How show this implication? 
Let $f$ be a strictly increasing continuous function defined on $(0, 1)$. Prove that
  $f'
 > 0$ almost everywhere if and only if 
  $f^{-1}$
  is absolutely continuous.

What I did
$(\Leftarrow)$ for $[a,b] \subset (0,1)$ $f$ is of bounded variation and then $f'$ exists a.e., as $f$ is strictly increasing then $f'\geq 0$. $f: [a,b] \rightarrow f([a,b])$ is bijection and $f([a,b])$ is closed interval. 
$f^{-1}:f([a,b])\rightarrow [a,b]$ is of absolutely continuous then $(f^{-1})'$ exists a.e. Therefore $f' > 0$ a.e. (in this case $(f^{-1})(y)' = \dfrac{1}{f'(x)}$ where $f(x) = y$)
As $[a,b]$ is arbritary then $f'> 0$ a. e. on $(0,1)$
$\blacksquare$
I would be very grateful for some help to prove $(\Rightarrow)$
 A: Let $(a,b)=f((0,1))$. If $f^{-1}:(a,b)\rightarrow(0,1)$ is not absolutely
continuous, then there exists a set $E\subset(a,b)$ of Lebesgue measure zero
such that $F:=f^{-1}(E)$ has positive measure (have you proved this? It's
called the (N) property). Since $f^{\prime}>0$ for $\mathcal{L}^{1}$ a.e.
$x\in F$, we can write
$$
F=F_{0}\cup\bigcup_{n=1}^{\infty}F_{n},
$$
where $F_{n}:=\{x\in F:\,f^{\prime}(x)\geq\frac{1}{n}\}$ and $F_{0}$ has
measure zero. It follows that%
$$
\sum_{n=1}^{\infty}\mathcal{L}^{1}(F_{n})\geq\mathcal{L}^{1}(F)>0,
$$
and so for some $n$, $\mathcal{L}^{1}(F_{n})>0$. Consider an open set $U$ that
covers $f(F_{n})$. Without loss of generality we can assume that
$U\subseteq(a,b)$. Then $U$ is a disjoint union of intervals. Let $J$ be any
such interval. Since $J\subseteq(a,b)=f((0,1))$, we can write $J=(f(c),f(d))$
where $0\leq c<d\leq1$. Then
\begin{align*}
\mathcal{L}^{1}(J)  & =f(d)-f(c)\geq\int_{c}^{d}f^{\prime}(t)\,dt\geq
\int_{(c,d)\cap F_{n}}f^{\prime}(t)\,dt\\
& \geq\frac{1}{n}\mathcal{L}^{1}((c,d)\cap F_{n}).
\end{align*}
It follows that $\mathcal{L}^{1}(U)\geq\frac{1}{n}\mathcal{L}^{1}(F_{n})>0$,
but $f(F_{n})\subseteq f(F)=E$ which has Lebesgue measure zero and so we can
find an open set $U$ covering $F_{n}$ with measure less than $\frac{1}
{n}\mathcal{L}^{1}(F_{n})$, which gives a contradiction. 
