Vitali's theorem

Note

A family of functions $\{f_i{\}}_{i\in I}\subseteq L^1(X,\mu )$ is called uniformly integrable if for every $ϵ>0$ there exists a $\delta >0$ such that for every small enough set $\mu (E)<\delta$

$${\int }_E|f_i|d\mu <ϵ\forall i\in I$$

This is the same as

$$\underset{\mu (E)\to 0}{\lim }\underset{i\in I}{\sup }{\int }_E|f_i|d\mu =0$$

Note

A family of functions $\{f_i{\}}_{i\in I}\subseteq L^1(X,\mu )$ is called tight if for every $ϵ>0$ there exists a subset $E\subseteq X$ with $\mu (E)<\infty$ such that

$${\int }_{X\setminus E}|f_i|d\mu <ϵ\forall i\in I$$

Note

Let $f\in L^1$, then for every $ϵ>0$ there exists a finite set $E$ such that

$${\int }_{X\setminus E}|f|d\mu <ϵ$$

Note A_n := \{ x\,:\, |f(x)| \geq \frac 1n\}. Then

Define sets

$$A:=\bigcup _{n\ge 1}{A}_n=\{x:|f(x)|>0\}$$

so that

$${\int }_A|f|d\mu ={\int }_X|f|d\mu$$

Define the sequence of functions $f_n:=|f|{\mathcal{A}}_n$, since $A_n\subseteq A_{n+1}$ by the [[Monotone Convergence Theorem|]]

$$\underset{n\to \infty }{\lim }{\int }_X|f|{\mathcal{A}}_{n}d\mu ={\int }_A|f|d\mu$$

This means that for every $ϵ>0$ there exists $N\ge 1$ such that

$${\int }_X|f|d\mu -{\int }_{A_n}|f|d\mu <ϵ,\forall n\ge N\square$$

Note

Let $\{f_n\}$ be a sequence that is uniformly integrable and tight over $E$. If $f_n\to f$ pointwise a.e. on $E$, then

$$\underset{n\to \infty }{\lim }{\int }_E|f-f_n|d\mu =0$$

Note \epsilon > 0, we work on the finite set E using thighness assumption, now we can use Egorov's theorem, f_n \rightrightarrows f on a set E' \subseteq E with \mu(E\setminus E') < \epsilon. If we split the integral

(Sketch) Fix

$${\int }_X|f-f_n|d\mu ={\int }_{X\setminus E}|f-f_n|d\mu +{\int }_{E\setminus E^{\prime }}|f-f_n|d\mu +{\int }_{E^{\prime }}|f-f_n|d\mu <2ϵ{\int }_{E^{\prime }}|f-f_n|d\mu$$

we can take the limit inside where there is uniform convergence, and since $ϵ$ was arbitrary the theorem follows. $\square$

Note

Let $\{h_n\}$ be a sequence of non-negative functions on $E$ that converges pointwise a.e. to $0$. Then

$$\underset{n\to \infty }{\lim }{\int }_E{h}_{n}d\mu =0$$

if and only if $\{h_n\}$ is uniformly integrable and tight over $E$.