Beppo Levi's Theorem

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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 $\displaystyle \sup_{n \mathop \in \N} f_n: X \to \overline \R$ be the pointwise supremum of $\sequence {f_n}_{n \mathop \in \N}$, where $\overline \R$ denotes the extended real numbers.


Then:

$\displaystyle \int \sup_{n \mathop \in \N} f_n \rd \mu = \sup_{n \mathop \in \N} \int f_n \rd \mu$

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


Proof

Since by definition $\displaystyle \sup _{n \mathop \in \N} f_n \ge f_m$ for all $m$, we have:

$\displaystyle \int \sup_{n \mathop \in \N} f_n \rd \mu \ge \int f_m \rd \mu$

and hence the inequality holds for the supremum as well:

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

To show the reverse inequality, since the integral of $\sup\limits_{n \mathop \in \N}f_n$ is defined as the supremum of the integrals of positive simple functions $s \le \sup\limits_{n \mathop \in \N}f_n$, we show that given a simple function $\displaystyle s = \sum_{i \mathop = 1}^k \lambda_i \chi_{E_i} \le \sup_{n \mathop \in \N} f_n$ (where $\lambda_i \in \closedint 0 {+\infty}$ and $E_i \in \Sigma$) we have:

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

To show this, we use the fact that $\nu_s: \Sigma \to \closedint 0 {+\infty}$ defined by $\map {\nu_s} E = \displaystyle \int \chi_Es \rd \mu$ 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 $X$ ($X = \displaystyle \bigcup_{m \mathop \in \N} A_m$) by definition of the supremum, because $\displaystyle s \le \sup_{n \mathop \in \N} f_n$.

Furthermore, $A_m \uparrow X$ ($m \to \infty$).



By definition of the $A_m$ we have that:

$\displaystyle \int f_m \rd \mu \ge \int \chi_{A_m} f_m \rd \mu \ge \paren {1 - \epsilon} \int \chi_{A_m} s \rd \mu = \paren {1 - \epsilon} \map {\nu_s} {A_m}$

where the first inequality follows from the fact that the $f_m$ are positive.

Now, since the sequence $\sequence {f_n}_{n \mathop \in \N}$ increases monotonically, by the increasing sequence of sets property given in any measure space, we have that taking the supremum of both sides yields:

\(\displaystyle \sup_{m \mathop \in \N} \int f_m \rd \mu\) \(=\) \(\displaystyle \lim_{m \mathop \to +\infty} \int f_m \rd \mu\)
\(\displaystyle \) \(\ge\) \(\displaystyle \lim_{m \mathop \to +\infty} \paren {1 - \epsilon} \map {\nu_s} {A_m}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {1 - \epsilon} \map {\nu_s} X\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {1 - \epsilon} \int s \rd \mu\)

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

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

$\blacksquare$


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.


Sources