Beppo Levi's Theorem

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 is in the ordering on $\overline \R$.

Proof
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:
 * $\ds \int \sup_{n \mathop \in \N} f_n \rd \mu \ge \sup_{m \mathop \in \N} \int f_m \rd \mu$

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:

Since $\epsilon$ was selected arbitrarily, we have that the desired inequality holds:
 * $\ds \sup_{m \mathop \in \N} \int f_m \rd \mu \ge \int s \rd \mu$

Also known as
Some authors refer to this result as Beppo Levi's lemma, while others call it the monotone convergence theorem.

On the latter name is reserved for the general result: Monotone Convergence Theorem (Measure Theory).