$m^*(f(E))\leq\int_E|g'(x)|dx$ 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?
 A: 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.
A: Let $f : [0, 1] \to \mathbb{R}$ be absolutely continuous.
Step 1. Suppose $E$ is a compact subinterval of $[0, 1]$, and pick $c, d \in E$ such that $f(c)=\min_E f$ and $f(d) = \max_E f$. Then $f(E)$ is also a compact interval with endpoints $f(c)$ and $f(d)$, and so,
$$ m^*(f(E)) = \left| f(d) - f(c) \right| \leq \int_{\min\{c,d\}}^{\max\{c,d\}} \left|f'(x)\right| \, \mathrm{d}x \leq \int_{E} \left|f'(x)\right| \, \mathrm{d}x. \tag{1} $$
It is also clear that the equality extends to any type of subintervals $E$ of $[0,1]$.
Step 2. Next, let $E$ be an open subset of $[0,1]$. Then $E$ can be written as disjoint union of at most countably many subintervals of $[0,1]$, say $E = \cup_n I_n$. Then by the countable subadditivity of $m^*$,
$$ m^*(f(E)) = m^*(\cup_n f(I_n)) \leq \sum_n m^*(f(I_n)) \stackrel{\text{(1)}}\leq \sum_n \int_{I_n} \left|f'(x)\right| \, \mathrm{d}x = \int_{E} \left|f'(x)\right| \, \mathrm{d}x. \tag{2} $$
Step 3. Finally, let $E$ be an arbitrary measurable subset of $[0, 1]$. Also, choose a sequence of open subsets $E_n \subseteq [0, 1]$ containing $E$ such that $m^*(E_n) \downarrow m^*(E)$. Then
$$ m^*(f(E)) \leq m^*(f(E_n)) \stackrel{\text{(2)}}\leq \int_{E_n} \left| f'(x) \right| \, \mathrm{d}x = \int_{[0,1]} \left| f'(x) \right|\mathbf{1}_{E_n}(x) \, \mathrm{d}x. \tag{3} $$
Since $E$ is measurable, $m^*(E_n\setminus E) \to 0$ holds, and this in turn implies that $\left| f' \right| \mathbf{1}_{E_n} \to \left| f' \right| \mathbf{1}_{E}$ in $L^1$. So by taking limit as $n\to\infty$ to $\text{(3)}$, we obtain
$$ m^*(f(E)) \leq \int_{[0,1]} \left| f'(x) \right|\mathbf{1}_{E}(x) \, \mathrm{d}x = \int_{E} \left| f'(x) \right| \, \mathrm{d}x. $$
