proof of Vitali convergence theorem
Theorem.
Let be -integrable functions ona measure space , for .The following conditions are necessary and sufficient for to be a Cauchy sequence
in the norm:
- (i)
the sequence
is Cauchy in measure;
- (ii)
the functions are uniformly integrable; and
- (iii)
for each , there is a set of finite measure, with for all .
Proof.
We abbreviate by .
- Necessity of (i).
Fix , and let.Then
- Necessity of (ii).
Select such that when .The family is uniformly integrablebecause it consists of only finitely many integrable functions.
So for every ,there is such that implies for .On the other hand, for ,
for the same sets ,and thus the entire infinite
sequence isuniformly integrable too.
- Necessity of (iii).
Select such that for all .Let be a simple function
approximating in norm up to .Then for all .Let be the support
of , which musthave finite measure.It follows that
For each , we can similarly construct sets of finite measure,such that .If we set , a finite union,then has finite measure, and clearly for any .
- Sufficiency.
We show to be small for large by a multi-step estimate:
Use condition (iii) to choose of finite measuresuch that for every .Then .
Let ,and .By condition (ii) choose so that whenever.By condition (i), take such that if ,then;it follows immediately that .
Finally, , since on the complement of .Hence for .∎
Remark. In the statement of the theorem, insteadof dealing with Cauchy sequences,we can directly speak of convergence of to in and in measure.This variation of the theoremis easily proved,for:
- •
a sequence converges
in if and only if it isCauchy in ;
- •
a sequence that converges in measure is automatically Cauchy in measure;
- •
a simple adaptation of the argument
shows that in implies in measure; and
- •
the limit in measure is unique.