(New page: 6. First we show <math>f_n\rightarrow f </math> in <math> L^p</math>. Let <math>\delta > 0</math>. Let <math> \epsilon = \frac {\delta }{1+2^{p+1}} </math> Then by Egorov <math>\exists ...)
 
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</math>
 
</math>
 
Thus <math>\int_X f_ng \rightarrow \int_X fg</math>.
 
Thus <math>\int_X f_ng \rightarrow \int_X fg</math>.
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--[[User:Wardbc|Wardbc]] 16:56, 10 July 2008 (EDT)
 
--[[User:Wardbc|Wardbc]] 16:56, 10 July 2008 (EDT)

Revision as of 15:56, 10 July 2008

6.

First we show $ f_n\rightarrow f $ in $ L^p $.

Let $ \delta > 0 $. Let $ \epsilon = \frac {\delta }{1+2^{p+1}} $ Then by Egorov $ \exists E \subset X $ such that

$ m(E) > m(X)-\epsilon $ and $ |f_n-f|<\frac{\epsilon}{m(X)} $ on $ E \ \forall \ n>N $, some N.

Also, by Fatou, $ \int_X |f^p| \leq lim_n \int |f_n^p| \leq 1 $ (by hypothesis)

So $ \int_X |f-f_n|^p=\int_E |f-f_n|^p+\int_{X-E} |f-f_n|^p\leq m(X)\frac{\epsilon}{m(x)}+\int_{X-E} (|f|+|f_n|)^p\leq \epsilon + 2^p(\int_{X-E} (|f^p|+|f_n^p|) \leq \epsilon + 2^p(2m(X-E) $) (Holder)

So $ \int_X |f-f_n|^p \leq \epsilon (1+2^{p+1})=\delta $ if $ n>N \Rightarrow f_n\rightarrow f in L^p $.

Now,

$ |\int_X fg-f_ng|=|\int_X g(f-f_n)| \leq \|{g}\|_q \|{f-f_n}\|_p \rightarrow 0 $ Thus $ \int_X f_ng \rightarrow \int_X fg $.


--Wardbc 16:56, 10 July 2008 (EDT)

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