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A measure space $(X, \Sigma, \mu)$ is finite if $\mu(X)<\infty$.

It is equivalent to saying that $(X, \Sigma, \mu)$ is finite if $\mu(E)<\infty$ for all $E \in \Sigma$

A measure space $(X, \Sigma, \mu)$ is $\sigma$-finite if X is a countable union of sets with finite measure.

  1. Does $\sigma$-finiteness imply that $\mu(E)<\infty$ for all $E \in \Sigma$?
  2. If $\mu(E)<\infty$ for all $E \in \Sigma$, dose it imply $\sigma$-finiteness or finiteness of a measure space?
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  • $\begingroup$ Do you know any examples of $\sigma$-finite measure spaces? $\endgroup$
    – user147263
    Apr 17 '15 at 4:03
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    $\begingroup$ I know that Real numbers with Lebesgue measure is sigma finite, but not finite just by taking all unit intervals [k, k+1] etc $\endgroup$
    – user232063
    Apr 17 '15 at 4:08
  • $\begingroup$ Not finite because $\mu(\mathbb{R})=\infty$. So that's your answer to 1. $\endgroup$
    – user147263
    Apr 17 '15 at 4:08
  • $\begingroup$ I think both of your questions would be resolved at once if you recalled that $X$ itself is an element of $\Sigma$. $\endgroup$
    – user147263
    Apr 17 '15 at 4:09
  • $\begingroup$ oh yea, I see it. Thanks pizza! $\endgroup$
    – user232063
    Apr 17 '15 at 4:11
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Probably the best example of a finite measure space is $[0, 1]$ with its usual structure, and the best example of a $\sigma$-finite measure space is $\mathbb{R}$ with its usual structure. So, are all the measurable subsets of $\mathbb{R}$ finite in measure? That should answer your first.

For your second, consider what $\mu(X) < \infty$ implies.

edit to add: and I think pizza has said it much better than me.

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    $\begingroup$ Thanks user24142! $\endgroup$
    – user232063
    Apr 17 '15 at 4:12

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