PlanetMath: proof of Vitali convergence theorem

Proof. We abbreviate $\abs{f_n - f_m}$ by $f_{mn}$ .

Necessity of (i).

Fix $t > 0$ , and let $E_{mn} = \{ f_{mn} \geq t \}$ . Then $$ \mu(E_{mn})^{1/p} = \frac{1}{t} \norm{ t \, \indc(E_{mn}) } \leq \frac{1}{t} \norm{ f_{mn} } \to 0\,, \quad \text{as $m, n \to \infty$.} $$

Necessity of (ii).

Select $N$ such that $\norm{f_n - f_N} < \epsilon$ when $n \geq N$ . The family $\{ \lvert f_1\rvert ^p, \dotsc, \lvert f_{N-1}\rvert ^p, \lvert f_N\rvert ^p \}$ 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 $\norm{f_n \indc(E)} < \epsilon$ for $n \leq N$ . On the other hand, for $n > N$ , $$ \norm{f_n \indc(E)} \leq \norm{ (f_n - f_N) \indc(E)} + \norm{f_N \indc(E)} < 2\epsilon $$ for the same sets $E$ , and thus the entire infinite sequence $\{ \abs{f_n}^p \}$ is uniformly integrable too.

Necessity of (iii).

Select $N$ such that $\norm{f_n - f_N} < \epsilon$ for all $n \geq N$ . Let $\varphi$ be a simple function approximating $f_N$ in $\Le^p$ norm up to $\epsilon$ . Then $\norm{f_n - \varphi} < 2\epsilon$ for all $n \geq N$ . Let $A_N = \{ \varphi \neq 0 \}$ be the support of $\varphi$ , which must have finite measure. It follows that

$% latex2html id marker 304 $\displaystyle \lVert f_n \mathbf{1}(X \setminus A_N)\rVert = \lVert f_n - f_n \mathbf{1}(A_N)\rVert$$	$% latex2html id marker 305 $\displaystyle \leq \lVert f_n - \varphi\rVert + \lVert\varphi - f_n \mathbf{1}(A_N)\rVert$$
	$% latex2html id marker 306 $\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 $\norm{f_n \indc(X \setminus A_n)} < 4\epsilon$ . If we set $A = A_1 \cup \dotsb \cup A_{N-1} \cup A_N$ , a finite union, then $A$ has finite measure, and clearly $\norm{f_n \indc(X \setminus A)} < 4\epsilon$ for any $n$ .

Sufficiency.

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

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

$% latex2html id marker 311 $\displaystyle \leq \lVert f_{mn} \mathbf{1}(A \setm... ...n} \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 $\norm{f_n \indc(X \setminus A)} < \epsilon$ for every $n$ . Then $\norm{f_{mn} \indc(X \setminus A)} < 2 \epsilon$ .

Let $t = \epsilon/\mu(A)^{1/p} > 0$ , and $E_{mn} = \{ f_{mn} \geq t \}$ . By condition (ii) choose $\delta > 0$ so that $\norm{f_n \indc(E)} < \epsilon$ 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 $\norm{ f_{mn} \indc(E_{mn}) } < 2\epsilon$ .

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

$\qedsymbol$

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 $\Le^p$ and in measure. This variation of the theorem is easily proved, for: