Let continuous $f:X\to\mathbb R$. $X$ is an interval of $\mathbb R$.
Are the following two statements equivalent?
1) For almost every $x\in X$ $\exists$ open interval (neighborhood) $I\ni x$ s.t. $f$ is either locally convex or locally concave on $I$. (i.e. only countable number of $x\in X$ does not have such neighborhood).
$f$ is locally convex on $I$ if $\forall x,y\in I$, we have $f(\lambda x+(1-\lambda)y)\leq f(\lambda x)+f((1-\lambda)y)$
(In another word, epi$f(I)$ is a convex set.)
2) $\forall x\in X$ $\exists$ (non-singleton) closed interval $I\ni x$ s.t. $f$ is either locally convex or locally concave over $I$.
If they are equivalent, then, in general, why people tend to use open set rather than closed set to define a neighborhood?
$2\Rightarrow 1$. Without loss of generality we can assume that $X$ is an open subset of $\Bbb R$. Let $X_{conv}$ (resp. $X_{conc}$) bet a set of all $x\in X$ such that there exists an open interval $I\ni x$ s.t. $f$ is locally convex (resp. locally concave) over $I$. Assume to the contrary that (1) does not hold, that is a set $X’=X\setminus(X_{conv}\cup X_{conc})$ is uncountable. For a natural $n$ we call a point $x\in X’$ an $\frac 1n$-corner provided $(x-\frac 1n,x)\cap X’=\varnothing$ or $(x,x+\frac 1n)\cap X’=\varnothing$. It is easy to check that among any three distinct $\frac 1n$-corners we can find two with the distance between them bigger than $\frac 1n$. This implies that the set of $\frac 1n$-corners is countable, so there exists a point $x\in X’$ which is not an $\frac 1n$-corner for any natural $n$. By (2), there exists a (non-singleton) closed interval $I’\ni x$ s.t. $f$ is either locally convex or locally concave over $I’$. Then $f$ is either locally convex or locally concave over an open interval $I=\operatorname{int} I’$. But $X’\cap I\ne\varnothing$, a contradiction.
$1\not\Rightarrow 2$. Define a function $f:\Bbb R\to\Bbb R$ as follows. Put $ f(x)=0$, if $x=0$ or $x=\frac 1{2n}$, with $n\in\Bbb Z\setminus\{0\}$ or $|x|\ge 1$. For $k\in\Bbb N\cup{0}$ put $ f\left(\pm\frac 1{4k+1}\right)= \frac 1{4k+1}$ and $ f\left(\pm\frac 1{4k+3}\right)=-\frac 1{4k+3}$. Extend $ f(x)$ piesewise-linearly to the remaining $x\in\Bbb R$. Since $f$ is linear at $\Bbb R$ but a countable closed set, it satifies (1). On the other hand, on any open interval $I\ni 0$, $f$ has countably infinite many zeros, so $f$ is neither convex, nor concave on $I$.
I don’t see a relation of this with the previous. For me a choice to work with open or closed neighborhoods usually is a matter of convenience. Maybe, to work with open neighborhoods is often more convenient than with closed these.
By the way, a closed interval containing $x$ is not its neighborhood, because a neighborhood of $x$ must contain $x$ in its interior.