I want to prove the monotone convergence theorem using Fatou's lemma (and its reverse) as exercise, and I need a check; I will use also the following properties of limit inferior and limit superior:
Let $f,g: D \to \mathbb{R}$ be functions. Then if $\lim_{x \to c} f(x)$ exists in $\tilde{\mathbb{R}}=\mathbb{R} \cup \{-\infty, + \infty \}$ we have $$\liminf_{x \to c} (f(x) +g(x))=\lim_{x \to c} f(x) + \liminf_{x \to c} g(x)$$ and the same for limit superior.
Statement. Let $(X, \mathcal{M}, \mu)$ be a measure space. Assume that $f_0 \le f_1 \le f_2 \le \dots$ is an increasing sequence of functions in $L^{+}(X)$ ($=$ the set of all extended real valued positive measurable functions), such that $f_n \uparrow f$ pointwise. Then $$ \int_X f = \lim_{n \to \infty} \int_X f_n $$
Proof. $g_n=(f-f_n)$ is a sequence of positive measurable functions, and then I can apply Fatou's lemma; it is $$\int_X \liminf_n (f-f_n) d\mu \le \liminf_n \int_X (f-f_n) d \mu$$Using the property above we have that $$\int_X \liminf_n (f-f_n) d\mu =\int_X (f - \liminf_n f_n) d \mu=0$$ and $$\liminf_n \int_X (f-f_n) d \mu=\liminf_n \left[ \int_X f d\mu - \int_X f_n d\mu \right]=\int_X f d \mu - \liminf_n \int_X f_n d \mu$$ So it is $$\int_X f d \mu \ge \liminf_n \int_X f_ d \mu \ge \int_X \liminf_n f_n d \mu=\int_X f d \mu$$using the lemma on $f_n$, which implies $$\int_X f d \mu = \liminf_n \int_X f_n d \mu$$Now: if $$\begin{split} \int_X f d \mu=\infty & \ \longrightarrow \ \underbrace{\liminf_n \int_X f_n d \mu}_{=\infty} \le \limsup_n \int_X f_n d \mu = \infty \\ & \ \longrightarrow \ \int_X f d \mu = \lim_n \int_X f_n d \mu = \infty \end{split}$$ and if $$\int_X f d \mu < \infty$$ We can apply, in a similar way as above, the reverse of Fatou's lemma considering that $f-f_n \le f - f_0$ and that $f-f_0$ is integrable ( - here I've used the fundamental hypothesis of $(f_n)_{n \in \mathbb{N}}$ increasing). We obtain $$ \int_X f d \mu=\limsup_n \int_X f_n d \mu $$and then the thesis.
What do you think about it?
Thank you very much.