$\varphi$ monotone,continuous on $[0,1]$. Show $\varphi$ is AC, iff for any borel $f_n\to f$ in $L^1$,$\{\varphi(f_n)\}$ converges in measure. Let $\varphi$ be monotonic and continuous on $[0,1]$. Show it is AC, iff for any borel sequence $f_n$ that converges in $L^1$ the sequences $\varphi(f_n)$ converges in measure.
The forward direction follows from the fact that $f_n \to f$ in $L^1$ implies convergence in measure and continuity preserves convergence in measure on finite measures. The backwards direction is the one that is giving me a little more trouble.
Since $\varphi$ is monotonic (WLOG its increasing) all we have to show is that  $\nexists x$ s. t. $\int_{0}^{x}\varphi'\lt\varphi(x)-\varphi(0)$. However I do not really know how to do that.
Any hints would be highly appreciated.
 A: What you showed is that the claim is always true, even without assuming that $\varphi$ is AC. Therefore, the converse direction is not true (take your favourite non AC monotonic continuous function, e.g. the devil's staircase).

A correct characterization (at least for the case that $\varphi$ is strictly increasing) is as follows:
Assume that $\varphi : [0,1] \to [a,b]$ is strictly increasing,
so that $\varphi^{-1} : [a,b] \to [0,1]$ is well-defined.
Then $\varphi$ is AC if and only if whenever $f_n \to f$ in $L^1$,
then $f_n \circ \varphi^{-1} \to f \circ \varphi^{-1}$ in measure.
Indeed, if $f_n \to f$ in measure, then it is not too hard to see
(with $\lambda$ denoting the Lebesgue measure) that
\begin{align*}
  \lambda \bigl(\{ x : |f_n \circ \varphi^{-1}(x) - f \circ \varphi^{-1} (x)| > \epsilon \}\bigr)
  & = \lambda\bigl( \varphi ( \{ y : |f_n (y) - f (y)| > \epsilon \} ) \bigr) \\
  & = \int_0^1
      \varphi'(t) \cdot 1_{|f_n(t) - f(t)| > \epsilon}
    \, d t
  & \xrightarrow[n\to\infty]{} 0
\end{align*}
for arbitrary $\epsilon > 0$ by the dominated convergence theorem,
since the indicator function converges to zero in measure.
Conversely, suppose towards a contradiction that $\varphi$ is not absolutely continuous.
By definition, this means that there is some $\epsilon > 0$ such that for each $n \in \Bbb{N}$
there is a pairwise disjoint family of intervals $(a_i^{(n)}, b_i^{(n)}) \subset [0,1]$
($i = 1,\dots,m_n$) such that $\sum_{i=1}^{m_n} (b_i^{(n)} - a_i^{(n)}) < \frac{1}{n}$,
but $\sum_{i=1}^{m_n} \bigl(\varphi(b_i^{(n)}) - \varphi(a_i^{(n)})\bigr) \geq \epsilon$.
This means if we define $f_n := \sum_{i=1}^{m_n} 1_{(a_i^{(n)}, b_i^{(n)})}$, then $f_n \to 0$
in $L^1$, but $f_n \circ \varphi^{-1} = \sum_{i=1}^{m_n} 1_{\bigl(\varphi(a_i^{(n)}), \varphi(b_i^{(n)})\bigr)}$
does not converge to zero in measure.
