# Subsequence such that integrals converge over any Borel set in $[0,1]$

I was reading this question: Existence of subsequence such that integration converge

The idea is this. I have a sequence of uniformly bounded measurable functions $$\{f_{n}\}$$ on $$[0,1]$$ and I want to find a subsequence $$f_{n_{j}}$$ such that $$\lim_{n \to \infty} \int_{A} f_{n_{j}}$$ exists for all Borel sets $$A$$. I can show the following:

(1) If $$\{S_{i}\}_{i}$$ is a countable collection of Borel sets, then we can find a subsequence so that $$\int_{S_{i}} f_{n_{j}}$$ has a limit for all $$S_{i}$$.

(2) That this holds for all half-open half-closed intervals $$(a_{i}, b_{i}]$$ with rational endpoints.

e know that the collection of half-open half-closed intervals with rational endpoints is countable and generates the Borel $$\sigma$$-algebra, so the idea is now to approximate every Borel set using sets in this algebra and show that the result holds for them. In particular if $$A \subset [0,1]$$ is a Borel subset then we can find a sequence $$I_{i}$$ of half-open half-closed intervals with rational endpoints such that $$I_{i} \downarrow A$$, but I'm not able to proceed further. Is it true that if $$\int f_{n_{j}}$$ has a limit on each $$I_{i}$$, and $$I_{i}$$ is a decreasing sequence of sets, then $$\int f_{n_{j}}$$ has a limit on $$\bigcap_{i} I_{i}$$?

So what you have is a (countable) family $$(A_m)$$ of subsets of $$[0,1]$$ and a subsequence $$g_n$$ of the $$(f_n)$$ such that $$(\int_{A_m} g_n)_n$$ is convergent for every $$m$$.

( you got that $$g_n$$ with the diagonal process).

Now, you can take $$(A_m)$$ such that for every $$A$$ measurable and $$\epsilon > 0$$ there exists $$A_m$$ such that $$\mu(A\Delta A_m) < \epsilon/3$$. For instance, take $$A_m$$'s to be the finite union of intervals with rational ends.

Since $$\int_{A_m} g_n$$ is convergent there exists $$N$$ such that $$|\int_{A_m} g_n - \int_{A_m} g_{n'}| < \epsilon/3$$ for all $$n,n'> N$$.

Note that since $$|g_n|\le 1$$ for all $$n$$ we have $$\int_{A_m} g_n - \int_{A} g_n| < \epsilon/3$$

Now use the triangle inequality to conclude $$|\int_{A} g_n - \int_{A} g_{n'}| < \epsilon$$ for all $$n,n'> N$$.

We conclude $$\int_A g_n$$ is convergent.

Note that this implies $$g_n$$ converges in measure, so there exists a subsequence of it that converges a.e.to a function.