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Suppose $f$ is measurable. For each natural number $n$, define $ X_n = \{x \in X : f(x) \geq 1/n\}.$ Why is $\bigcup_{n=1}^\infty X_n = \{x\in X : f(x) \gt 0\}?$ Suppose $f$ is measurable and integrable. How can one write the measurable set $\{x: f(x) \lt 0\}$ as a countable union of measurable sets?

Can these be generalized?

Thanks.

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    Can you show that the left-hand-side is contained in the right-hand-side?2012-03-09

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The set equality follows from the set inclusion in both directions. If $ x \in \cup_{n=1}^{\infty} X_n $ then $ x \in X_n $ for some $n$, so $ f(x) \geq 1/n >0 $ and thus it is in the right hand set as well. Conversely, if $x\in X$ is such that $f(x)>0 $ then due to the Archimedian property we can find $n$ large enough so that $ f(x) \geq 1/n $ so $ x\in X_n $ and thus in the left hand set.

For your second question, the result is obtained by switching the inequality signs in every step of the first example.

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    yes, right. Thanks again.2012-03-09