How do I see that if $A$ is a Borel subset of $[0, 1]$, then there exists a subsequence $n_j$ such that $\int_A f_{n_j}(x)\,dx$ converges? Let $\{f_n\}$ be a sequence of measurable real-valued functions on $[0, 1]$ that is uniformly bounded. How do I see that if $A$ is a Borel subset of $[0, 1]$, then there exists a subsequence $n_j$ such that $\int_A f_{n_j}(x)\,dx$ converges?
 A: The sequence $(x_n)$  where $x_n = \int_{A} f_{n}(x)dx$ is bounded, and every bounded sequence has a convergent subsequence.
A: Assuming that with uniformly bounded you mean $\exists M>0: \|f_n\| \le M$ for $n \in \mathbb{N}$, we have
$$|\int_A f_n\, d\lambda| \le \lambda(A) M$$
Therefore the sequence of the integrals is bounded in $\mathbb{R}$ and the claim follows with Bolzano Weierstrass theorem.
A: If $X$ is a separable normed linear space, and $\varphi_n$ is a norm bounded sequence in $X^*,$ then there is a subsequence $\varphi_{n_k}$ that converges to some $\varphi$ in the weak-* sense. This is a theorem due to Banach. Apply this with $X=L^1[0,1]$ and your given bounded sequence $f_n$ in $L^\infty[0,1] = (L^1[0,1])^*.$  You then get a subsequence $f_{n_k}$ such that for every $g\in L^1,$ $\int_0^1f_{n_k}g$ converges. In particular, for this one subsequence, we see that $\int_0^1f_{n_k}\chi_A$ converges for every Borel $A-$ a much stronger conclusion that I think is likely to have been the intent of the problem.
