Counting measure

Template:Short description In mathematics, specifically measure theory, the counting measure is an intuitive way to put a measure on any set – the "size" of a subset is taken to be the number of elements in the subset if the subset has finitely many elements, and infinity $\mathrm{\infty }$ if the subset is infinite.

The counting measure can be defined on any measurable space (that is, any set $X$ along with a sigma-algebra) but is mostly used on countable sets.

In formal notation, we can turn any set $X$ into a measurable space by taking the power set of $X$ as the sigma-algebra $\mathrm{\Sigma };$ that is, all subsets of $X$ are measurable sets. Then the counting measure $\mu$ on this measurable space $(X,\mathrm{\Sigma })$ is the positive measure $\mathrm{\Sigma }\to [0,+\mathrm{\infty }]$ defined by $$\mu (A)=\{\begin{array}{cc}|A|& \text{if }A\text{ is finite}\\ +\mathrm{\infty }& \text{if }A\text{ is infinite}\end{array}$$ for all $A\in \mathrm{\Sigma },$ where $|A|$ denotes the cardinality of the set $A.$

The counting measure on $(X,\mathrm{\Sigma })$ is σ-finite if and only if the space $X$ is countable.

Integration on the set of natural numbers with counting measure

Take the measure space $(\mathbb{N},2^{\mathbb{N}},\mu )$, where $2^{\mathbb{N}}$ is the set of all subsets of the naturals and $\mu$ the counting measure. Take any measurable $f:\mathbb{N}\to [0,\mathrm{\infty }]$. As it is defined on $\mathbb{N}$, $f$ can be represented pointwise as $$f(x)=\sum _{n=1}^{\mathrm{\infty }}f(n)1_{\{n\}}(x)=\underset{M\to \mathrm{\infty }}{lim}\underset{{\varphi }_M(x)}{\underbrace{\sum _{n=1}^{M}f(n)1_{\{n\}}(x)}}=\underset{M\to \mathrm{\infty }}{lim}{\varphi }_M(x)$$

Each ${\varphi }_M$ is measurable. Moreover ${\varphi }_{M+1}(x)={\varphi }_M(x)+f(M+1)\cdot 1_{\{M+1\}}(x)\ge {\varphi }_M(x)$. Still further, as each ${\varphi }_M$ is a simple function $${\int }_{\mathbb{N}}{\varphi }_{M}d\mu ={\int }_{\mathbb{N}}(\sum _{n=1}^{M}f(n)1_{\{n\}}(x))d\mu =\sum _{n=1}^{M}f(n)\mu (\{n\})=\sum _{n=1}^{M}f(n)\cdot 1=\sum _{n=1}^{M}f(n)$$Hence by the monotone convergence theorem $${\int }_{\mathbb{N}}fd\mu =\underset{M\to \mathrm{\infty }}{lim}{\int }_{\mathbb{N}}{\varphi }_{M}d\mu =\underset{M\to \mathrm{\infty }}{lim}\sum _{n=1}^{M}f(n)=\sum _{n=1}^{\mathrm{\infty }}f(n)$$

Discussion

The counting measure is a special case of a more general construction. With the notation as above, any function $f:X\to [0,\mathrm{\infty })$ defines a measure $\mu$ on $(X,\mathrm{\Sigma })$ via $$\mu (A):=\sum _{a\in A}f(a)\text{ for all }A\subseteq X,$$ where the possibly uncountable sum of real numbers is defined to be the supremum of the sums over all finite subsets, that is, $$\sum _{y\in Y\subseteq ℝ}y:=\underset{F\subseteq Y,|F|<\mathrm{\infty }}{sup}\left\{\sum _{y\in F}y\right\}.$$ Taking $f(x)=1$ for all $x\in X$ gives the counting measure.

See also

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