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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. – leslie townes Mar 3 '12 at 5:41
@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"? – Tim Mar 4 '12 at 13:25
  1. If $\mu$ is a locally finite counting measure then every compact set has finite measure as we can cover the compact set with finitely many open sets of finite measure. Thus, $\Lambda\cap K$ must be finite for all compact $K$.
  2. $\sum_{x\in\Lambda}\delta_x$ it is always well-defined as $\sup\sum_{x\in F}\delta_x$, where $F$ ranges over all finite subsets of $\Lambda$. This definition of sum just works for nonnegative terms. In general, we consider the the net of finite subsets of the index set ordered by inclusion then the sum is defined as the limit of the net.
  3. It is true if we assume that the space is locally compact since any precompact neighbourhood would have finite measure, hence $\mu$ would be locally finite measure.
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+1 Thanks! Is the sume of a family of real numbers $\{ x_a: a \in A \}$ also defined via $\sup$ over all finite subcollection of the family? – Tim Mar 3 '12 at 1:54
I'll edit my post to include a more general version of the sum. – azarel Mar 3 '12 at 1:58
Thanks! I was wondering if local compactness is necessary for the converse to be true? – Tim Mar 4 '12 at 6:03

I am adding this as an answer because it is too large to fit as a comment. But mostly its purpose is to clarify one of my comments (and answer a reply to that comment).

One of the questions raised by the original post concerns the definition of "counting measure" that is implicitly being used in the statements linked above.

My guess is that the definition of "counting measure" used there is equivalent to the following:

A measure $m$ on a measure space $(X, \mathcal{B})$ is said to be a counting measure if and only if there is some $S \subseteq X$ satisfying $$ m(E) = \#(E \cap S), \qquad E \in \mathcal{B}, $$ where $\#$ denotes the set cardinality (either a nonnegative integer or $\infty$).

In my comment on the original question, I stated that this condition was equivalent to the condition that $m(\{x\}) \in \{0,1\}$ for all $x \in X$, but on further thought I realized that this is not true. For one thing, there is nothing in the above definition which requires or implies that the singleton subsets of $X$ are measurable. If we assume this, the above definition implies that $m(\{x\}) \in \{0,1\}$ for all $x$ in $X$, but there are measures satisfying this weaker condition which are not counting measures in the above sense. (For example, fix any uncountable $X$, take $\mathcal{B}$ to be the power set of $X$, fix $p \in X$, and take the measure $m$ to be $\delta_p + \nu$, where $\delta_p$ is the Dirac mass at $p$ and $\nu$ is the measure given by declaring that $\nu(E) = 0$ if $E$ is countable and $\nu(E) = \infty$ otherwise.) One can, of course, add additional hypotheses to this weaker condition to make it equivalent to the above definition. But I think the above definition is more fundamental in capturing what a "counting measure" ought to be.

In any case, it is clear that if $m$ is a counting measure in the above sense, then the set $S$ appearing in the definition is uniquely determined by $m$: it must be $\{x \in X: m(\{x\}) = 1\}$. It is also clear, straight from the definitions, that a counting measure $m$ is finite if and only if its set $S$ is finite. This is immediate from the equality $m(X) = \#(S)$.

Now suppose that $X$ is a second countable, locally compact Hausdorff space, and that $\mathcal{B}$ is the Borel $\sigma$-algebra of $X$. Suppose $m$ is a counting measure on $X$, and let $S = \{x \in X: m(\{x\}) = 1\}$. I claim that the measure $m$ is locally finite (in the sense given in the OP) if and only if the set $S$ is locally finite (in the sense given in the OP).

This is true, but not immediately obvious. What is immediately obvious is that the measure $m$ is locally finite if and only if the set $S$ has the following property:

For all $x \in X$, there is an open subset $U$ of $X$ containing $x$ with the property that $U \cap S$ is finite.

Indeed, since $m(U)$ is finite if and only if $U \cap S$ is finite, this is a direct transcription of "every point of the space has an open neighborhood of finite measure". But "$S$ is locally finite", a priori, means something slightly different. So we need to prove the equivalence of the following properties of a subset $S$ of $X$:

  1. For all $x \in X$, there is an open subset $U$ of $X$ containing $x$ with the property that $U \cap S$ is finite.

  2. For all compact $K \subseteq X$, one has that $K \cap S$ is finite.

If condition (2) fails then we can choose a compact $K \subseteq X$ and an infinite sequence $(k_n)_{n=1}^{\infty}$ of distinct points of $K \cap S$. Since $X$ is second countable, we know that $K$ is sequentially compact, and so we may assume (by passing to a subsequence as needed) that $(k_n)_{n=1}^{\infty}$ is convergent to some $p \in X$. But then, for any open subset $U$ of $X$ containing $p$, there is some positive integer $N$ with the property that the infinite set $\{k_n: n \geq N\}$ is contained in $U$. So $U \cap S$ is infinite for every open subset $U$ of $X$ containing $p$, and condition (1) fails.

Conversely, if condition (1) fails there is some $p \in X$ with the property that every open subset $U$ of $X$ containing $p$ has the property that $U \cap S$ is infinite. Since $X$ is locally compact, we can find an open subset $U$ of $X$ containing $p$ with the property that there is a compact subset $K$ of $X$ satisfying $K \supseteq U$. But then $K \cap S$ is infinite (since it has the infinite subset $U \cap S$) and the second condition fails.

This proves that conditions (1) and (2) are indeed equivalent. So, assuming those topological hypotheses on $X$, one does know that a counting measure $m$ on $(X, \mathcal{B})$ is locally finite if and only if its set $S$ is locally finite.

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