$\newcommand{\R}{\mathbb R}$ Let $f$ be a function from $\R$ to $\R$. It is said that $f$ is midpoint-convex if for any real $x$ and $y$ we have $f(x-y)+f(x+y)\ge2f(x)$ or, equivalently, $f(x+y)-f(x)\ge f(x)-f(x-y)$. So, $f$ is midpoint-convex iff for any real $x$ and and any real $y>0$
$$\frac{f(x+y)-f(x)}y\ge\frac{f(x)-f(x-y)}y.$$ One may call the ratio $\frac{f(x+y)-f(x)}y$ the average rate of change of the function $f$ over the interval $[x,x+y]$. Thus, $f$ is midpoint-convex iff for any real $x$ and and any real $y>0$ the average rate of change of $f$ over $[x,x+y]$ is no less than the average rate of change of $f$ over $[x-y,x]$.

By Sierpiński's theorem, every Lebesgue measurable midpoint-convex function from $\R$ to $\R$ is convex. However (assuming the axiom of choice), one can easily construct a (necessarily non-Lebesgue measurable) midpoint-convex function from $\R$ to $\R$ which is not convex -- see e.g. this, where $f(i_1)$ should be replaced by $f(b_{i_1})$.

Suppose now that $f$ has non-decreasing average rate of change in the sense that $$\frac{f(x_2+y)-f(x_2)}y\ge\frac{f(x_1+y)-f(x_1)}y$$ for any real $x_1,x_2,y$ such that $x_1\le x_2$ and $y>0$. This is equivalent to the following: $$\frac{f(x-y_2)+f(x+y_2)}2\ge\frac{f(x-y_1)+f(x+y_1)}2$$ for any real $x,y_1,y_2$ such that $y_1\le y_2$. One may note here that every function with non-decreasing average rate of change is obviously midpoint convex.

The question is then the one in the title: Is there a (necessarily non-Lebesgue measurable) non-convex function with non-decreasing average rate of change?