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Please correct me if I am not correct. On a Borel sigma algebra on a topological space $S$,

  • a locally finite measure is defined as a measure for which every point of the measure space has a open neighbourhood of finite measure.
  • A local finite subset is defined as a subset with which the intersection of every compact subset is a finite set. (I think it is a pure topological concept?)

I wonder

  1. why is it that every locally finite counting measure on $S$ is of the form $\sum_{x\in \Lambda}\delta_x$ where $\Lambda$ is a locally finite subset of $S$?
  2. For $\sum_{x\in \Lambda}$ to be well-defined, is $\Lambda$ required to be countable, i.e. to avoid being uncountable?
  3. Is the converse true, i.e. is it true that a counting measure on $S$ is locally finite if and only if it is of the form $\sum_{x\in \Lambda}\delta_x$ where $\Lambda$ is a locally finite subset of $S$?
  4. In general, is a counting measure on a measurable space $X$ defined as $\mu(B) := \#\{A \cap B\}$ for some fixed and arbitrary subset $A$ of $X$? This is how I infer from "locally finite counting measure", but Wikipedia says a counting measure is defined as the cardinality function of measurable subsets.

    Is a measure counting measure if and only if it is of the form $\sum_{x\in A}\delta_x$ where $A$ is a fixed and arbitrary subset of $X$? (Similar to a previous question, does $A$ need to be countable for $\sum_{x\in A}$ to be well-defined?)

Thanks and regards!

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    The definition of "counting measure" used here is more general than Wikipedia's. The following conditions on a measure $m$ on a space $X$ are equivalent: (1) for all $x \in X$ one has $m(\{x\}) \in \{0,1\}$, (2) there is $S \subseteq X$ satisfying $m(E) = \#(E \cap S)$ for all $E \subseteq X$, where $\#$ is the cardinality (nonnegative integer or $\infty$). (To see (1) implies (2), take $S = \{x \in X: m(\{x\}) = 1\}$.) A counting measure is a measure satisfying (1) or (2), such a measure is finite iff the set $S$ in (2) is finite, and locally finite iff the set $S$ in (2) is locally finite.2012-03-03
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    @leslietownes: Thanks! That is clearer! Does it need any other condition for "such a measure is locally finite iff the set S in (2) is locally finite"?2012-03-04

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