Lebesgue measurability of subset of the preimage of a measurable set under increasing absolutely continuous function

Question

For an increasing function F absolutely continuous on $[a,b]$, $E$ be a subset of $[F(a),F(b)]$. The preimage of $E$, $F^{-1}(E)$, is not neccessarily measurable, however, $F^{-1}(E)\cap\{F'(x)>0\}$ is.

This is problem 20 from chapter 3 of Stein's Real Analysis and I'm having problem showing the intersection is measurable. There is a hint suggesting to show $m(O)=\int_{F^{-1}(O)}F'$ for open sets, which I can prove but have no idea how to use.

Answer 1

Currently I am also working on this problem. I have the following ideas:

1. It is easy to show that $\{ F' (x)>0\}$ is open whenever $F$ is an increasing absolutely continuous function. Hence we can write $\{ F' (x)>0\}=\bigcup_{n} I_n$, where $I_n =(\alpha_n ,\beta_n )$ is disjoint from each other;
2. Since every L-measurable set can be written as the difference of a $G_{\delta}$ set and a measure zero set, we denote that $E=(\bigcap_n E_n )- E_0$, where $E_n$ is open and $E_0$ is of zero measure;
3. Due to the monotoncity of F, we have $F^{-1} (E) \bigcap \{ F' (x)>0\} = ( \bigcap_n (F^{-1} (E_n ) \bigcap F^{-1} (E) )- (\bigcup_n F^{-1}(E_0 \bigcap I_n ))$. Since the continuity of $F$, we will conclude our proof as soon as we prove that $F^{-1}(E_0 \bigcap I_n )$ is measureable (in fact, it is of zero measure for any $n$). Hereafter, we denote $E_0 \bigcap I_n$ as $E_n$;
4. WLOG, we only need to consider $F^{-1} (E_1 )$. Let $O_n$ is a series of open set covering $E_1$ and satisfies $m (O_n -E_1) <\frac{1}{n}$ (note that we can prove that $O_n \subset I_1$ when $n$ is big enough). Hence we have $$ \frac{1}{n} > m(O_n ) =\int_{F^{-1} (O_n )} F'(x) dx \geq \int_{F^{-1} (\bigcap_n O_n )} F'(x) dx $$ Thus, we have $\int_{F^{-1} (\bigcap_n O_n )} F'(x) dx =0$. Since $F'$ is positive on $\bigcap_n O_n$, we have $m(F^{-1}(\bigcap_n O_n ))=0$ and hence, $m(F^{-1} (E_1 ))=0$ since $F^{-1} (E_1 ) \subset F^{-1}(\bigcap_n O_n )$. Here we conclude our proof.

Answer 2

Here is another approach presented in Real Analysis by Royden, 3rd Edition as Problem 21 in Chapter 5. In the following I will use $$ [c, d] = [F(A), F(B)] \quad \text{and} \quad H = \{ F'>0\} $$ The problem goes by the following steps:

1. $m(O) = \int _{F^{-1}(O)} F'$ for open sets $O \subset [c, d]$.
2. If $E \subset [c, d]$ with $m(E) = 0$ then $m(F^{-1}(E) \cap H) = 0$.
3. If $E \subset [c, d]$ is measurable then $F^{-1}(E) \cap H$ is measurable.

As you can handle (1), I will start by showing my thoughts from (2). Let $E$ be such a set, and we can assume $c, d \not \in E$ since if so, then $F$ will be a constant on $F^{-1}(\{c\}) \cup F^{-1}(\{d\})$, meaning $F' = 0$ which is out of our interest. $E$ having measure 0 means that there is a $G_\delta$ set $G \subset (c, d)$ which $E \subset G$ and $m(G - E) = 0$. This further implies $$ m(G) = m(E) + m(G-E) = 0 $$ As a $G_\delta$ set, we can write $G$ as $$ G = \bigcap _{k = 1} ^\infty O_k = \bigcap _{n = 1} ^\infty G_n \, , \quad \text{where all of } O_k \text{ are open and }G_n = \bigcap _{k = 1} ^n O_k $$ such that $G_n$ is open for all $n$ and $\{ G_n \}$ is a descending sequence of sets. Further observe that $$ F^{-1} (G) = F^{-1} \left( \bigcap _{n = 1} ^\infty G_n \right) = \bigcap _{n = 1} ^\infty F^{-1} (G_n) $$ which the right hand side is also a descending sequence of open sets as $F$ is continuous. Thus, by the continuity of integration, part (1) and the continuity of measure, \begin{align*} \int _{F^{-1}(G)} F' = \lim _{n \to \infty} \int _{F^{-1}(G_n)} F' = \lim _{n \to \infty} m(G_n) = m(G) = 0 \end{align*} Because $F$ is increasing, $F' \geq 0$ on $(a, b)$. Applying the Chebychev's inequality gives

\begin{align*} m^*(F^{-1}(E) \cap H) &\leq m(F^{-1}(G) \cap H) \\ &= m \left( \bigcup _{n = 1} ^\infty \left\{ x \in F^{-1}(G): F'(x) \geq \frac{1}{n}\right\} \right) \\ &= \lim _{n \to \infty} m \left( \left\{ x \in F^{-1}(G): F'(x) \geq \frac{1}{n}\right\} \right) \\ &\leq \lim _{n \to \infty} n \cdot \int _{F^{-1}(G)} F' \\ &= 0 \end{align*} This finishes the proof of (2).

For part (3), let $E \subset (c, d)$ be measurable, where we exclude $\{c, d\}$ for the same reason in (2). As both $E$ and $H$ are measurable set, there are $G_\delta$ set $A \subset (c, d)$ and $K \subset (a ,b)$ for which $$ E \subset A , \, m(A - E) = 0 \quad \text{and} \quad H \subset K , \, m(K - H) = 0 $$ Note that $F^{-1}(A) \cap K$ is still $G_\delta$, as by writing, $$ A = \bigcap _{n = 1} ^\infty A_n \quad \text{and} \quad K = \bigcap _{m = 1} ^\infty K_n $$ we have \begin{align*} F^{-1}(A) \cap K &= \left( \bigcap _{n = 1} ^\infty F^{-1}(A_n) \right) \cap \left( \bigcap _{m = 1} ^\infty K_m \right) \\ &= \bigcap _{n, m = 1} ^\infty [F^{-1}(A_n) \cap K_m] \\ &= \bigcap _{n = 1} ^\infty [F^{-1}(A_n) \cap K_n] \end{align*} We will then show $[F^{-1}(G) \cap K] - [F^{-1}(E) \cap H]$ has measure 0, which we can do so using set operations, \begin{align*} &\phantom{{}={}} [F^{-1}(G) \cap K] - [F^{-1}(E) \cap H] \\ &= [F^{-1}(G) \cap K] \cap [F^{-1}(E^c) \cup H^c] \\ &= F^{-1}(G) \cap [(F^{-1}(E^c) \cap K) \cup (K - H)] \\ &= [F^{-1}(G) \cap F^{-1}(E^c) \cap K] \cup [F^{-1}(G) \cap (K - H)] \\ &= [F^{-1}(G - E) \cap K] \cup [F^{-1}(G) \cap (K - H)] \\ &= [F^{-1}(G - E) \cap (H \cup (K-H))] \cup [F^{-1}(G) \cap (K - H)] \\ &= [F^{-1}(G - E) \cap H] \cup [F^{-1}(G - E) \cap (K-H)] \cup [F^{-1}(G) \cap (K - H)] \\ &= [F^{-1}(G - E) \cap H] \cup [F^{-1}(G) \cap (K - H)] \\ &\subset [F^{-1}(G - E) \cap H] \cup (K - H) \end{align*} From (2), $m(F^{-1}(G - E) \cap H) = 0$, so by monotonicity, $$ m^*([F^{-1}(G) \cap K] - [F^{-1}(E) \cap H]) \leq m(F^{-1}(G - E) \cap H) + m(K - H) = 0 $$ Therefore there is a $G_\delta$ set covering $F^{-1}(E) \cap H$ with their difference having zero measure, and hence $F^{-1}(E) \cap H$ is measurable.