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One can easily construct an example of a measurable function $f:(a,b)\to \mathbb{R}$ which satisfies the following property:

$$\label{p}\tag{P} f\notin L^1(I),\ \mbox{for each interval}\ I\subset (a,b). $$

We know that a convex function $g:(a,b)\to \mathbb{R}$ is locally Lipschitz and its second derivative $g''$ exists a.e. $x\in (a,b)$.

Can we find a increasing convex function $g$, such that $g''$ satisfies the property \ref{p}?

If the answer is negative

What can we say about the set of intervals $I$ for which $g''\notin L^1(I)$?

Remark: I've asked a similar question on Math.stack with no answer.

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    $\begingroup$ Hint: If $g$ is a convex function, then $g'$ is an increasing function. So an equivalent question is: How bad can be the derivative of an increasing function? $\endgroup$ – Gerald Edgar Mar 22 '17 at 11:19
  • $\begingroup$ @GeraldEdgar Following your question The derivative of an increasing function is a Borel measure $\mu$ on the interval, so the question would be how bad such a measure could be? I think that $g''$ would be the density of the absolutely continuous part of $\mu$. This would be $L^1$ on each interval $I$ such that $\mu(I)<\infty$. $\endgroup$ – Liviu Nicolaescu Mar 22 '17 at 12:16
  • $\begingroup$ @LiviuNicolaescu ... correct. $g''$ (in the sense of a function) is in $L^1$ on every closed interval contained in the domain of $g$. If the OP means $g''$ is the distributional sense, he/she should see the answer of Denis. $\endgroup$ – Gerald Edgar Mar 22 '17 at 13:53
  • $\begingroup$ I mean g'' in the sense of a function @GeraldEdgar. $\endgroup$ – Tomás Mar 22 '17 at 14:41
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The second derivative of a convex function, in the distributional sense, is a non-negative bounded measure. And conversely. If this measure $\mu$ contains a sum $\sum_na_n\delta_{x=x_n}$, where $(x_n)_n$ is dense and $a_n>0$, $\sum_na_n<\infty$, then $\mu|_I\not\in L^1(I)$ for every non-void open interval $I$. The corresponding $g$ is an example for the first question.

On the other hand $g''$ can be decomposed uniquely into a singular part $\mu_s$ and a part $\mu_a<\!<{\cal Leb}$. Because $\mu_s,\mu_a$ are non-negative, we have $$\int_c^dd\mu_a\le|g''|<\infty,$$ at least if $a<c<d<b$. Therefore $\mu_a\in L^1(I)$ for every $I\subset\!\subset(a,b)$.

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