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 $.
--Wardbc 20:17, 1 July 2008 (EDT)Ben Ward
