# proof of Vitali convergence theorem

###### Theorem.

Let $f_{1},f_{2},\ldots$ be $\mathbf{L}^{p}$-integrable functions on a 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:

1. (i)

$f_{n}$ is Cauchy in measure;

2. (ii)

the functions $\{\lvert f_{n}\rvert^{p}\}$ are uniformly integrable; and

3. (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 integrable because 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 $\{\lvert f_{n}\rvert^{p}\}$ is uniformly 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 $A_{N}=\{\varphi\neq 0\}$ be the support  of $\varphi$, which must have 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, 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 measure such 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} 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, instead of 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 theorem is easily proved, for:

Title proof of Vitali convergence theorem ProofOfVitaliConvergenceTheorem 2013-03-22 17:31:04 2013-03-22 17:31:04 stevecheng (10074) stevecheng (10074) 5 stevecheng (10074) Proof msc 28A20