I'm doing the following exercise:
Let $\{ f_n \}$ be a sequence of measurable real-valued functions on $[0, 1]$ that is uniformly bounded. Show for each Borel subset there's a subsequence $k_n$ such that $\int_A f_{k_n}$ converges. Do the same for a countable family of Borel sets, then for all Borel sets.
The first two are pretty straightforward: uniform boundedness and $[0, 1]$ as the domain show the integrals are inside an interval, so by compactness we have a convergent subsequence.
Next, I can do the "diagonal method": enumerate the sets, find a subsequence for the first and replace the sequence by it. Next, for the second, find a subsequence and replace the sequence by it, and so on... Then collect all diagonal indexes.
I'm stuck on the third item. My idea: use second item on all rational intervals (semi-closed, open, etc.), then try to show the set of Borel subsets for which the integral converges is a $\sigma$-algebra. However, I'm having some trouble. I know it has all complements, but I don't know how to deduce the intersection case. I can't see any convergence relation between $\int_A f_n$, $\int_B f_n$ and $\int_{A \cap B} f_n$. Without this I can't proceed to unions. Actually, I even don't know if it is enough to choose the subsequence with only the rational intervals in mind.
Any help is appreciated!
The basic approximation theorem of Measure Theory (Ref. Theorem D, p. 56 of Halmos' Measure Theory) shows that given any Borel set $B$ in $[0,1]$ there exists a seqeuence of sets $(B_n)$ such that each $B_n$ is a finite disjoint union of intervals and $m (B\Delta B_n) \to 0$. ($m$ denotes Lebesgue measure). By further approximation we may take these intervals to have rational end points. You can now finish the proof using the fact that $|\int_B f_{k_n} -\int_{B_m} f_{k_n}| \leq C m(B\Delta B_m)$ where $C$ is a bound for $f_n$'s.