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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 $\bigcap_{i} I_{i}$?

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  • $\begingroup$ Your question is answered here. $\endgroup$ Nov 29, 2020 at 20:57

1 Answer 1

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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.

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