Convergence of integral on every measurable subset Let $\{f_n\}$ be sequence in $L^2[0,1]$ and $f$ be a Lebesgue measurable function such that for every Lebesgue measurable set $E$ in $[0,1]$, $\lim_{n \to \infty} \int_{E}f_n dx = \int_{E}fdx$. Assume also that $\sup_{n}\int_{0}^{1} \lvert f_n \rvert^2 dx < \infty$. I want to show that $f \in L^2[0,1]$.
My thought so far: If we manage to show there is a subsequence of $f_n$ converging pointwise a.e. on $[0,1]$ to $f$, we're done. Indeed, if this is true, then by Fatou's lemma, we have $$\int_{0}^{1} \lvert f \rvert^2 dx \leqslant \liminf \int_{0}^{1} \lvert f_{n_k}\rvert^2 dx \leqslant \sup_{n}\int_{0}^{1} \lvert f_n \rvert^2 dx < \infty$$ But I don't know how to achieve that. I tried to show $f_n \to f$ in $L^1$ or in measure but I failed. Any insight would be appreciated.
 A: From the assumption that $\lim_{n \to \infty} \int_{E}f_n dx =\int_{E}fdx$ for all measurable sets $E\subset[0,1]$, we have
$$\lim_{n \to \infty} \int_0^1f_n \varphi \,dx = \int_0^1f\varphi \,dx,$$
for all simple functions $\varphi$ on $[0,1]$. Since $C^2:=\sup_{n}\int_{0}^{1} \lvert f_n \rvert^2 dx < \infty$, by Cauchy-Schwartz we have
$$\left|\int_0^1f_n\varphi \,dx\right|\leq C\|\varphi\|_{L^2}\Longrightarrow\left|\int_0^1f\varphi \,dx\right|\leq C\|\varphi\|_{L^2},\ \ \ \text{for all simple functions }\varphi.\tag{$*$}$$
Now it follows from duality that $f\in L^2$.
Here in the last step we used Theorem 6.14 in Folland's Real Analysis. Here is a (simplified) proof from Folland that $(*)$ implies $f\in L^2$: Let $\phi_n$ be a sequence of simple functions such that $\phi_n\to f$ a.e. and $|\phi_n|\leq|f|$, and let
$$\varphi_n=\frac{|\phi_n|\mathrm{sgn}f}{\|\phi_n\|_{L^2}}.$$
Then $\varphi_n$ is a sequence of simple functions with $\|\varphi_n\|_{L^2}=1$. By Fatou's lemma and all these definitions we have
\begin{align*}
\|f\|_{L^2}&\leq\liminf_{n\to\infty}\|\phi_n\|_{L^2}=\liminf_{n\to\infty}\int_0^1 |\varphi_n\phi_n|\,dx\\
&\leq\liminf_{n\to\infty}\int_0^1 |\varphi_nf|\,dx=\liminf_{n\to\infty}\int_0^1 \varphi_nf\,dx\\
&\stackrel{(*)}{\leq} C\|\varphi_n\|_{L^2}=C<\infty.
\end{align*}
