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2. Let $ (X,\mathcal{F},\mu) $ be a measure space and $ f \in L(\mu) $. Show that $ \{f \ne 0\} $ is $ \sigma $-finite.


Solution

Since $ f \in L(\mu) $, we know that $ \int_X |f|d\mu < \infty $. Let $ F_{n} \equiv \{|f| > 1/n\} $. Then we have $ \{f \ne 0\} = \cup_{n=1}^\infty F_n. $

Clearly all of the $ F_n $ must have finite measure, else $ \int_X |f|d\mu $ would be infinite. Thus, $ \{f \ne 0\} $ is $ \sigma $-finite.

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva