Questions about absolutely continuous function It could be a silly question but the definition of absolutely continuous function says that
"A real valued function $f:[a,b]\rightarrow\mathbb{R}$ is absolutely continuous on $[a,b]$ if for all $\epsilon&gt;0$, there is a $\delta&gt;0$ such that for every finite disjoint collection $\{(a_k,b_k)\}_{k=1}^n$ of $(a,b)$, $\sum_{k=1}^n[b_k-a_k]&lt;\delta$ implies $\sum_{k=1}^n|f(b_k)-f(a_k)|&lt;\epsilon$."
As I know every absolutely continuous function on $[a,b]$ is continuous on $[a,b]$. But with this definition it seems not hold if we take strange values at $x=a,b$ because the definition only depends on open intervals in $(a,b)$. Is it right that this statement does not hold with above definition?
You're overthinking this. Let $\varepsilon&gt;0$, and choose the corresponding $\delta$. Then for any $\eta&lt;\delta$ such that $a+\eta\leqslant b$, we have
$$|f(a+\eta)-f(a)|&lt;\varepsilon, $$
and hence $f$ is continuous at $a$. $f$ is continuous at $b$ by a similar argument.
There is no 'continuity' in a closed interval, mainly because of the +/-(δ>0) criteria at the upper/lower bounds.