Suppose $\{f_k(x)\}$ is a sequence of nonnegative integrable functions on $[0,1]$, and $f_k(x)$ converges to $f(x)$ in measure on $[0,1]$. Suppose, in addition, that $$ \lim_{k\to\infty}\int_0^1 f_k(x)dx=\int_0^1 f(x)dx .$$ Prove that for any measurable subset $E$ of $[0,1]$, $$ \lim_{k\to\infty}\int_E f_k(x)dx=\int_E f(x)dx .$$
My attempt:
Since $f_k(x)$ converges to $f(x)$ in measure, we can select a subsequence $\{f_{k_i}\}$ such that $f_{k_i}$ converges to $f(x)$ almost everywhere. Now by Fatou's lemma, we have \begin{align} \int_0^1 f(x)dx&=\int_0^1\lim_{k_i\to\infty}f_{k_i}(x)dx\\ &\le\varliminf_{k_i\to\infty}\int_E f_{k_i}(x)dx+\varliminf_{k_i\to\infty}\int_{[0,1]\setminus E} f_{k_i}(x)dx\\ &\le\varliminf_{k_i\to\infty}\int_0^1 f_{k_i}(x)dx\\ &=\int_0^1 f(x)dx \end{align} by the condition that $ \lim_{k\to\infty}\int_0^1 f_k(x)dx=\int_0^1 f(x)dx $. Hence, it follows that $$ \lim_{k_i\to\infty}\int_E f_{k_i}(x)dx=\int_E f(x)dx .$$ But now I get stuck in proving the limit is also correct for $f_k$, not just a subsequence of it. Any idea is appreciated. Thanks.
Edit:
I think the question missed a hypothesis that $f\in L^1([0,1])$ which plays a vital role in the proof. Otherwise, we can construct a nice counterexample like the one discussed in the comment provided by Sangchul Lee.
Assume that for some measurable $E \subseteq[0,1]$ we have
$$\int_E f_k \not\to \int_E f$$
Then there is $\varepsilon > 0$ and a subsequence $(f_{p(k)})_k$ such that $$\left|\int_E f - \int_E f_{p(k)}\right| \ge \varepsilon, \forall k \in \mathbb{N}$$
$f_{p(k)} \to f$ in measure so there exists a further subsequence $(f_{p(q(k))})_k$ which converges to $f$ a.e.
However, in the comments you claimed that you proved the statement when you have convergence a.e. so it follows
$$\int_E f_{p(q(k))} \to \int_E f$$
which is a contradiction with the fact that the difference is bounded from below by $\varepsilon$.