Prove that $ \lim_{k\rightarrow\infty}\int_E |f_k(x)-f(x)|\mathrm{d}x=0. $ 
Let $ f_k(x) (k\geq 1)$ be non-negetive integrable functions on a measurable set $E$, $f\in L(E)$. If $f_k$ converges to $f(x)$ in measure in $E$, and
$$
\lim_{k\rightarrow\infty}\int_E f_k(x)\mathrm{d}x=\int_E f(x)\mathrm{d}x,
$$
prove that
$$
\lim_{k\rightarrow\infty}\int_E |f_k(x)-f(x)|\mathrm{d}x=0.
$$

By the convergence in measure condition, $\forall\epsilon>0, \forall\delta>0, \exists N, \forall k>N$, we have
$$
m(E[|f_k-f|>\epsilon])<\delta.
$$
Thus, I write
$
\int_E |f_k(x)-f(x)|\mathrm{d}x
$
as
$$
\int_E |f_k(x)-f(x)|\mathrm{d}x=\int_{E[|f_k-f|>\epsilon]} |f_k(x)-f(x)|\mathrm{d}x+\int_{E[|f_k-f|\leq\epsilon]} |f_k(x)-f(x)|\mathrm{d}x,
$$
but I stuck in the estimation of the right two terms, especialy the first term.
Appreciate any help or hint!
 A: Let me summarize everything and elaborate.
Proposition 1: Let $(X,\mathcal{F},\mu)$ be a measure space. Let
$(f_{n})$ be a sequence of integrable functions and $f$ an integrable
function. If $f_{n}\rightarrow f$ a.e., then $||f_{n}||_{1}\rightarrow||f||$
$\Longleftrightarrow$ $||f_{n}-f||_{1}\rightarrow0$.
Proof of Proposition 1: $\Leftarrow$ is easy because $\left||f_{n}|-|f|\right|\leq|f_{n}-f|$.
It follows that $\int\left||f_{n}|-|f|\right|d\mu\leq\int|f_{n}-f|d\mu\rightarrow0$
and hence $\int|f_{n}|d\mu\rightarrow\int|f|d\mu$.
$\Rightarrow:$ We go to show that $|f_{n}|-|f|-|f_{n}-f|$ is dominated
by an integrable function. Note that
\begin{eqnarray*}
|f_{n}| & = & |f_{n}-f+f|\\
 & \leq & |f_{n}-f|+|f|,
\end{eqnarray*}
so $|f_{n}|-|f|-|f_{n}-f|\leq0$. On the other hand, $|f_{n}-f|\leq|f_{n}|+|f|$,
so $|f_{n}|-|f|-|f_{n}-f|\geq-2|f|$. Hence, $\left||f_{n}|-|f|-|f_{n}-f|\right|\leq2|f|$.
Observe that $|f_{n}|-|f|-|f_{n}-f|\rightarrow0$ a.e.. By Lebesgue
Dominated Convergence Theorem, it follows that $\int\left\{ |f_{n}|-|f|-|f_{n}-f|\right\} d\mu\rightarrow0$.
Rearranging terms yields $||f_{n}-f||_{1}\rightarrow0$.

Proposition 2 (due to Vitali): Let $(X,\mathcal{F},\mu)$ be a measure
space. Let $(f_{n})$ be a sequence of measurable functions and $f$
a measurable function. If $f_{n}\rightarrow f$ in measure, in the
sense that for each $a>0$, $\mu\left(x\mid|f_{n}(x)-f(x)|\geq a\right)\rightarrow0$,
then there exists a subsequence $(f_{n_{k}})$ such that $f_{n_{k}}\rightarrow f$
a.e.
Proof of Proposition 2: We choose sequence of integers $1\leq n_{1}<n_{2}<\ldots$
by recursion. Since $\mu\left(x\mid|f_{n}(x)-f(x)|\geq1\right)\rightarrow0$
as $n\rightarrow\infty$, we can choose $n_{1}$ such that $\mu\left(x\mid|f_{n_{1}}(x)-f(x)|\geq1\right)\leq\frac{1}{2^{1}}$.
Suppose that $1\leq n_{1}<n_{2}<\ldots<n_{k}$ have been chosen. Note
that $\mu\left(x\mid|f_{n}(x)-f(x)|\geq\frac{1}{k+1}\right)\rightarrow0$
as $n\rightarrow\infty$, so $\mu\left(x\mid|f_{n}(x)-f(x)|\geq\frac{1}{k+1}\right)\leq\frac{1}{2^{k+1}}$ when $n$ is sufficiently large. Therefore we can choose $n_{k+1}>n_{k}$
such that $\mu\left(x\mid|f_{n_{k+1}}(x)-f(x)|\geq\frac{1}{k+1}\right)\leq\frac{1}{2^{k+1}}$.
We go to prove that $f_{n_{k}}\rightarrow f$ a.e.
Let $A=\{x\mid f_{n_{k}}(x)\not\rightarrow f(x)\}$. Note that $A$
can be described as
$$
A=\bigcup_{r\in\mathbb{Q}\cap(0,\infty)}\bigcap_{K=1}^{\infty}\bigcup_{k=K}^{\infty}\{x\mid|f_{n_{k}}(x)-f(x)|\geq r\}.
$$
We go to show that for each $r\in\mathbb{Q}\cap(0,\infty)$, $\mu\left(\bigcap_{K=1}^{\infty}\bigcup_{k=K}^{\infty}\{x\mid|f_{n_{k}}(x)-f(x)|\geq r\}\right)=0$.
Fix $r\in\mathbb{Q}\cap(0,\infty)$. Denote $B_{K}=\cup_{k\geq K}\{x\mid|f_{n_{k}}(x)-f(x)|\geq r\}$.
Choose $K_{0}\in\mathbb{N}$ such that $\frac{1}{K_{0}}<r$. For any
$k\geq K_{0}$, we have that $\{x\mid|f_{n_{k}}(x)-f(x)|\geq\frac{1}{k}\}\supseteq\{x\mid|f_{n_{k}}(x)-f(x)|\geq r\}$,
so $\mu\{x\mid|f_{n_{k}}(x)-f(x)|\geq r\}\leq\mu\{x\mid|f_{n_{k}}(x)-f(x)|\geq\frac{1}{k}\}\leq\frac{1}{2^{k}}$.
Therefore, for any $K\geq K_{0}$, we have that
\begin{eqnarray*}
 &  & \mu(B_{K})\\
 & \leq & \sum_{k=K}^{\infty}\mu\{x\mid|f_{n_{k}}(x)-f(x)|\geq r\}\\
 & \leq & \sum_{k=K}^{\infty}\frac{1}{2^{k}}\\
 & = & \frac{1}{2^{K-1}}.
\end{eqnarray*}
Note that $B_{1}\supseteq B_{2}\supseteq\ldots$ and $\mu(B_{K_{0}})<\infty$.
By continuity of measure, we have that $\mu\left(\cap_{K}B_{K}\right)=\lim_{K\rightarrow\infty}\mu(B_{K})=0$.
Hence $\mu(A)=0$. In another word, $f_{n_{k}}\rightarrow f$ a.e.

We go back to your question. Let $(f_{n})$ be a sequence of non-negative integrable
functions and $f$ an integrable function. Suppose that $f_{n}\rightarrow f$
in measure and $\int f_{n}\rightarrow\int f$. We prove that $\int|f_{n}-f|\rightarrow0$
by contradiction. Suppose not, then there exists $a>0$ and a subsequence
$(f_{n_{k}})$ such that $\int|f_{n_{k}}-f|\geq a$ for all $k$.
Note that we still have $f_{n_{k}}\rightarrow f$ in measure. By Vitali
Theorem (Prop. 2), we can further choose a subsequence $(f_{n_{k_{l}}})$
of $(f_{n_{k}})$ such that $f_{n_{k_{l}}}\rightarrow f$ a.e. as
$l\rightarrow\infty$. In particular, $f$ is non-negative (a.e.). Since $f_{n}$ and $f$ are non-negative, $\int f_{n}\rightarrow\int f$
implies that $\int|f_{n}|\rightarrow\int|f|$. In particular, $||f_{n_{k_{l}}}||_{1}\rightarrow||f||_{1}$.
By Prop. 1, we have $||f_{n_{k_{l}}}-f||_{1}\rightarrow0$. However,
this is impossible because $||f_{n_{k_{l}}}-f||_{1}\geq a>0$ for
each $l$.
A: Using a generalization of the dominated convergence theorem, it can be shown that if $f_n, f \in L^1$ and $f_n \to f$ a.e., then $\lVert f_n - f \rVert_{L^1} \to 0$ if and only if $\lVert f_n \rVert_{L^1} \to \lVert f \rVert_{L^1}$.
In your case, you have convergence in measure, not convergence a.e. But convergence in measure implies some subsequence converges a.e. You can combine this with the fact that a sequence $f_n$ in a topological space converges to $f$ if and only if every subsequence of $f_n$ has a further subsequence that converges to $f$ to prove the claim.
A: This answer is only valid in the case where $m(E) < \infty$.
We have
\begin{align*}
\int_{E}|f_k-f|\,dx
&= \int_E (f_k-f)\,dx - 2\int_{f_k-f<0}(f_k-f)\,dx\\
&= \int_E (f_k-f)\,dx + 2\int_{0<f-f_k<\epsilon}(f-f_k)\,dx + 2\int_{f-f_k>\epsilon}(f-f_k)\,dx\\
&\le \int_E (f_k-f)\,dx + 2m(E)\epsilon + 2\int_{f-f_k>\epsilon}f\,dx\\
&\le \int_E (f_k-f)\,dx + 2m(E)\epsilon + 2\int_{|f-f_k|>\epsilon}f\,dx.
\end{align*}
The claim now follows from the following lemma and its corollary.

Lemma. If $f\in L(E)$ and $A_n\subset E$ such that $m(A_n)\to 0$, then $\int_{A_n}f\,dx\to 0$.

Proof. We have $\int_{A_n}f\,dx = \int_E 1_{A_n}\cdot f\,dx$. Suppose the claim is false. Then there exists a subsequence $(A_n')$ of $(A_n)$ such that $\int_E 1_{A_n'}\cdot f\,dx\ge a > 0$ for all $n$. But we know that $1_{A_n'}\to 0$ in $L^1$. Hence, there exists a subsequence $(A_n'')$ of $(A_n')$ such that $1_{A_n''}\to 0$ $m$-almost everywhere. Hence, $1_{A_n''}\cdot f\to 0$ $m$-a.e., which implies $\int_E 1_{A_n''}\cdot f\,dx\to 0$ by the dominated convergence theorem. But this contradicts $\int_E 1_{A_n'}\cdot f\,dx\ge a > 0$.

Corollary. If $f\in L(E)$, then for every $\epsilon>0$ there exists $\delta>0$ such that $m(A)<\delta$ implies $\int_{A}f\,dx<\epsilon$.

Proof. Follows from the lemma by contradiction.
