I have this question:
Let be $ f_n $ a sequence of integrable functions on $E$ and $f$ measurable such that $ f_n \overset{m}{\to} f $ on E. If there is an integrable function $g$ on $E$ such that
$$|f_n|\leq g$$ on E, then $f$ integrable and $$\tag{1} \lim_{x \to \infty} \int_{E} |f_n(x)-f(x)|dx=0.$$
I could prove that $f$ is integrable, using the existence of a subsequence of $f_n$ that converges to $f$ in AEP of $E$. But I could not prove (1). Should I use the dominated convergence theorem?
$f_n\xrightarrow{\text{measure}}f$ implies there is a subsequence $\{f_{n_k}\}$ such that $f_{n_k}\xrightarrow{\text{ a.e.}}f$. Now, $\big|f_{n_k}\big|\leq g$ for all $k$ implies $|f|\leq g$ a.e. By Dominated Convergence Theorem $f$ is integrable.
Now, if possible let $\lim\int|f_n-f|\not=0$. So, we have a subsequence $\{f_{n_k}\}$ and $\epsilon>0$ with $\int|f_{n_k}-f|\geq \epsilon$ for all $k$. But, $f_{n_k}\xrightarrow{\text{measure}}f$ implies we again have a subsequence $\big\{f_{n_{k_l}}\big\}$ such that $f_{n_{k_l}}\xrightarrow{\text{a.e.}}f$. Now, $\big|f_{n_{k_l}}-f\big|\leq 2|g|$ a.e. for all $l$. Now, dominated convergence theorem implies $\lim_{l\to \infty}\int\big|f_{n_{k_l}}-f\big|=0$ as $\lim_{l\to \infty}\big|f_{n_{k_l}}-f\big|=0$ a.e., a contradiction.