proof of Vitali convergence theorem

Theorem.

Let $f_{1},f_{2},\\ldots$ be $\\mathbf{L}^{p}$ -integrable functions ona measure space $(X,\\mu)$ , for $1\\leq p<\\infty$ .The following conditions are necessary and sufficient for $f_{n}$ to be a Cauchy sequence in the $\\mathbf{L}^{p}(X,\\mu)$ norm:

(i)
the sequence $f_{n}$ is Cauchy in measure;
(ii)
the functions $\\{\\lvert f_{n}\\rvert^{p}\\}$ are uniformly integrable; and
(iii)
for each $\\epsilon>0$ , there is a set $A$ of finite measure, with $\\lVert f_{n}\\mathbf{1}(X\\setminus A)\\rVert<\\epsilon$ for all $n$ .

Proof.

We abbreviate $\\lvert f_{n}-f_{m}\\rvert$ by $f_{mn}$ .

Necessity of (i).

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

\\mu(E_{mn})^{1/p}=\\frac{1}{t}\\lVert t\\,\\mathbf{1}(E_{mn})\\rVert\\leq\\frac{1}{t}%\\lVert f_{mn}\\rVert\\to 0\\,,\\quad\\text{as $m,n\\to\\infty$.}

Necessity of (ii).

Select $N$ such that $\\lVert f_{n}-f_{N}\\rVert<\\epsilon$ when $n\\geq N$ .The family $\\{\\lvert f_{1}\\rvert^{p},\\ldots,\\lvert f_{N-1}\\rvert^{p},\\lvert f_{N}\\rvert^{p}\\}$ is uniformly integrablebecause it consists of only finitely many integrable functions.

So for every $\\epsilon>0$ ,there is $\\delta>0$ such that $\\mu(E)<\\delta$ implies $\\lVert f_{n}\\mathbf{1}(E)\\rVert<\\epsilon$ for $n\\leq N$ .On the other hand, for $n>N$ ,

\\lVert f_{n}\\mathbf{1}(E)\\rVert\\leq\\lVert(f_{n}-f_{N})\\mathbf{1}(E)\\rVert+%\\lVert f_{N}\\mathbf{1}(E)\\rVert<2\\epsilon

for the same sets $E$ ,and thus the entire infinite sequence $\\{\\lvert f_{n}\\rvert^{p}\\}$ isuniformly integrable too.

Necessity of (iii).

Select $N$ such that $\\lVert f_{n}-f_{N}\\rVert<\\epsilon$ for all $n\\geq N$ .Let $\\varphi$ be a simple function approximating $f_{N}$ in $\\mathbf{L}^{p}$ norm up to $\\epsilon$ .Then $\\lVert f_{n}-\\varphi\\rVert<2\\epsilon$ for all $n\\geq N$ .Let $A_{N}=\\{\\varphi\eq 0\\}$ be the support of $\\varphi$ , which musthave finite measure.It follows that

	$\\displaystyle\\lVert f_{n}\\mathbf{1}(X\\setminus A_{N})\\rVert=\\lVert f_{n}-f_{n}%\\mathbf{1}(A_{N})\\rVert$	$\\displaystyle\\leq\\lVert f_{n}-\\varphi\\rVert+\\lVert\\varphi-f_{n}\\mathbf{1}(A_{N%})\\rVert$
		$\\displaystyle=\\lVert f_{n}-\\varphi\\rVert+\\lVert(\\varphi-f_{n})\\mathbf{1}(A_{N})\\rVert$
		$\\displaystyle<2\\epsilon+2\\epsilon\\,.$

For each $n<N$ , we can similarly construct sets $A_{n}$ of finite measure,such that $\\lVert f_{n}\\mathbf{1}(X\\setminus A_{n})\\rVert<4\\epsilon$ .If we set $A=A_{1}\\cup\\cdots\\cup A_{N-1}\\cup A_{N}$ , a finite union,then $A$ has finite measure, and clearly $\\lVert f_{n}\\mathbf{1}(X\\setminus A)\\rVert<4\\epsilon$ for any $n$ .

Sufficiency.

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

\\displaystyle\\lVert f_{mn}\\rVert

\\displaystyle\\leq\\lVert f_{mn}\\mathbf{1}(A\\setminus E_{mn})\\rVert+\\lVert f_{mn%}\\mathbf{1}(E_{mn})\\rVert+\\lVert f_{mn}\\mathbf{1}(X\\setminus A)\\rVert\\,.

Use condition (iii) to choose $A$ of finite measuresuch that $\\lVert f_{n}\\mathbf{1}(X\\setminus A)\\rVert<\\epsilon$ for every $n$ .Then $\\lVert f_{mn}\\mathbf{1}(X\\setminus A)\\rVert<2\\epsilon$ .

Let $t=\\epsilon/\\mu(A)^{1/p}>0$ ,and $E_{mn}=\\{f_{mn}\\geq t\\}$ .By condition (ii) choose $\\delta>0$ so that $\\lVert f_{n}\\mathbf{1}(E)\\rVert<\\epsilon$ whenever $\\mu(E)<\\delta$ .By condition (i), take $N$ such that if $m,n\\geq N$ ,then $\\mu(E_{mn})<\\delta$ ;it follows immediately that $\\lVert f_{mn}\\mathbf{1}(E_{mn})\\rVert<2\\epsilon$ .

Finally, $\\lVert f_{mn}\\mathbf{1}(A\\setminus E_{mn})\\rVert\\leq t\\mu(A)^{1/p}=\\epsilon$ , since $f_{mn}<t$ on the complement of $E_{mn}$ .Hence $\\lVert f_{mn}\\rVert<5\\epsilon$ for $m,n\\geq N$ .∎

Remark. In the statement of the theorem, insteadof dealing with Cauchy sequences,we can directly speak of convergence of $f_{n}$ to $f$ in $\\mathbf{L}^{p}$ and in measure.This variation of the theoremis easily proved,for:

•
a sequence converges in $\\mathbf{L}^{p}$ if and only if it isCauchy in $\\mathbf{L}^{p}$ ;
•
a sequence that converges in measure is automatically Cauchy in measure;
•
a simple adaptation of the argumentshows that $f_{n}\\to f$ in $\\mathbf{L}^{p}$ implies $f_{n}\\to f$ in measure; and
•
the limit in measure is unique.