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The subsequent statement can be regarded as a follow-up to

Let $f:[a,b]\to\mathbb{R}$ be Lebesgue integrable. Furthemore, let $ g:[a,b]\ni x\mapsto\int_a^x f(t)\,\mathrm{d}t\in\mathbb{R} $ be convex. Then $f$ is non-decreasing almost everywhere.

Let $a\le x_0. Since $f$ is convex, we have $ \frac{g(x_2)-g(x_1)}{x_2-x_1}-\frac{g(x_1)-g(x_0)}{x_1-x_0}\ge 0\text{.} $ This can be reduced to $ \int_{x_1}^{x_2} \frac{f(t)}{x_2-x_1}\,\mathrm{d}t \ge\int_{x_0}^{x_1} \frac{f(t)}{x_1-x_0}\,\mathrm{d}t\text{.} $ The last formula roughly shows that the 'average' $f$ on $[x_0,x_1]$ does not exceed the 'average' of $f$ on $[x_1,x_2]$. Do you know a rigorous argument showing that $f$ is non-decreasing a.e.?

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    @TonyK It means that there exists a set $X\subseteq [a,b]$ of measure $b-a$ such that for all $x_0,x_1\in X$ with $x_0\le x_1$ it holds $f(x_0)\le f(x_1)$.2012-01-20

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A standard result about convex functions is that they have nondecreasing one-sided derivatives everywhere. Since g'(x)=f(x) almost everywhere, $f$ is equal almost everywhere to the nondecreasing right-hand derivative of $g$.

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    @Jonas: Thank you. This reference is similar to what I was looking for. In the meantime, I also found *[Convex Functions](http://books.google.de/books?id=cqyHkkCxVtcC)* by Roberts and Varberg. Maybe 'deep' was said too much. On the other hand, these results are striking even if they seem to be folklore.2012-01-21