Is this “limit” of a sequence of $L^2$ functions in $L^2$? Suppose we have a sequence $\{f_n\}$ in $L^2([0,1])$ and a Lebesgue measurable $f$ such that $$\int_E f_ndx\rightarrow\int_E fdx$$ as $n\rightarrow\infty$ for every Lebesgue measurable subset $E\subseteq[0,1]$. If $\sup_n\|f_n\|_2<\infty$, then do we necessarily have $f\in L^2([0,1])$?
I’m not seeing how to show this, one way or the other. I tried constructing a counterexample along the lines of $f_n(x)=x^{\frac{1}{n}-\frac{1}{2}}$, since then $f(x)=x^{-1/2}$ isn’t square-integrable, but this doesn’t satisfy either condition. I’m inclined to think that we do have $f\in L^2([0,1])$, since it doesn’t seem like you can make $\int_{[0,1]}|f|^2dx$ blow up without either of $\int_{[0,1]}|f_n|^2dx$ or $\int f dx$ blowing up, but I don’t see a way to formalize or justify this instinct.
 A: Disclaimer: I prefer the "functional analysis" approach, as it is more natural to me; but here is a pure measure-theory argument. All it uses is the definition of Lebesgue integral and the Cauchy-Schwarz inequality.
The key observation is that, by linearity, the property can be written as
$$
\int_F f\,s=\lim_n\int_F f_n\,s
$$
for any measurable set $F$ and any simple function $s$.
I will assume $f$ real; if it is complex, one can work with real and imaginary parts, so no loss of generality.
Suppose that $f\not\in L^2[0,1]$. That is, $$\int_{[0,1]}f^2=\infty.$$
By working on the set $F=\{f\geq0\}$, we may assume $\int_F f^2=\infty$ (otherwise, we work with $-f$). By definition of Lebesgue integral, this means that there exists a sequence of simple functions $\{s_m\}$ with $0\leq s_m\leq f^2$ and $\lim_m\int_F s_m=\infty$. By choosing a subsequence if necessary, we may assume that the numbers $\int_Fs_m$ increase monotonically.
Choose positive integers $k(m)$ with $k(m)\leq \int_Fs_m< k(m)+1$. Then $k(m)\nearrow\infty$. We have, with $C=\sup_n\|f_n\|^2$,
\begin{align}
 k(m)&\leq \int_F s_m^{1/2}\,s_m^{1/2}\leq\int_F f\,s_m^{1/2}=\lim_n\int_Ff_n\,s_m^{1/2}\\[0.3cm]
&\leq \sup_n\|f_n\|_2\,\bigg(\int_Fs_m\bigg)^{1/2}\\[0.3cm]
&\leq C\,(k(m)+1)^{1/2}\leq 2C\,k(m)^{1/2}.
\end{align}
This implies that $k(m)$ is bounded, a contradiction.
