Lebesgue measurability of subset of the preimage of a measurable set under increasing absolutely continuous function 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.
 A: Currently I am also working on this problem. I have the following ideas:

*

*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;

*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;

*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 $E_n$;

*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.

A: 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:


*

*$m(O) = \int _{F^{-1}(O)} F'$ for open sets $O \subset [c, d]$.

*If $E \subset [c, d]$ with $m(E) = 0$ then $m(F^{-1}(E) \cap H) = 0$.

*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.
