Prove Lebesgue integrability

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.

• Why does the result follow from $f_{n_k}$ being arbitrary?
– GuPe
Commented Dec 25, 2018 at 19:29
• @Gaucho If every subsequence of a sequence of real numbers converges to some number $A$, then the sequence converges to $A$. In this case, the sequence under consideration is $\int_X f_n\, d\mu$.
– kobe
Commented Dec 25, 2018 at 19:30
• But it is only proved that a subsequence of every subsequence converges to $\int f$.
– GuPe
Commented Dec 25, 2018 at 19:45
• @GuachoPerez no, it has been shown that every subsequence of $\int_X f_n\, d\mu$ has a further subsequence which converges to $\int_X f\, d\mu$, so $\int_X f_n\, d\mu \to \int_X f\, d\mu$. Indeed, since $\int_X f_{n_j}\, d\mu$ has a subsequence which converges to $\int_X f\, d\mu$, then $\int_X f_{n_j}\, d\mu$ to converges to the same number. But since $\int_X f_{n_j}\, d\mu$ is an arbitrary subsequence of $\int_X f_n\, d\mu$, $\lim_n \int_X f_n\, d\mu = \int_X f\, d\mu$.
– kobe
Commented Dec 25, 2018 at 19:53
• @Gaucho take a look here: math.stackexchange.com/questions/397978/….
– kobe
Commented Dec 25, 2018 at 19:55