Prove that convex function on $[a,b]$ is absolutely continuous

Question

In the book Roberts, Varberg, Convex functions, on pp.9-10 is proved that

If $f:(a,b)\rightarrow \mathbb R$ is continuous and convex then $f$ is absolutely continuous on each $[c,d]\subset (a,b)$.

How can this be extended to the case when the domain of $f$ is instead $[a,b]$?

Edit

I wish to prove the following theorem:

If $f:[a,b] \rightarrow \mathbb R$ is continuous and convex then $f(b)-f(a)=\int_a^b f'(t)dt$.

Thanks

Answer 1

Hint. It should be enough to use the case from the book, together with:

If $\phi$ is nondecreasing, then $$ \int_a^b \phi(t)\,dt = \lim_{\alpha\searrow a, \beta \nearrow b} \int_\alpha^\beta \phi(t),dt $$

Answer 2

Suppose that $f\colon (a,b)\to \mathbb{R}$ is convex. Consider the function

$$F(x,t) = \frac{f(x+t) - f(x)}{t}\quad(t\neq 0).$$

Then $F$ is increasing in each variable. So $F$ is uniformly bounded on each compact subinterval $[c,d]\subset (a,b)$. In particular, $f$ is Lipschitz on $[c,d]$ hence absolutely continuous.