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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.

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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.

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  • $\begingroup$ you say "equivalent characterization" but isn't that the usual definition? $\endgroup$ – zhw. May 15 '18 at 18:57
  • $\begingroup$ It may as well be, I took the definition to be that $f(x) = f(a) + \int_a^xg(t)dt$ with $g$ Lebesgue-integrable. I should have probably clarified it, though. $\endgroup$ – Guido A. May 15 '18 at 20:52

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