem)).

It is easy to see that $M =F^{-1}([c,d])$ is a compact interval $M =[x,y]$.

Hence, $\chi_I (F(t)) F'(t) =\chi_{[x,y]}(t) F'(t)$ is measurable with

$$ \mu(I) := \int_{[a,b]} \chi_I (F(t)) F'(t)\, dt = \int_x^y F'(t)\, dt = F(y)-F(x)=d-c, $$

where we used the choice of $x,y$ in the last step.

Side Remark: Using Dynkins $\pi$-$\lambda$ theorem, one can now deduce that $\mu(A)=\lambda(A)$ holds for all Borel sets $A\subset [A,B]$, where $\lambda$ is Lebesgue measure. But we do not need that here.

As $F$ is continuous (hence Borel measurable) and $F'$ is measurable, it is easy to see that $f(F(t))F'(t)$ is measurable for $F=\chi_A$, where $A$ is a Borel set.

Every Lebesgue measurable $A$ set can be written as $A=A' \cup N$, where the union is disjoint, $A'$ is Borel measurable and $N$ is a null set. Hence, it suffices to show the claim for $f=\chi_N$, the general case then follows by expressing $f$ as a limit of simple functions (how exactly?)

But for every $n\in \Bbb{N}$, there is a covering $N\subset \bigcup I_j$ of $N$ by (compact) intervals $I_j$ with $\sum \lambda(I_j)<1/n$, where $\lambda $ denotes Lebesgue measure.

Then $0\leq \chi_N (F(t)) F'(t) \leq \sum \chi_{I_j}(F(t)) F'(t)$ with

$$ \int \sum \chi_{I_j}(F(t)) F'(t) =\sum \int \chi_{I_j}(F(t))F'(t) =\sum \lambda(I_j)<1/n, $$

where we used the calculation at the beginning of the proof.

This easily entails $ \chi_N (F(t)) F'(t)=0$ almost everywhere, so that this function is in particular Lebesge measurable.

kayak
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PhoemueX
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  • To show that $\chi_N (F(t)) F'(t)$ is measurable, you seemed to use the following argument: $\int_E \chi_N (F(t)) F'(t)<\frac{1}{n}$ $\forall n\implies $ $\chi_N (F(t)) F'(t)=0$ a.e and is thus measurable. But $\int_E \chi_N (F(t)) F'(t)$ relies on the measurability of $ \chi_N (F(t)) F'(t)$. – Asigan May 27 '23 at 13:43
  • @Asigan: What I am showing is that for each $n$, you can find a (measurable) function $f_n$ that satisfies $0\leq g(t) := \chi_N(F(t))F'(t)\leq f_n(t)$ and such that $\int f_n dx \leq 1/n$. This then easily implies that $g=0$ almost everywhere (how?!), and thus $g$ is measurable. – PhoemueX May 27 '23 at 17:25
  • (Yesterday I gave a wrong proof, so I deleted the old comment.) Thanks. I thought about it again and now understand it. This is implied by the following lemma: $E\subset\mathbb{R}$ is measurable. If $f:E\to\mathbb{R}$ is a non-negative function s.t. $\forall\epsilon>0$, $\exists\ g$ measurable s.t. $g\ge f$ on $E$ and $\int_E g<\epsilon$, then $f=0$ a.e. $x\in E$. I give two proof: – Asigan May 29 '23 at 14:54
  • proof 1: Construct a sequence of functions ${f_n}$ s.t. $f_n\ge f$ and $\int_E f_n\to 0$. Define $\psi_n=\min{g_1,\cdots, g_n}$. Then $\int_E\psi_n\to 0$, $\psi_n\ge \psi_{n+1}$, $\psi_n\ge f$. Since $\psi_n\ge \psi_{n+1}\ge 0$, let $\psi_n\to \psi$ pointwise. $\int_E\psi_n\to 0$, $\psi_1\ge \psi_n\ge 0\implies $ (Dominated Convergence Theorem) $\int_E \psi=0\implies \psi=0$ a.e. $x\in E$. But $\psi\ge f\implies$ $f=0$ a.e. $x\in E$. – Asigan May 29 '23 at 14:55
  • proof 2: Assume $m(f>0)>0$, where $m$ is the exterior measure.. Then $\exists\ n$, $m(f>\frac{1}{n})>0$. (Otherwise $m(f>\frac{1}{n})=0$ $\forall\ n$, i.e. $m(f>\frac{1}{n})=0$, and $(f>0)=\cup (f>\frac{1}{n})$ is union of set of measure zero, and is of measure zero.) Suppose $m(f>\frac{1}{N})=A>0$. If $g\ge f$ is measurable, then $m(g>\frac{1}{N})=m(g>\frac{1}{N})\ge m*(f>\frac{1}{N})\implies\ \int_E g\ge\frac{1}{N}A $ is a fixed number, a contradiction. – Asigan May 29 '23 at 14:55
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There is another proof that uses the result of Exercise 20(c) in the same book. Its proof can be found here.

(c) Prove, however, that for any increasing absolutely continuous $F$, and $E$ a measurable subset of $[A, B]$, the set $F^{−1} (E) \cap \{ F′ (x) > 0 \}$ is measurable.

By the definition of measurable functions, we only need to show that the set $I_a \equiv \{ x | f(F(x))F′(x) > a \}$ is measurable for any $a$.

For $a>0$, we have $$I_a = \bigcup_{q\in \mathbb{Q}^+}^\infty \big\{ x | f(F(x)) > \frac{a}{q} \big\} \cap \big\{ x| F′(x) > q \big\} = \bigcup_{n=1}^\infty \Big[ \big\{ x | f(F(x)) > \frac{a}{q} \big\} \cap \big\{ x| F′(x) > 0 \big\} \Big] \cap \big\{ x| F′(x) > q \big\}. $$

For each positive rational number $q$, the term in the brackets is measurable by 20(c), the other one is also measurable because $F'$ is measurable by Corollary 3.7. So $I_a$ is a countable union of measurable sets and therefore measurable.

For $a<0$, we can look at the complement of $I_a$ and get the same result.

So we only need to prove that $\big\{ x | f(F(x))F'(x) = 0 \big\}$ is measurable. But this set can be written as $$\Big[\big\{ x | f(F(x)) = 0 \big\} \cap \big\{x|F'(x) > 0\big\} \Big] \cup \big\{x|F'(x) = 0\big\}, $$ which is also measurable by the same argument.

Petra Axolotl
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  • Could you tell me why we can consider the complement of $I_a $to conclude the case $a<0$ – lee Dec 31 '23 at 12:09