I am going to show that if f is absolutely continuous it implies that f is continuous, the test seems somewhat trivial to me of the definition, however I would like to know if I am doing it well.
If $f$ is absolutely continuos on [a,b], let $\varepsilon >0$, there exist $\delta>0$ such that $\displaystyle\sum_{k=1}^{n}|f(b_{k})-f(a_{k})|<\varepsilon$, if $\sum_{k=1}^{n}(b_{k}-a_{k})<\delta$, for some collection finite $\lbrace (a_{k},b_{k}),1\leq k \leq n\rbrace$.
If $n=1$, $|b_{1}-a_{1}|<\delta$ implies $|f(b_{1})-f(a_{1})|<\varepsilon$, then $f$ is uniformly continous on $[a,b]$
Best Answer
Let $ c\in [a,b]$ and $\epsilon>0$.
$f $ is absolutely continuous at $ [a,b]$ $$\implies$$
$$\exists \delta>0\;\; : \; \forall (a_1,b_1)\in [a,b]^2\;$$ $$ |a_1-b_1|<\delta \implies |f(a_1)-f(b_1)|<\epsilon$$
$$\implies$$ $$\exists \delta>0\;\; :\;\; \forall x\in[a,b]$$ $$|x-c|<\delta \;\;\implies \;\;|f(x)-f(c)|<\epsilon$$ This proves that $ f$ is continuous at any point $ c\in [a,b].$