Let $f: I\to \mathbb{R}$ be a convex function.
Why does the second derivative of $f$ exist almost everywhere?
By searching I knew that is Alexandrov theorem, but I didn't find the proof...
My try
For $x_1<x<x_2$ where function is defined , we have $$\frac{f(x)-f(x_1)}{x-x_1}\le\frac{f(x_2)-f(x_1)}{x_2-x_1}\le\frac{f(x_2)-f(x)}{x_2-x}$$ Considering $x_1\to x^-$ and $x_2\to x^+$, we can conclude that $f_-'(x)$ and $f_+'(x)$ exist and $f_-'(x)\le f_+'(x)$. (in this process we only need $\frac{f(x)-f(x_1)}{x-x_1}\le\frac{f(x_2)-f(x)}{x_2-x}$)
Considering $x\to x_1$ and $x\to x_2$, we can conclude that $$f_+'(x_1)\le \frac{f(x_2)-f(x_1)}{x_2-x_1} \le f_-'(x_2)$$
So we get that $f_-(x)$ and $f_+(x)$ exist and both are monotonically increasing.
In addition, $f_-'(x)\le f_+'(x)$.
Because monotone functions only have jump discontinuities and are continuous everywhere except countably many points, we can conclude that $f'$ exists everywhere except countably many points.
How to move on? Any hints? Thank you in advance!