According to Rudin's Real and Complex Analysis, a complex-valued function $ f $ on $ [a,b] $ is said to be absolutely continuous if $\forall $$\epsilon >0 \ $, $\exists\delta>0\ $ such that $$ \sum^n_{k=1}|f(b_k) -f(a_k)|< \epsilon $$

for every $n$ disjoint subintervals $ \ (a_k,b_k) $ of $ \ [a,b] $, $k=1,\cdots,n$, such that $ \sum^n_{k=1}|b_k -a_k|< \delta $.

Rudin says that $f$ is continuous on $[a,b]$. Clearly $f$ is continuous on $(a,b)$, but how can I show that $f$ is continuous on the endpoints?

  • $\begingroup$ What's the problem with the endpoints? I don't see how the proof of continuity on $[a,b]$ is any different from the proof for $(a,b)$. What goes wrong when you try just using the same argument? You're making an error or misunderstanding something; can't be more specific until you explain what the problem is... $\endgroup$ Commented Feb 12, 2020 at 14:13

1 Answer 1


It's simply obvious that the definition implies continuity on all of $[a,b]$; I can't see what the problem is, because I literally can't think of a proof that $f$ is continuous on $(a,b)$ that doesn't work at the endpoints.

Suppose $\epsilon>0$. Choose $\delta>0$ as above. Suppose $s,t\in[a,b]$ and $|s-t|<\delta$. Wlog $s<t$. Let $n=1$, $a_1=s$, $b_1=t$. Note that $(a_1,b_1)\subset[a,b]$ and $|b_1-a_1|<\delta$. So the condition above implies $|f(t)-f(s)|=\sum|f(b_j)-f(a_j)|<\epsilon$, qed.

(Yes, I think "obvious" is right...)

  • $\begingroup$ I made a silly comment earlier. When I saw this answer I upvoted it. It appears that someone later downvoted the answer. I wonder why this was downvoted. I am very glad that David Ulrich pointed out the mistake in comment. $\endgroup$ Commented Feb 13, 2020 at 7:22
  • $\begingroup$ @KaviRamaMurthy There's no point to trying to figure out why people downvote things. (If anything it seems to me the OP has some downvotes coming, not for the question per se, but for not answering my question about how one gives a proof of continuity on $(a,b)$ that does not work exactly the same for continuity on $[a,b]$...) $\endgroup$ Commented Feb 13, 2020 at 13:08

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