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Which implications are true (if any) for a measure $\mu$:

  1. $\sigma$- finite $\implies$ locally finite
  2. locally finite $\implies$ $\sigma$-finite

My guess would be that both are false, but no counterexamples come to mind, since I am only used to working with $\sigma-$finite measure spaces, hence do not know many (if any) measures which do not satisfy that condition.

I was thinking about this because I was wondering why local finiteness is used in the definition of Radon measures instead of $\sigma$-finiteness, and I figured that the counterexamples to the above question might give me my answer.

Specifically, if there is a counterexample for the second implication which is a Radon measure, that would be greatly preferred.

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To see that the first implication is false, let $X=\mathbb{R}$ with the standard topology, and define a Borel measure $\mu$ by $$ \mu=\sum_{n=1}^{\infty}\delta_{\frac{1}{n}}$$ Then $\mu$ is $\sigma$-finite because $\mathbb{R}$ is the union of $(-\infty,0]$ and the collection $\{(\frac{1}{n},\infty):n\geq 1\}$, each of which has finite measure. However $\mu$ is not locally finite because every open neighborhood of $0$ has infinite measure.

To see that the second implication is false, let $X$ be an uncountable set with the discrete topology and $\mu$ the counting measure on $X$, then $\mu$ is locally finite but not $\sigma$-finite. Note that $\mu$ is a Radon measure.

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  • $\begingroup$ Very elegant examples! Thank you so much $\endgroup$ – Chill2Macht Jun 2 '16 at 3:39

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