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Example 1:
 
Example 1:
  
For <math>n</math> odds, <math>f_{n}(x)=1</math> if <math>x=\frac{1}{n}</math>, <math>0</math> otherwise.
+
For <math>n</math> odds, <math>f_{n}(x)=1\frac{}{}</math> if <math>x=\frac{1}{n}</math>, <math>0</math> otherwise.
  
For <math>n</math> even, <math>f_{n}(x)=3</math> if <math>x=\frac{1}{n}</math>, <math>0</math> otherwise.
+
For <math>n</math> even, <math>f_{n}(x)=3\frac{}{}</math> if <math>x=\frac{1}{n}</math>, <math>0</math> otherwise.

Revision as of 11:00, 22 July 2008

By Fatou's Lemma, we get the upper bound is 1 and since all the functions $ f_{n}\frac{}{} $ are positive, we get the lower bound is 0. This is as good as it get. Examples:

Let $ \Omega=[0,1]\frac{}{} $, the $ \sigma- $algebra is the power set and counting measure.

Example 1:

For $ n $ odds, $ f_{n}(x)=1\frac{}{} $ if $ x=\frac{1}{n} $, $ 0 $ otherwise.

For $ n $ even, $ f_{n}(x)=3\frac{}{} $ if $ x=\frac{1}{n} $, $ 0 $ otherwise.

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