# mid-point convex but not a.e. equal to a convex function

I recently stumbled upon Wikipedia's page on convexity (http://en.wikipedia.org/wiki/Convex_function#Properties) and there's reference to Sierpinski's theorem from which we can deduce that for Lebesgue measurable functions mid-point convexity implies convexity. For reference: mid-point convexity means $$f\left(\frac{1}{2} x + \frac{1}{2} y\right) \leq \frac{f(x) + f(y)}{2} \ldotp$$

Now, the question is: how pathological has to be an example of a mid-convex function that is not convex? The usual examples usually do something like 'put one value on rationals, different on non-rationals' but that won't work here, since such a function is a.e. equal to a convex, constant function.

I understand that since the function can't be measurable it has to be pretty weird, but how weird exactly in this case?

The only related question I've found is this one: Mid-point convexity does not imply convexity but it doesn't contain any examples, there's just a remark mentioning Sierpinski's result in the comments.

• You'd need a function that is not Lebesgue measurable, which I think requires the axiom of choice. So any "examples" would be non-constructive. – felipeh Jun 25 '14 at 14:39
• See also this related question. Maybe some references given in the answers to this post and other questions linked there might be of interest. – Martin Sleziak Dec 8 '15 at 15:12
• I was about to give an example but the post has been deleted. – DanielWainfleet Apr 23 '17 at 22:31