Question: Find a subsequence $(f_{n_j})_j$ of uniformly integrable sequence $(f_n:[0,1] \rightarrow \mathbb{R})$, such $\int_0^1 f_{nj}g$ converges for all bounded, measurable, real-valued $g$.
Attempt: I thought to do this 'building up', as follows:
- Consider the indicators $g = \chi_A$.
- If $A$ is an interval, we note that $(\int_0^1 f_{n} \chi_A)_n$ is a bounded sequence, so must have a convergent subsequence of form $(\int_0^1 f_{nj} \chi_A)_n$
- Now, let $(I_k)_k =$ sub-intervals of $[0,1]$ with rational endpoints. We construct a subsequence $(f_{n_j})_j$ such that for all $g = \chi_{I_k}$, we have $\int_0^1 f_{n_j} \chi_{I_k}$ converge (as a sequence of $j$). This follows by a diagonalization argument: for fixed $k$, we may have $(\int_0^1 f_{n_j(k)} \chi_{I_k})_{n_j(k)}$ such that it converges, and then we may extract a subsequence $(f_{n_j(k+1)})$ such that $(\int_0^1 f_{n_j(k+1)} \chi_{I_k+1})_{n_j(k+1)}$ converges. Finally, we take $(f_{n_k(k)})_k$ as our subsequence, for which $\int_0^1 f_{n_k(k)} \chi_{I}$ converges for any interval $I$ with rational endpoints, by construction.
- Now, we will prove that $\{A \subset [0,1]: \int_0^1 f_{n_k(k)} \chi_A \text{converges}\} $ is a $\sigma-$algebra. Complements are easy. To see countable unions, note that
$$ \int_0^1 f_{n_k(k)} \chi_{\sqcup_i A_i} = \int_0^1 f_{n_k(k)} \sum_i \chi_{ A_i} =_{\text{DCT}} \sum_i \int_0^1 f_{n_k(k)} \chi_{ A_i} $$
Question: (1) Now, this is where I have trouble. I'm not sure how to prove that $\lim_k \sum_i \int_0^1 f_{n_k(k)} \chi_{A_i}$ exists. It is certainly true that the other iterated limit, $\sum_i \lim_k \int_0^1 f_{n_k(k)} \chi_{A_i}$, exists.
(2) Is there another, perhaps less lengthy approach to this question (one that preferably doesn't use the existence of an a.s. convergent subsequence of the $(f_n)_n$?)
Note I found this asked previously, but the answer only provides a subsequence that depends on $g$. I also find this related to the style of another question.
