Let $f_n \in L_0(X), |f_n|\leqslant\phi \in L_1(X), n \in N$ and $f_n \rightarrow f$ in measure. Prove that $f \in L_1(X)$ and $$\lim_{n\to\infty}\int_{X}{f_n}d\mu = \int_{X} fd\mu$$
Here $L_0(X)$ stands for Lebesgue measurability, and $L_1(X)$ — for Lebesgue integrability on $X$.
As far as I understand, we should use the fact that $f_n$ is bounded by a Lebesgue integrable function being itself measurable, so $f$ is also Lebesgue integrable. Is that correct? I remember similar reasoning being used in the classroom. If the first part is correct, how do I prove the equality?
Let $\{f_{n_k}\}_{k=1}^\infty$ be an arbitrary subsequence of $\{f_n\}_{n=1}^\infty$. Then $f_{n_k} \to f$ in measure, so one may extract a further subsequence $\{f_{n_{k_j}}\}_{j=1}^\infty$ of $\{f_{n_k}\}_{k=1}^\infty$ such that $f_{n_{k_j}}\to f$ pointwise a.e. in $X$. By Fatou's lemma $f\in L_1(X)$; by the dominated convergence theorem $\int_X f\, d\mu = \lim\limits_{j\to \infty} \int_X f_{n_{k_j}}\, d\mu$. Since $\{f_{n_k}\}_{k=1}^\infty$ was arbitrary, the result follows.