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.
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.