proof of Vitali convergence theorem
We abbreviate by .
- Necessity of (i).
Fix , and let . Then
- Necessity of (ii).
Select such that when . The family is uniformly integrable because 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 is uniformly integrable too.
- Necessity of (iii).
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 .
We show to be small for large by a multi-step estimate:
Use condition (iii) to choose of finite measure such 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, instead of dealing with Cauchy sequences, we can directly speak of convergence of to in and in measure. This variation of the theorem is easily proved, for:
|Title||proof of Vitali convergence theorem|
|Date of creation||2013-03-22 17:31:04|
|Last modified on||2013-03-22 17:31:04|
|Last modified by||stevecheng (10074)|