Limit of integral exists imply a sequence of function converges in $L^1$ 
Let $\{f_n\}$ be a non-negative sequence in $L^1(X,\mathcal{M},\mu)$ with $\mu(X) < \infty$ satisfying, for each $E\in \mathcal{M}$,
$$
\lim_{n\to\infty}\int_E f_n d\mu
$$
exists, and $\{f_n\}$ converges to some $f$ almost everywhere.
Show that $f_n \to f$ in $L^1$.

I try to use Vitali Convergence theorem to solve the problem by first showing $\{f_n\}$ is uniformly integrable. Therefore, I truncated $f_n$ in the integral, but the limit is taking on $n$, so it fails. I also used Radon-Nikodym theorem by defining $d\hat{\mu} =\chi_E d\mu$, this did not help.
I tried weak convergence. So in this case,
$$
\lim_{n\to\infty}\int f_n \chi_E d\mu \to \int f \chi_E d\mu
$$
I am stuck here.
Any help would be appreciated.
 A: First observe that the choice $E=X$ gives boundedness of $\left(\lVert f_n\rVert_1\right)_{n\geqslant 1}$ and from Fatou's lemma, it follows that $f$ belongs to $\mathbb L^1$.
The only missed thing to conclude convergence in $\mathbb L^1$ is uniform integrability. Since $(f_n)$   is bounded in $L^1$, it suffices to prove that for each positive $\varepsilon$, there exists a positive $\delta$ such that for all $n\geqslant 1$ and each $E\in\mathcal M$ such that $\mu(E)<\delta$, $\int_Ef_nd\mu<\varepsilon$.
Suppose not: there exists $\varepsilon_0$ such that for each positive $\delta$, there exists $E\in\mathcal M$ for which $\mu(E)<\delta$ and $\int_Ef_nd\mu>\varepsilon_0$. Using the fact that for each integrable function $g$, $\lim_{\delta\to 0}\sup_{E\in\mathcal M,\mu(E)<\delta}\int_Ehd\mu=0$, we can construct an increasing sequence $(n_k)$ of positive integers and a sequence of sets $(E_k)$ such that $E_k\in\mathcal M$, $\mu(E_k)<2^{-k-1}$ and $\int_{E_k}f_{n_k}d\mu>\varepsilon_0$. Let $F_j=\bigcup_{k\geqslant j}E_k$. Then $\mu(F_j)<2^{-j}$ and for $k\geqslant j$,
$$
\varepsilon_0<\int_{E_k}f_{n_k}d\mu\leqslant \int_{F_j}f_{n_k}d\mu.
$$
Letting $k$ going to infinity (where $j$ is fixed) gives
$$
\varepsilon_0\leqslant \int_{F_j}fd\mu,
$$
which contradicts integrability of $f$.
