# 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 $\epsilon>0$, there is a $\delta>0$ such that for every finite disjoint collection $\{(a_k,b_k)\}_{k=1}^n$ of $(a,b)$, $\sum_{k=1}^n[b_k-a_k]<\delta$ implies $\sum_{k=1}^n|f(b_k)-f(a_k)|<\epsilon$." 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?

• Are you sure the definition says $(a,b)$ (and not $[a,b]$)? If you look, for example, at en.wikipedia.org/wiki/Absolute_continuity, you'll see that the interval $I$ in the definition is the same as the interval from which the $a_k$ are chosen. – Thomas Jan 23 '15 at 14:48
• Yes. I'm sure. I read the definition in Royden's real analysis and many other books. – user156043 Jan 24 '15 at 2:31
• I just looked it up in Royden. Actually in his definition he does not make it precise from where the $x_i$ (as he denotes them) are taken, but in particular he does not request to take them from the open interval, so what you said is not correct. I, to the contrary, am absolutely sure that if you define absolute continuity on an interval $I$ (regardless whether open or close) you have to allow the $x_i$ to be taken from all of $I$. – Thomas Jan 24 '15 at 6:07

You're overthinking this. Let $\varepsilon>0$, and choose the corresponding $\delta$. Then for any $\eta<\delta$ such that $a+\eta\leqslant b$, we have $$|f(a+\eta)-f(a)|<\varepsilon,$$ and hence $f$ is continuous at $a$. $f$ is continuous at $b$ by a similar argument.
• $f: I\rightarrow \mathbb{R}$ is continuous in $x\in I$ if for every $\varepsilon >0$ you can find $\delta >0$ such that $|f(x)-f(y)| < \varepsilon$ for every $y\in I$ such that $|y-x|< \delta$. By requesting $y\in I$ it does not matter whether $x$ is a boundary point of $I$. The reason to define right continuity is more that there are functions which have a discontinuity in the interior of an interval but behave still nicely on side of that point. – Thomas Jan 23 '15 at 15:42