Beppo Levi's Lemma

Another application of the monotone convergence theorem: $\textit{Beppo Levi's Lemma}$

Given a sequence of measurable functions $\{f_n\}_{n=1}^\infty$ where $f_n: E \subseteq X \rightarrow [0, \infty] $, $f_n \leq f_{n+1}$ and $\left\lbrace \int_E f_n \right\rbrace_{n=1}^\infty$ is a bounded sequence of (non-negative) real numbers, then

$f_n \longrightarrow f$ a.e., $f$ is finite a.e. and $\int_E f_n \longrightarrow \int_E f < \infty $

$\textit{Proof} $ : By the monotinicity of integration, $f_n \leq f_{n+1} \Longrightarrow \int_E f_n \leq \int_E f_{n+1} $ and so $\left\lbrace \int_E f_n \right\rbrace_{n=1}^\infty$ is an increasing sequence in $[0,\infty)$ with a limit in $[0,\infty)$. Since $\{f_n\}$ is increasing, then for each fixed $x \in{E}$, $$f(x) = \lim_{n \rightarrow \infty}f_n(x)$$ exists in $[0,\infty]$. Also, $f$ is measurable since it is the pointwise limit of a sequence of measurable functions. Now, the monotone convergence theorem applies and gives $$\lim_{n \rightarrow \infty} \int_E f_n = \int_E f $$ Since $\left\lbrace \int_E f_n \right\rbrace_{n=1}^\infty$ is bounded, so is its limit, $ \int_E f$. Thus, $f$ must be finite a.e. To actually show $f$ is finite a.e. we can use Chebychev's inequality:
Define $E_n = \{x: f(x) \geq n\} = \{x: \frac{f(x)}{n} \geq 1 \}$, $n\in{\mathbb{N}}$. Then, by the linearity and monotinicity of integration, $$m(E_n) = \int_{E_n}1 \leq \int_{E_n} \frac{f}{n} = \frac{1}{n} \int_{E_n} f \leq \frac{1}{n} \int_E f < \infty$$ Thus, $m(E_n) \overset{n \rightarrow \infty}{\longrightarrow} 0$ so in particular, $m(\{x: f(x) = \infty \}) = 0$ i.e. $f$ is finite a.e. $\blacksquare$