Let $g:[a,b] \rightarrow \mathbb{R}$ a strictly increasing and absolutly continous function with $g(a)=c$ and $g(b)=d$. Let $H=\{x: g'(x) \neq 0\}$ and $E \subseteq [c,d]$ with $|E|=0$. Show that $|g^{-1}(E) \cap H|=0$.
I want to check if this is fine or not
The previous exercise was to show that if $G \subseteq[c,d]$ was open then $\int_{g^{-1}(G)}g'(x)dx=|G|$ and I've done it.
So, what I was thinking is that given some $\epsilon > 0$ I can take a set $G=\cup_{n\in \mathbb{N}}G_n$ with $G_n=(c_n, d_n)$ such that $E \subseteq G$ and $|G| < \epsilon$, so, since $g$ is absolutely continous and biyective (because is strictly incresing) then exists a $\delta_\epsilon >0$ such that $\sum g^{-1}(d_n)-g^{-1}(c_n)<\delta_\epsilon$ and $\delta_\epsilon \rightarrow 0$ when $\epsilon \rightarrow 0$ because is a strictly incresing function.
Is this ok? I'm bothered by the fact that I don't use the set $H$ at all and a little bit because I don't use the previous part either.
Thanks
What your wrote at your argument its not always true, even if both $g$ and $g^{-1}$ are absolutely continuous functions. For example take $g(x)=x^2$ defined on the interval $[0,1]$. Then both $g$ and $g^{-1}$ are strictly increasing, hence both are absolutely continuous functions on $[0,1]$. Now consider the intervals $$[\alpha_n,\,\beta_n]=\biggl[\frac{1}{2n},\,\frac{1}{n}\biggr]$$
Then, $$\sum_{n=1}^{N}f(\beta_n)-f(\alpha_n)=\sum_{n=1}^{N}f(\frac{1}{n})-f(\frac{1}{2n})=\sum_{n=1}^{N}\frac{1}{n^2}-\frac{1}{4n^2}\sim \sum_{n=1}^{N}\frac{1}{n^2}$$
Hence, the tails, $\sum_{k=N}^{\infty}f(\beta_k)-f(\alpha_k)\sim \sum_{k=N}^{\infty}\frac{1}{k^2}\to 0$ as $N\to \infty$. But on the other hand, $$\sum_{k=N}^{\infty}\beta_n-\alpha_n=\sum_{k=N}^{\infty}\frac{1}{n}-\frac{1}{2n}=\sum_{k=N}^{\infty}\frac{1}{2n}=\infty$$ Which means that for every $\epsilon>0$ you can find large $N$ such that $$\sum_{k=N}^{\infty}f(\beta_n)-f(\alpha_n)<\epsilon$$ but you always have $\sum_{k=N}^{\infty}(\beta_n-\alpha_n)=\infty$. Now, to answer your question since you have proved that $$\tag{1} \int_{g^{-1}(G)}g'(x)\,dx=|G|$$ you can use this to prove that $|g^{-1}(E)\cap H|=0$. First observe that since $g$ is strictly increasing, then $g'(x)\geq 0$ almost everywhere. Using $(1)$ for $G=E$ we get $$\int_{g^{-1}(E)}g'(x)\,dx=|E|=0$$ On $H^c$ we have $g'(x)=0$, hence \begin{align} \int_{g^{-1}(E)}g'(x)\,dx&=\int_{g^{-1}(E)\cap H}g'(x)\,dx+\int_{g^{-1}(E)\cap H^c}g'(x)\,dx\\ &=\int_{g^{-1}(E)\cap H}g'(x)\,dx=0 \end{align} Now since on $g^{-1}(E)\cap H$ we have that $g'(x)>0$ we deduce that $|g^{-1}(E)\cap H|=0$.