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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 $

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009