In a lecture notes about Variational Methdos, I found the following theorem:
THEOREM: Let $(f_n)$ a sequence in $L^{p}(\Omega)$ and $f \in L^{p}(\Omega)$, such that $f_{n} \rightarrow f$ in $L^{p}(\Omega)$. Then, there are a subsequence $(f_{n_{j}}) \subset (f_{n})$ and $g \in L^{p}(\Omega)$ such that:
'
$$|f_{n_{j}}(x)|\leq g(x) \textrm{ a.e } \in \Omega$$
and
$$ f_{n_{j}}(x) \rightarrow f(x) \textrm{ a.e } \in \Omega$$
The author say this is a well know theorem, called Vainberg Theorem.
For me, is well know that convergence in $L^{p}$ implies convergence in measure, and convergence in measure implies convergence a.e (for a subsequence). But I dont know about boundness a.e of the subsequence. Moreover, the name of this theorem is a bit strange, I search for "Vainberg Theorem" in the web and in some books of Real Analisys, but I not found anything.
If someone can show me a good reference about thi or outlines of the proof, I will be grateful.
PS 1: In the notes, $\Omega$ is a open subset of $R^{N}$.
PS 2: Throughout the text, this theorem is used to show that a functional satisfies the Palais-Smale condition.
Since $\|f - f_n \|_p \to 0$, we can extract a subsequence $f_{n_j}$ so that $\| f - f_{n_j} \|_p \le \frac{1}{2^j}$. Put $$g = \lvert f \rvert + \sum^\infty_{j=1} \lvert f - f_{n_j} \rvert.$$ Then $g \in L^p$ and by the triangle inequality we have $$\lvert f_{n_j} \rvert \le g \,\,\, (\text{almost everywhere}).$$ From this subsequence, you can extract a further subsequence which converges to $f$ a.e. as outlined here Convergence in $L^p$ and convergence almost everywhere.