is it true that if an absolutely continuous function defined on a closed interval in $\mathbb{R}$ has a non-negative derivative almost everywhere (non negative wherever the derivative is defined), then such function is monotone? (increasing, but not strictly)

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    $\begingroup$ Yes. Because $f(y)-f(x) = \int_x^y f'(t) dt$. $\endgroup$
    – copper.hat
    Nov 6, 2014 at 6:13

1 Answer 1


Since $f$ is absolutely continuous we have $$ f(x)-f(y)=\int_y^x f'(t)dt\geq 0, \qquad \text{ for }y<x. $$ In fact this shows that if $f'>0$ a.e. you get $f$ strictly increasing.


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