A function $f:[a,b]\rightarrow \mathbb{C}$ is said to be absolutely continuous if for each $\epsilon>0$, there exists $\delta>0$ such that for every mutually disjoint finite sequence of closed sub-intervals $\{[a_1,b_1],...,[a_n,b_n]\}$ of $[a,b]$ satisfying $\sum_{i=1}^n |b_i-a_i|<\delta$, $\sum_{i=1}^n |f(b_i)-f(a_i)|<\epsilon$ holds.

Let $f:[a,b]\rightarrow \mathbb{C}$ be an absolutely continuous function and $\epsilon>0$.

Then, how do I prove that there exists $\delta>0$ such that for every mutually disjoint finite sequence of open sub-intervals $\{(a_1,b_1),...,(a_n,b_n)\}$ of $[a,b]$ satisfying $\sum_{i=1}^n |b_i-a_i|<\delta$, $\sum_{i=1}^n |f(b_i)-f(a_i)|<\epsilon$ holds ?

  • $\begingroup$ @FaraadArmwood That is my question and the converse is trivial $\endgroup$ Commented Aug 4, 2016 at 0:14
  • $\begingroup$ Notice that if you turn parentheses into brackets, then the only case that can lead to you not getting a disjoint collection of closed intervals is to have $b_j=a_i$ for some indices. (If $a_i=a_j$ for two indices, then the open intervals $(a_i,b_i)$ and $(a_j,b_j)$ will have to intersect.) So, the question kind of reduces to analyzing what happens to the sum of variations when you "break up" an interval. That is, what happens to your sum if you introduce a new point in an interval $[a_i,b_i]$. If you can contain the amount that it will add, then you're done. $\endgroup$ Commented Aug 4, 2016 at 0:47

1 Answer 1


Hint: If $\{(a_k,b_k)\}$ are pairwise disjoint, then for small $t>0,$ $\{[a_k+t,b_k-t]\}$ are pairwise disjoint.

Added details, in light of comments: Let $\epsilon>0.$ Find a $\delta$ that works for closed intervals. Suppose $(a_k,b_k), k =1,\dots, n,$ are pairwise disjoint, with $\sum (b_k-a_k) < \delta.$ Then for small $t>0$, $[a_k+t,b_k-t]$ are pairwise disjoint, the sum of whose lengths is $< \delta.$ Then

$$\sum_{k=1}^{n} |f(b_k-t)-f(a_k+t)|<\epsilon$$

for small $t >0.$ Let $t\to 0^+$ to get

$$\sum_{k=1}^{n} |f(b_k)-f(a_k)|\le\epsilon.$$

Here we have used the continuity of $f,$ which is guaranteed because we are assuming $f \in AC.$

  • $\begingroup$ But then you'll have to deal with small, but many corrections to have the sum you want to bound. $\endgroup$ Commented Aug 4, 2016 at 0:37
  • $\begingroup$ @Behnam the stated condition implies $f$ is continuous. It suffices to derive an inequality and let $t \to 0^+$. $\endgroup$
    – Umberto P.
    Commented Aug 4, 2016 at 0:39
  • $\begingroup$ @UmbertoP. In fact we are assuming f is AC, so obviously f is C. $\endgroup$
    – zhw.
    Commented Aug 4, 2016 at 0:40
  • $\begingroup$ Let's look at the difference between $\sum_{i=1}^n |f(b_i)-f(a_i)|$, that we are trying to show is less that $\epsilon$, and $\sum_{i=1}^n |f(b_i-t)-f(a_i+t)|$, which you already know that is less than $\epsilon.$ At each $i$, you do get a small value by continuity of $f$, BUT don't forget that you're adding $n$ many of these for an arbitrary $n$. So, it is not too obvious to see the conclusion from your hint. (Although, I just found how to fix it!!) $\endgroup$ Commented Aug 4, 2016 at 0:53
  • $\begingroup$ @Behnam It is uniformly continuous, so I think zhw's hint works directly. $\endgroup$ Commented Aug 4, 2016 at 1:06

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