Absolutely continuous function is continuous

Question

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?

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...)