I need to prove this version of MCT:
Let $1\le p < +\infty$ and $(f_n) \subset L^p(\Omega)$ a non-decreasing sequence of non-negative functions. Suppose that: $$ \sup_{n\in \mathbb{N}}\int_{\Omega} f_n(x)^p dx < \infty$$
Then, there exists $f\in L^p(\Omega)$ such that $f_n(x) \to f(x)$ a.e. in $\Omega$
In Brezis's book, this version is enunciate but it does not display a proof or indicate a reference to refer to.
I know that, for monotonicity, for each $x\in \Omega$ $$ f(x) = \lim f_n(x) $$ is well defined. But, I need to guarantee that $f$ belongs to $L^p$.
You can use the classical monotone convergence theorem: since $0\leq f_n(x)\leq f_{n+1}(x)$ a.e. we have $0\leq f_n(x)^p\leq f_{n+1}^p(x)$ a.e. Thus $\{f_n^p\}$ is a non-decreasing sequence of positive functions. We also have that $f_n(x)^p\to f(x)^p$ a.e. By the monotone convergence theorem, we have $$\int_Xf(x)^pd\mu(x)=\lim_{n\to\infty}\int_Xf_n(x)^pd\mu(x)$$ Since $0\leq f_n^p\leq f_{n+1}^p$ for all $n$, we also have that the sequence $\{\int_Xf_n^pd\mu\}_{n=1}^\infty$ is an increasing sequence of positive numbers and thus converges to its supremum, so $\lim_{n\to\infty}\int_Xf_n(x)^pd\mu(x)=\sup_{n\in\mathbb{N}}\int_Xf_n(x)^pd\mu(x)$ which is finite, and thus $\int_Xf(x)^pd\mu(x)<\infty$.