Convexity implies absolute continuity? The following is taken from an exam:

$f:[a,b]\rightarrow\mathbb{R}$ is convex implies $f$ is absolutely continuous (recall $f'$ exists a.e.)

One has local Lipschitz-ness by convexity, but how to show absolute continuity without global Lipschitz-ness?
 A: Convex functions on the real line are expressible as integrals of one-sided derivatives.
The ration $k(x,y)=\frac{f(y)-f(x)}{y-x}$  is increasing in $y$ on $[x,b]$.
Hence, the right-hand derivative $D_+f(x)$ exists for all $x\in [a,b]$.
In a similar way we  conclude that  the left-hand derivative $D_-f(x)$ exists for all $x\in[a,b]$, and that  $D_-f(x)\leq D_+f(x)$.
Hence the set of points for which is  $D_-f(x)< D_+f(x)$ is countable.
If $f$ is convex  on $[a,b]$, then   then both
$D_+f(x)$  and   $D_-f(x)$
are integrable with respect to Lebesgue
measure on
$[a,b]$, and  $f(x)=f(a)+ \int_a^x D_+f(t) dt=f(a)+ \int_a^x D_-f(t) dt$.
More generally, suppose D is an increasing, real-valued function defined (at least) on $[a,b)$. Define $g(x) := \int^x_a D(t)dt$, for $a \leq x \leq  b$. (Possibly  $g(b) =\infty$.) Then $g$ is convex.
In the book Roberts, Varberg, Convex functions, on pp.9-10 is proved that
If $f:(a,b)\rightarrow R$
is continuous and convex then f
is absolutely continuous on each $[c,d]\subset (a,b)$.
A: $f$ is continuous since it is absolutely continuous. Moreover, $f'$ exists almost everwhere, and is increasing. Thus $f'' \geq 0$ whence $f$ is convex.
Notice that continuity is important since you could move a point on the graph of a convex function upward or downward rendering it nonconvex.
A: The assertion is not true as stated. In fact, the function $$f(x) = \begin{cases}1 & \text{if } x = 0 \text{ or } x = 1 \\ 0 & \text{else}\end{cases}$$
is convex on $[0,1]$, but not even continuous.
However, if we add continuity (which may only be violated at the end points), the assertion is true, here is a (brief) sketch of a possible proof.
Let $\varepsilon > 0$ be given. Then, by continuity at the end points and convexity of $f$ (which implies some monotonicy around the end points), there is $\delta > 0$, such that the sum of function value differences on $[a, a+\delta]$ and $[b-\delta,b]$ is less than $\varepsilon$ if the subintervals sum to less than $\delta$. On the remaining interval you can argue by Lipschitz continuity.
