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If $f\in M^{+}(X,$ X $)$ and $\int f\, d\mu<{\infty}$, then the set $N = \{x\in X: f(x)>0\}$ is $\sigma$-finite (that is, there exists a sequence $(F_n)$ in X such that $N \subset \bigcup F_{n}$ and $\mu(F_n)<\infty$).

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closed as off-topic by Fundamental, Tunk-Fey, RecklessReckoner, Mike Miller, Claude Leibovici Aug 9 '14 at 6:32

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Nice to know. And what is your question? –  Did Jan 15 '13 at 6:25

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Consider the sets $F_n = \{x \in X: f(x) > \frac{1}{n}\}$. Can you see why $F_n$ must be a set of finite measure for $\int f d\mu$ to be finite?

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