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Let <math>\phi = \lim_{n\rightarrow\infty}\phi_n</math>. | Let <math>\phi = \lim_{n\rightarrow\infty}\phi_n</math>. | ||
| − | <math>\phi_n</math> is monotone increasing so the | + | <math>\phi_n</math> is monotone increasing so the Monotone Convergence Theorem implies <math>\int{\phi d\mu} = \lim_{n\rightarrow\infty}\int{\phi_n d\mu}</math>. |
Since <math>\int{\phi_n d\mu}\le M</math>, we have <math>\int{\phi d\mu} \le \lim_{n\rightarrow\infty}M = M</math>. | Since <math>\int{\phi_n d\mu}\le M</math>, we have <math>\int{\phi d\mu} \le \lim_{n\rightarrow\infty}M = M</math>. | ||
Latest revision as of 11:22, 9 July 2008
Let $ \phi_n = \sup\{f_1,...,f_n\} $.
Let $ \phi = \lim_{n\rightarrow\infty}\phi_n $.
$ \phi_n $ is monotone increasing so the Monotone Convergence Theorem implies $ \int{\phi d\mu} = \lim_{n\rightarrow\infty}\int{\phi_n d\mu} $.
Since $ \int{\phi_n d\mu}\le M $, we have $ \int{\phi d\mu} \le \lim_{n\rightarrow\infty}M = M $.
Note that $ f_k \le \phi, \forall k $.
Now by the Dominated Convergence Theorem, $ \lim_{n\rightarrow\infty}\int{f_n d\mu}=\int{\lim_{n\rightarrow\infty}f_n d\mu} $
Since $ \lim_{n\rightarrow\infty}f_n = 0 $ a.e., $ \lim_{n\rightarrow\infty}\int{f_n d\mu}=0 $
--Bbartle 10:03, 9 July 2008 (EDT)
