# Concerning existence of subsequence of converging integrals on subsets of $[0,1]$ of a sequence $(f_n)\in[0,1]$

Problem Statement

Let $\{f_n\}$ be a sequence of real-valued, measurable functions on $[0,1]$ that is uniformly bounded.

1. Show that if $A$ is a Borel subset of $[0,1]$ then there exists subsequence $n_j$ such that $\int_A f_{n_j}(x) \ \mathrm{d}x$ converges.

2. Show that if $(A_i)$ is a countable collection of Borel measurable subsets of $[0,1]$, then there exists a subsequence $n_j$ such that $\int_{A_i} f_{n_j}(x) \ \mathrm{d}x$ converges for each $i$.

3. Show that there exists a subsequence $n_j$ such that $\int_A f_{n_j}(x)$ converges for each Borel subset of $A$.

Attempt

At first I started thinking of using a diagonalization argument and to approach this problem step by step. Instead, I am wondering what might be wrong with the following naive approach.

$f_n$ is uniformly bounded on $[0,1]$ so $\int_{[0,1]}f_n \ \mathrm{d}x$ is an infinite sequence of real numbers on the compact set $[-2k,2k]$, where $|f_n|\leq k$. So there is a convergent subsequence $\int_{[0,1]}f_{n_k}\ \mathrm{d}x$. This subsequence also converges for any borel subset of $[0,1]$ so we have the result for all three of the above problems.

Question

What major concept(s) am I missing here? Note that I am not asking for a full solution to the problem but rather some feedback on my attempt at solving it.

I'm sorry for this silly question but I find it hard to dig into a problem until I realize why my "initial naive attempt" fails.

• The convergence of $\int_{[0, 1]} f_{n_k}$ does not guarantee the convergence of say $\int_{[0, 1/2]} f_{n_k}$. I think you should prove (2) first and then apply (2) to the set of all dyadic intervals to get (3) using the fact that Borel sets can be approximated by finite unions of dyadic intervals up to arbitrary precision. – hot_queen Jun 18 '15 at 20:55
• A simple example: $f_n = 0$ when $n$ is even and $f_n = 1_{[0, 1/2]} - 1_{(1/2, 1]}$ when $n$ is odd. Then $\int_{[0, 1]} f_n = 0$ for all $n$ but $\int_{[0, 1/2]} f_n$ oscillates. – hot_queen Jun 18 '15 at 21:01
• I.e., you might have to choose a different subsequence for each (Borel) subset of $[0,1]$. For a countable collection of sets (as in (2)), this can be circumvented using a diagonalization argument, as you write yourself. – PhoemueX Jun 18 '15 at 21:31

I decided to use the comments to come up with the following solutions (please let me know how I did).

1. If $A\subset [0,1]$ is Borel, then $\bigg|\int_A f_n \ dx\bigg|$ is a bounded sequence of real numbers on $A$ by $2K$, where $|f_n(x)|\leq K$. Therefore there is a convergent subsequence $\int_A f_{n_j} \ dx$ as desired.
2. Let $(A_i)_{i=1}^{\infty}$ be a countable collection of Borel sets in $[0,1]$. By $(1)$ there is a convergent subsequence $(f_{1,k})_{k=1}^\infty$ so that $\int_{A_1} f_{1,k}$ converges as $k$ goes to $\infty$. Now $(f_{1,k})$ is a uniformly bounded sequence so again by $(1)$ there is a subsequence $(f_{2,k})_{k=1}^\infty$ such that $\int_{A_2} f_{2,k}$ converges. In this manner, each $\int_{A_n} f_{n,k}$ converges and $(f_{n+1,k})_{k=1}^\infty$ is a subsequence of $(f_{n,k})_{k=1}^\infty$. Thus we have the following diagram: \begin{align*} & \int_{A_1} f_{1,1} \ , \ \ \int_{A_1} f_{1,2} \ , \ \ \int_{A_1} f_{1,3} \ \ \ ... \\ & \int_{A_2} f_{2,1} \ , \ \ \int_{A_2} f_{2,2} \ , \ \ \int_{A_2} f_{2,3} \ \ \ ... \\ & \int_{A_3} f_{3,1} \ , \ \ \int_{A_3} f_{3,2} \ , \ \ \int_{A_3} f_{3,3} \ \ \ ... \\ \vdots \end{align*} In this diagram, each row is a subsequence of the row before it and the elements in the $k^{th}$ row appear in the same order as they would in the $k-1$ row, unless if they are deleted when shifting down a row. Note that by construction, the sequence of diagonal elements, $\left(\int_{A_n} f_{n,n}\right)_{n=1}^{\infty}$ is a subsequence of every $\left(\int_{A_n} f_{n,k}\right)_{k=1}^{\infty}$ (except possibly for the first $n-1$ terms). Thus the diagonal sequence converges on each $A_i$.
3. The set of all Dyadic intervals is countable so the desired result follows from $(2)$ and the facts that every Borel measurable subset of $[0,1]$ can be arbitrarily approximated by Dyadic intervals and that the countable union of finite sets is countable. (I think this needs more work...)
• Do you really need dyadic intervals or can you just look at all intervals with rational end points? – rem Mar 11 '19 at 0:48