If $f$ is midpoint-convex and $f$ is a Baire function (Borel measurable) then $f$ is convex Suppose $f$ is a function defined on some interval $\mathrm{J}$ such that
$$
f \left( \dfrac{t_1 + t_2}{2} \right) \leq \dfrac{f(t_1) + f(t_2)}{2},
$$
for all pairs $t_1, t_2$ in $\mathrm{J};$ this is sometimes referred to as "midpoint convex." Furthermore, assume $f$ is a Baire function (this is equivalent to $f$ being Borel measurable). Then, $f$ is convex, meaning that $f(\alpha_1 t_1 + \alpha_2 t_2) \leq \alpha_1 f(t_1) + \alpha_2 f(t_2)$ for all $t_1, t_2$ as before and $\alpha_1 + \alpha_2 = 1$ both non-negative.
I read this result while reading the book of William Feller "An Introduction to Probability Theory and Its Applications" without a hint on how to proceed.
I know how to prove the result when $f$ is continuous (e.g. Midpoint-Convex and Continuous Implies Convex). After a few attempts, I don't see a way around to the measurable case. Any hints or references appreciated.
 A: First step: We prove that if $f$ is midpoint-convex and bounded near a point $x$ then $f$ is continuous in $x$.
To do so, we assume WLOG that $f$ is bounded by $M$ in $(a,b)$, $f(0)=0$ and $f$ is not continuous at $0$ (because $f$ is midpoint-convex/continuous iff $f-c$ is). Let $\epsilon > 0$ be such that there is a sequence $x_n$ converging to $0$ with $|f(x_n)|>\epsilon$. Note that, due to the inequality $f(x) + f(-x) \geq 0$ we may assume, changing $x_n$ by $-x_n$, that $f(x_n)>0$ so $f(x_n) \geq \epsilon$. By midpoint-convexity, and since $f(0)=0$, $f(2x)>2f(x)$, so
$$
f(2^k x_n) \geq 2^k \epsilon
$$
which leads to a contradiction if we pick $k$ big enough such that $2^k \epsilon >M$ and $n$ big enough such that $2^k x_n \in (a,b)$.
Second step: Now suppose $f$ is measurable on an interval $(a,b)$ but it is not locally bounded near $x$, with $f(x)=0$. After restricting the interval we can assume that $(a,b) = (x-2 \delta , x+ 2 \delta)$ for some $\delta >0$. For each $k>0$ pick $y_k$ in $(x_0-\delta, x_0+\delta)$ such that $f(y_k)>k$ (we can take positive values as before, using $f(y_k)+f(2x-y_k) \geq 0$).
Then for any point $z \in (y_k-\delta, y_k+\delta) \subset (x-2 \delta , x+ 2 \delta)$ we have $f(z)+f(2y_k-z) \geq 2k$. Note that $2y_k-z$ is the symmetric of $z$ with respect to $y_k$ so it is contained in $(y_k-\delta, y_k+\delta)$. Therefore, since $f$ is measurable, the set of points in $(y_k-\delta, y_k+\delta)$ with $f(z)>k$ has measure at least $\delta$. In particular, the set $\{z \in (a,b): f(z) >k\}$is measurable and has measure at least $\delta$ for all $k$. This is a contradiction because taking $k$ to be natural numbers, you would get that
$$0<\delta \leq \lim_{k \to \infty}m\left(\{ z \in (a,b) : f(x) >k \}\right) = m \left(\bigcap_{k \in \mathbb N} \{ z \in (a,b) : f(x) >k \}\right)=m(\emptyset)=0$$
