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Problem: Let $(f_n)$ be a uniformly bounded sequence of real valued continuous functions on $[0,1]$. Prove that there is ONE subsequence $(f_{n_k})$ such that for every $0\le a < b \le 1$, we have $$\lim_{k\to\infty} \int_a^b \! f_{n_k}(t) dt $$ exists.

Context: Advanced Undergraduate Analysis. Familiar with Real Analysis by Carothers and Principles of Analysis by Rudin

I think it would be obvious to show this for all rationals inbetween a and b but I do not know how to start showing that there is a single subsequence. Any help would be appreciated, Thank you.

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I doubt there would be only one. It was most probably meant 'at least one'. Any subsequence of such a sequence will be suitable, too. – Berci Nov 26 '12 at 2:01
Hint: To prove it for rationals $a,b$ use a diagonal argument. – Jose27 Nov 26 '12 at 2:25
Thank for the replies, I was of the impression it was 'at least one' but 'ONE' was emphasized so I'm still confused as to what it meant. Also, Jose could you expand a bit on the application of a diagonal argument in this particular case? – jimmywho Nov 26 '12 at 2:43
up vote 5 down vote accepted

Hint: Apply Arzelà–Ascoli theorem to the sequence $\int_0^xf_n(t)dt$, $x\in[0,1]$.

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Thanks for your reply. I was able to prove that there exist a subsequence of $F_n = \int_0^x f_n(t)dt$ that converges uniformly for all $x \in [0,1]$. – jimmywho Nov 27 '12 at 0:43
My question is that since this is true then for $ a < b \in [0,1] \int_0^a f_n(t)dt$ and $\int_0^b f_n(t)dt$ have uniformly converging subsequences then we have that by subtraction $\int_a^b f_n(t)dt$ has a uniformly converging subsequence? – jimmywho Nov 27 '12 at 0:50
@jimmywho: Since you have proved that there exists $n_k\to\infty$, such that $F_{n_k}(x)=\int_0^xf_{n_k}(t)d t$ uniformly converges to some $F$ on $[0,1]$, then by the definition of uniformly convergence, $\int_a^b f_{n_k}(t)d t=F_{n_k}(b)-F_{n_k}(a)$ uniformly converges to $F(b)-F(a)$. – 23rd Nov 27 '12 at 3:27
Thanks, I believe that was what I was attempting to state in my comment. I believe that this is the result I was looking for but am I missing anything? – jimmywho Nov 27 '12 at 3:51
@jimmywho: Sorry, I cannot understand what you mean. Is there anything still unclear to you? – 23rd Nov 27 '12 at 3:58

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