# 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 right hand side is in the ordering on $\overline \R$.

## Proof

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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:

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

$\Box$

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:

\(\ds \sup_{m \mathop \in \N} \int f_m \rd \mu\) | \(=\) | \(\ds \lim_{m \mathop \to +\infty} \int f_m \rd \mu\) | ||||||||||||

\(\ds \) | \(\ge\) | \(\ds \lim_{m \mathop \to +\infty} \paren {1 - \epsilon} \map {\nu_s} {A_m}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \paren {1 - \epsilon} \map {\nu_s} X\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \paren {1 - \epsilon} \int s \rd \mu\) |

Since $\epsilon$ was selected arbitrarily, we have that the desired inequality holds:

- $\ds \sup_{m \mathop \in \N} \int f_m \rd \mu \ge \int s \rd \mu$

$\blacksquare$

## Also known as

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

On $\mathsf{Pr} \infty \mathsf{fWiki}$ the latter name is reserved for the general result: Monotone Convergence Theorem (Measure Theory).

## Source of Name

This entry was named for Beppo Levi.