About an absolutely continuous function Let be $f:[a,b] \longrightarrow \mathbb{R}$ an absolutely continuous function, consider $f^+(x)=\max\{f(x),0\}$ and $f^-(x)=\max\{-f(x),0\}$ .Note that $f=f^+-f^-$, then $f^+,f^-$ are absolutely continuous functions. I have been trying it a lot but I don't achieve it. I know the inverse is true. 
 A: Suppose that $f$ is absolutely continuous. It is sufficient to show that $\max\{f,0\}$ is absolutely continuous, since the other case is symmetric taking $\bar{f} = -f$. Since 
$$
\max\{f,0\} = \frac{|f| + f}{2}
$$
and the sum of absolutely continuous functions is absolutely continuous, we only need to prove that $|f|$ is absolutely continuous. 
For this we will use an equivalent characterization of absolute continuity. A function $g$ is absolutely continuous in $[a,b]$ if given $\varepsilon >0$, there exists $\delta > 0$ such that if $\{(x_i,y_i)\}_{1\leq i\leq k}$ are pairwise disjoint intervals in $[a,b]$ with 
$$
\sum_{i=1}^k (y_i - x_i) < \delta
$$
then 
$$
\sum_{i=1}^k |g(y_i) - g(x_i)| < \varepsilon
$$
Let $\varepsilon > 0$. Since $f$ is absolutely continuous, there exists $\delta > 0$ such that if $\{(x_i,y_i)\}_{1\leq i\leq k}$ are pairwise disjoint intervals in $[a,b]$ with 
$$
\sum_{i=1}^k (y_i - x_i) < \delta
$$
then 
$$
\sum_{i=1}^k |f(y_i) - f(x_i)| < \varepsilon
$$
With this $\delta$, and by the triangular inequality, $||a| -|b|| \leq |a-b|$, we have that
$$
\sum_{i=1}^k ||f(y_i)| - |f(x_i)|| < \varepsilon
$$ 
which proves that $|f|$ is, as well, absolutely continuous.
