Is the pointwise maximum absolutely continuous? Consider two absolutely continuous real-valued functions in an interval $I$, $f(x)$ and $g(x)$.
Is the pointwise maximum,  $x\mapsto \max(f(x),g(x))$, also absolutely continuous?
 A: First observe that for all $x,y\in I$ we have
$$ |\max\{ f,g\}(x) - \max\{f,g\} (y)| \leq \max\{ | f(x) - f(y)|, |g(x)- g(y)|\}. $$
This is clear, if the maximum in $x$ and $y$ is attained in the same function $f$ or $g$. Otherwise, without loss of generality, assume $f(x) \geq g(x)$ and $f(y) \leq g(y)$. Then $f(x) \geq g(y) \geq f(y)$ or $g(x) \leq f(x) \leq g(y)$. In the first case
$$ |\max\{ f,g\}(x) - \max\{f,g\} (y)| = |f(x) - g(y)| = f(x) - g(y) \leq f(x) - f(y) = |f(x) - f(y)|.$$
In the second case, similarly
$$ |\max\{ f,g\}(x) - \max\{f,g\} (y)| = |f(x) - g(y)| = g(y) - f(x) \leq g(x) - g(y) = |g(x) - g(y)|.$$
Now to prove the claim, let $f,g :I \to \mathbb{R}$ be absolutely continuous. Starting from the definition of absolute continuity, fix $\varepsilon >0$ and let $\delta>0$ be such that for any pairwise disjoint family of intervals $[a_i,b_i]$, $i=1,\ldots,n$ in $I$ with
$$ \sum_{i=1}^n b_i-a_i < \delta$$
we have
$$ \sum_{i=1}^n |f(b_i) - f(a_i)| < \frac{\varepsilon}{2} \quad \text{and} \quad \sum_{i=1}^n |g(b_i) - g(a_i)|< \frac{\varepsilon}{2}.$$
Using our initial observation, we have for such a family of intervals that
$$ \sum_{i=1}^n |\max\{f,g\}(b_i) - \max\{f,g\}(a_i)| \leq \sum_{i=1}^n\max\{ | f(b_i) - f(a_i)|, |g(b_i)- g(a_i)|\}$$
$$ \leq \sum_{i=1}^n |f(b_i) - f(a_i)| + \sum_{i=1}^n |g(b_i) - g(a_i)| < \varepsilon.$$
This shows absolute continuity of $\max\{f,g\}$.
Version 2:
A more elegant solution can be obtained, if we use a bit more theory. Recall that the absolute value is globally Lipschitz continuous, that sums of absolutely continuous functions are absolutely continuous, and that the concatenation of a globally Lipschitz continuous function with an absolutely continuous function is absolutely continuous, see here (composition Lipschitz and absolutely continuous function). Consequently, the functions
$$ f+g  \quad \text{and} \quad |f-g| $$
are absolutely continuous.  Then on $I$
$$ \max\{ f,g\}  = \frac{1}{2}(f+g + |f-g|).$$
This is a linear combination of absolutely continuous functions and thus absolutely continuous.
