Show that $f(F(x))F'(x)$ is measurable. 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. 
 A: 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. 
A: 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.
