Difference between revisions of "Solution to number 5 Old Kiwi" - Rhea

Revision as of 20:08, 1 July 2008

8.

Fix $ \epsilon > 0 $.

Then $ \exists \ N $ such that $ n>N \Rightarrow \int_X|f_n-f|^p <\epsilon/3 $.

Define $ f_0=f $.

Then for $ i=0,1,...,N \ \exists \ \delta_i>0 $ such that $ m(A) <\delta_i \Rightarrow \int_A|f_i|^p<\epsilon/3 $, since $ f_i \in L^p $.

Define $ \delta=min\{\delta_0, \delta_1,...,\delta_N, \epsilon/3\} $.

Fix A with $ m(A)<\delta $.

Note: $ \int_A|f_n| = \int_{\{x\in A:|f_n|\leq1\}}|f_n|+\int_{\{x\in A:|f_n|>1\}}|f_n| \leq m(A) + \int_A|f_n|^p \leq \epsilon/3 + \int_A|f_n|^p \ \ \forall \ n=0,1,... $.

Then if n = 1,...,N

$ \int_A|f_n|\leq 2\epsilon/3 < \epsilon $ (by our choice of $ \delta $).

and if

n > N

then

$ \int_A|f_n|\leq \epsilon/3 + \int_A|f_n|^p\leq \epsilon/3 + \int_A|f|^p+\int_A|f_n-f|^p\leq \epsilon/3 + \int_A|f|^p+\int_X|f_n-f|^p\leq \epsilon $.

So $ m(A)<\delta \Rightarrow \int_A|f_n| <\epsilon \ \forall \ n $.