A function ${f: {\mathbb R} \rightarrow {\mathbb R}}$ is said to be convex if one has ${f((1-t) x + ty) \leq (1-t) f(x) + t f(y)}$ for all ${x < y}$ and ${0 < t < 1}$. Show that if ${f}$ is convex, then it is continuous and almost everywhere differentiable, and its derivative ${f'}$ is equal almost everywhere to a monotone non-decreasing function, and so is itself almost everywhere differentiable. (Hint: Drawing the graph of ${f}$, together with a number of chords and tangent lines, is likely to be very helpful in providing visual intuition.) Thus we see that in some sense, convex functions are “almost everywhere twice differentiable”. Similar claims also hold for concave functions, of course.
Attempt: First we show that secant lines of $f$ are steeper to the right. i.e. $\forall x_1 \leq x_2 \leq x_3 \in \mathbb{R}$, we have $\displaystyle \frac{f(x_2) - f(x_1)}{x_2 - x_1} \leq \frac{f(x_3) - f(x_1)}{x_3 - x_1} \leq \frac{f(x_3) - f(x_2)}{x_3 - x_2}. \tag{1}$
By convexity, the secant line through $(x_1, f(x_1))$ and $(x_3, f(x_3))$ is above $(x_2, f(x_2))$ at $x_2$, which gives $(1)$ by definition of the slope.
By $(1)$, and the fact that bounded monotone sequence converges, we have $\forall z \in \mathbb{R}$, $D^-f(z) \leq D^+f(z) \in \mathbb{R}$ exist, where $D^+f(z) := \lim_{h \to 0^+}\frac{f(z+h) - f(z)}{h}$ and $D^-f(z) := \lim_{h \to 0^-}\frac{f(z+h) - f(z)}{h}$ are right and left derivative of $f$ at $z$ respectively. In particular we have $\lim_{h \to 0^+}f(z + h) - f(z) = \lim_{h \to 0^-}f(z + h) - f(z) = 0$, so $\lim_{h \to 0}|f(x+h) - f(x)| = 0$ and $f$ is continuous. Also, if $x \leq y \in \mathbb{R}$, again by $(1)$ we have $D^-(x) \leq D^+(x) \leq D^-(y) \leq D^+(y)$. Finally, whenever there is a gap $D^-(x) < D^+(x)$, there is at least one rational number $q_x$ such that $D^-(x) < q_x < D^+(x)$, and from monotonicity, each rational number can be assigned to at most one gap, and since countable set has measure zero, this completes the proof.
I'm not sure if there are ingredients still missing or not?