$m^*(f(E))\leq\int_E|g'(x)|dx$ for absolutely continuous function $f$ Suppose $f$ is an absolutely continuous function on $[0,1]$, and suppose $E\subset (0,1)$ is any measurable set. I'd like to show that  $m^*(f(E))\leq\int_E|f'(x)|dx$.
I know that since $f$ is AC on $[0,1]$, we can write $f(x)=\int_0^xf'(t)dt+f(0)$. However, I'm not sure if/how this helps. What can I try?

See Measure Theory, Vol I by Bogachev, Proposition 5.5.4, p. 348 for the following:
If $f$ is differentiable at each point of a measurable set $E$ then $m^{*}(f(E)) \leq \int_E |f'(x)|dx$.
Your result follows follows from this since absolute continuity of $f$  implies differentiabilty of $f$ at almost all points and also implies that $f$ maps null sets to null sets.
