Theorem: Beppo Levi's Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\sequence {f_n}_{n \mathop \in \N} \in \MM_{\overline \R}^+$ be an increasing sequence of positive $\Sigma$-measurable functions.
Let $\ds \sup_{n \mathop \in \N} f_n: X \to \overline \R$ be the pointwise supremum of $\sequence {f_n}_{n \mathop \in \N}$, where $\overline \R$ denotes the extended real numbers.
Then:
:$\ds \int \sup_{n \mathop \in \N} f_n \rd \mu = \sup_{n \mathop \in \N} \int f_n \rd \mu$
where the supremum on the {{RHS}} is in the ordering on $\overline \R$.

Proof:
{{tidy|Break down some of the long complex sentences into simple ones}}
{{MissingLinks}}
Since by definition $\ds \sup _{n \mathop \in \N} f_n \ge f_m$ for all $m$, we have:
:$\ds \int \sup_{n \mathop \in \N} f_n \rd \mu \ge \int f_m \rd \mu$
and hence the inequality holds for the supremum as well:
:$\ds \int \sup_{n \mathop \in \N} f_n \rd \mu \ge \sup_{m \mathop \in \N} \int f_m \rd \mu$
{{qed|lemma}}
It remains to show the reverse inequality.
We have that the integral of $\sup \limits_{n \mathop \in \N}f_n$ is defined as the supremum of the integrals of positive simple functions:
:$s \le \sup \limits_{n \mathop \in \N} f_n$
Let $s$ be a simple function:
:$\ds s = \sum_{i \mathop = 1}^k \lambda_i \chi_{E_i} \le \sup_{n \mathop \in \N} f_n$
where $\lambda_i \in \closedint 0 {+\infty}$ and $E_i \in \Sigma$.
We are to show that:
:$\ds \sup_{m \mathop \in \N} \int f_m \rd \mu \ge \int s \rd \mu$
To show this, we use the fact that $\nu_s: \Sigma \to \closedint 0 {+\infty}$ defined by $\map {\nu_s} E = \ds \int \chi_Es \rd \mu$ clearly defines a measure over $X$, because it is simply a linear combination (with positive coefficients) of the measures $\bigvalueat \mu {E_i}$.
Now, if we fix $1 > \epsilon > 0$, we have that the sets:
:$A_m = \set {x \in X: \map {f_m} x \ge \paren {1 - \epsilon} \map s x}$
form a cover of $X$ ($X = \ds \bigcup_{m \mathop \in \N} A_m$) by definition of the supremum, because $\ds s \le \sup_{n \mathop \in \N} f_n$.
Furthermore, the increasing sequence $\sequence {A_m}$ has the limit $A_m \uparrow X$ ($m \to \infty$).
By definition of the $A_m$ we have that:
:$\ds \int f_m \rd \mu \ge \int \chi_{A_m} f_m \rd \mu \ge \paren {1 - \epsilon} \int \chi_{A_m} s \rd \mu = \paren {1 - \epsilon} \map {\nu_s} {A_m}$
where the first inequality follows from the fact that the $f_m$ are positive.
We have that the sequence $\sequence {f_n}_{n \mathop \in \N}$ increases monotonically.
By Measure of Limit of Increasing Sequence of Measurable Sets given in any measure space, we have that taking the supremum of both sides yields:
{{begin-eqn}}
{{eqn | l = \sup_{m \mathop \in \N} \int f_m \rd \mu
      | r = \lim_{m \mathop \to +\infty} \int f_m \rd \mu
      | c = 
}}
{{eqn | o = \ge
      | r = \lim_{m \mathop \to +\infty} \paren {1 - \epsilon} \map {\nu_s} {A_m}
      | c = 
}}
{{eqn | r = \paren {1 - \epsilon} \map {\nu_s} X
      | c = 
}}
{{eqn | r = \paren {1 - \epsilon} \int s \rd \mu
      | c = 
}}
{{end-eqn}}
Since $\epsilon$ was selected arbitrarily, we have that the desired inequality holds:
:$\ds \sup_{m \mathop \in \N} \int f_m \rd \mu \ge \int s \rd \mu$
{{qed}}