Concerning existence of subsequence of converging integrals on subsets of $[0,1]$ of a sequence $(f_n)\in[0,1]$ Problem Statement
Let $\{f_n\}$ be a sequence of real-valued, measurable functions on $[0,1]$ that is uniformly bounded.


*

*Show that if $A$ is a Borel subset of $[0,1]$ then there exists
subsequence $n_j$ such that $\int_A f_{n_j}(x) \ \mathrm{d}x$
converges.

*Show that if $(A_i)$ is a countable collection of Borel measurable
subsets of $[0,1]$, then there exists a subsequence $n_j$ such that
$\int_{A_i} f_{n_j}(x) \ \mathrm{d}x$ converges for each $i$.

*Show that there exists a subsequence $n_j$ such that $\int_A
    f_{n_j}(x)$ converges for each Borel subset of $A$.
Attempt
At first I started thinking of using a diagonalization argument and to approach this problem step by step. Instead, I am wondering what might be wrong with the following naive approach.
$f_n$ is uniformly bounded on $[0,1]$ so $\int_{[0,1]}f_n  \ \mathrm{d}x$ is an infinite sequence of real numbers on the compact set $[-2k,2k]$, where $|f_n|\leq k$. So there is a convergent subsequence $\int_{[0,1]}f_{n_k}\ \mathrm{d}x$. This subsequence also converges for any borel subset of $[0,1]$ so we have the result for all three of the above problems.
Question
What major concept(s) am I missing here? Note that I am not asking for a full solution to the problem but rather some feedback on my attempt at solving it.
I'm sorry for this silly question but I find it hard to dig into a problem until I realize why my "initial naive attempt" fails.
 A: I decided to use the comments to come up with the following solutions (please let me know how I did).


*

*If $A\subset [0,1]$ is Borel, then $\bigg|\int_A f_n \ dx\bigg|$ is a bounded sequence of real numbers on $A$ by $2K$, where $|f_n(x)|\leq K$. Therefore there is a convergent subsequence $\int_A f_{n_j} \ dx$ as desired.

*Let $(A_i)_{i=1}^{\infty}$ be a countable collection of Borel sets in $[0,1]$. By $(1)$ there is a convergent subsequence $(f_{1,k})_{k=1}^\infty$ so that $\int_{A_1} f_{1,k}$ converges as $k$ goes to $\infty$. Now $(f_{1,k})$ is a uniformly bounded sequence so again by $(1)$ there is a subsequence $(f_{2,k})_{k=1}^\infty$  such that $\int_{A_2} f_{2,k}$ converges. In this manner, each $\int_{A_n} f_{n,k}$ converges and $(f_{n+1,k})_{k=1}^\infty$ is a subsequence of $(f_{n,k})_{k=1}^\infty$. Thus we have the following diagram:
\begin{align*}
& \int_{A_1} f_{1,1} \ , \ \ \int_{A_1} f_{1,2}  \ , \ \ \int_{A_1} f_{1,3} \ \ \ ... \\
& \int_{A_2} f_{2,1} \ , \ \ \int_{A_2} f_{2,2}  \ , \ \ \int_{A_2} f_{2,3} \ \ \ ... \\
& \int_{A_3} f_{3,1} \ , \ \ \int_{A_3} f_{3,2}  \ , \ \ \int_{A_3} f_{3,3} \ \ \ ... \\
 \vdots
\end{align*}
In this diagram, each row is a subsequence of the row before it and the elements in the $k^{th}$ row appear in the same order as they would in the $k-1$ row, unless if they are deleted when shifting down a row. Note that by construction, the sequence of diagonal elements, $\left(\int_{A_n} f_{n,n}\right)_{n=1}^{\infty}$ is a subsequence of every $\left(\int_{A_n} f_{n,k}\right)_{k=1}^{\infty}$ (except possibly for the first $n-1$ terms). Thus the diagonal sequence converges on each $A_i$.

*The set of all Dyadic intervals is countable so the desired result follows from $(2)$ and the facts that every Borel measurable subset of $[0,1]$ can be arbitrarily approximated by Dyadic intervals and that the countable union of finite sets is countable. (I think this needs more work...)

