proof of Vitali convergence theorem

Theorem.

Let $f_{\mathrm{1}}\mathrm{,}{f}_{\mathrm{2}}\mathrm{,}\mathrm{\dots }$ be ${\mathrm{L}}^p$-integrable functions on a measure space $\mathrm{(}X\mathrm{,}\mu \mathrm{)}$, for . The following conditions are necessary and sufficient for $f_n$ to be a Cauchy sequence in the ${\mathrm{L}}^p\mathit{}\mathrm{(}X\mathrm{,}\mu \mathrm{)}$ norm:

1. the sequence $f_n$ is Cauchy in measure;

2. the functions $\{{|f_n|}^p\}$ are uniformly integrable; and

3. for each $ϵ>0$, there is a set $A$ of finite measure, with for all $n$.

Proof.

We abbreviate $|f_n-f_m|$ by $f_{mn}$.

Necessity of (i).

Fix $t>0$, and let $E_{mn}=\{f_{mn}\ge t\}$. Then

$$\mu {(E_{mn})}^{1/p}=\frac{1}{t}\parallel t\mathbf{\hspace{0.17em}1}(E_{mn})\parallel \le \frac{1}{t}\parallel f_{mn}\parallel \to 0,\text{as }m,n\to \mathrm{\infty }\text{.}$$

Necessity of (ii).

Select $N$ such that when $n\ge N$. The family $\{{|f_1|}^p,\mathrm{\dots },{|f_{N-1}|}^p,{|f_N|}^p\}$ is uniformly integrable because it consists of only finitely many integrable functions.

So for every $ϵ>0$, there is $\delta >0$ such that implies for $n\le N$. On the other hand, for $n>N$,

for the same sets $E$, and thus the entire infinite sequence $\{{|f_n|}^p\}$ is uniformly integrable too.

Necessity of (iii).

Select $N$ such that for all $n\ge N$. Let $\phi$ be a simple function approximating $f_N$ in ${𝐋}^p$ norm up to $ϵ$. Then for all $n\ge N$. Let $A_N=\{\phi \ne 0\}$ be the support of $\phi$, which must have finite measure. It follows that

$\parallel f_n\mathrm{𝟏}(X\setminus A_N)\parallel =\parallel f_n-f_n\mathrm{𝟏}(A_N)\parallel$ $\le \parallel f_n-\phi \parallel +\parallel \phi -f_n\mathrm{𝟏}(A_N)\parallel$

$=\parallel f_n-\phi \parallel +\parallel (\phi -f_n)\mathrm{𝟏}(A_N)\parallel$

For each , we can similarly construct sets $A_n$ of finite measure, such that . If we set $A=A_1\cup \mathrm{\cdots }\cup A_{N-1}\cup A_N$, a finite union, then $A$ has finite measure, and clearly for any $n$.

Sufficiency.

We show $f_{mn}$ to be small for large $m,n$ by a multi-step estimate:

$\parallel f_{mn}\parallel$ $\le \parallel f_{mn}\mathrm{𝟏}(A\setminus E_{mn})\parallel +\parallel f_{mn}\mathrm{𝟏}(E_{mn})\parallel +\parallel f_{mn}\mathrm{𝟏}(X\setminus A)\parallel .$

Use condition (iii) to choose $A$ of finite measure such that for every $n$. Then .

Let $t=ϵ/\mu {(A)}^{1/p}>0$, and $E_{mn}=\{f_{mn}\ge t\}$. By condition (ii) choose $\delta >0$ so that whenever . By condition (i), take $N$ such that if $m,n\ge N$, then ; it follows immediately that .

Finally, $\parallel f_{mn}\mathrm{𝟏}(A\setminus E_{mn})\parallel \le t\mu {(A)}^{1/p}=ϵ$, since on the complement of $E_{mn}$. Hence for $m,n\ge N$. ∎

Remark. In the statement of the theorem, instead of dealing with Cauchy sequences, we can directly speak of convergence of $f_n$ to $f$ in ${𝐋}^p$ and in measure. This variation of the theorem is easily proved, for:

• a sequence converges in ${𝐋}^p$ if and only if it is Cauchy in ${𝐋}^p$;

• a sequence that converges in measure is automatically Cauchy in measure;

• a simple adaptation of the argument shows that $f_n\to f$ in ${𝐋}^p$ implies $f_n\to f$ in measure; and

• the limit in measure is unique.

Title	proof of Vitali convergence theorem
Canonical name	ProofOfVitaliConvergenceTheorem
Date of creation	2013-03-22 17:31:04
Last modified on	2013-03-22 17:31:04
Owner	stevecheng (10074)
Last modified by	stevecheng (10074)
Numerical id	5
Author	stevecheng (10074)
Entry type	Proof
Classification	msc 28A20