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| − | For <math>n</math> odds, <math>f_{n}(x)=1\frac{}{}</math> if <math>x=1</math>, <math>0\frac{}{}</math> otherwise. | + | For <math>n</math> odds, <math>f_{n}(x)=1\frac{}{}</math> if <math>x=1\frac{}{}</math>, <math>0\frac{}{}</math> otherwise. |
| − | For <math>n</math> even, <math>f_{n}(x)=1\frac{}{}</math> if <math>x=1</math>, <math>f_{n}(x)= | + | For <math>n</math> even, <math>f_{n}(x)=1\frac{}{}</math> if <math>x=1\frac{}{}</math>, <math>f_{n}(x)=2\frac{}{}</math> if <math>x=\frac{1}{n}</math>, <math>0\frac{}{}</math> otherwise. |
Latest revision as of 11:03, 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\frac{}{} $ otherwise.
For $ n $ even, $ f_{n}(x)=3\frac{}{} $ if $ x=\frac{1}{n} $, $ 0\frac{}{} $ otherwise.
Example 2:
For $ n $ odds, $ f_{n}(x)=1\frac{}{} $ if $ x=1\frac{}{} $, $ 0\frac{}{} $ otherwise.
For $ n $ even, $ f_{n}(x)=1\frac{}{} $ if $ x=1\frac{}{} $, $ f_{n}(x)=2\frac{}{} $ if $ x=\frac{1}{n} $, $ 0\frac{}{} $ otherwise.
