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.



How bad can the second derivative of a convex function be?
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 $I\subset\!\subset(a,b)$.


