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Let $X$ be a measure space with measure $\mu$. Suppose that $\mu$ is not $\sigma$-finite, so that $X$ is not the countable union of measurable sets of finite measure. Does this imply that the measure of $X$ is infinite?

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up vote 4 down vote accepted

Yes, otherwise take $A_1:=X$ and $A_k=\emptyset$ for $k\geq 2$.

Actually, it seems reasonable that a finite measure space is $\sigma$-finite.

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Thank you. I just realized it myself. – Ricardo Buring Sep 22 '12 at 11:23

If $X$ has finite measure then $X$ can be written as $$X=X \cup \emptyset \cup \emptyset \cup ...$$ which is a countable union of finite measure sets. Therefore $X$ would be $\sigma$-finite.

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