Counting measure

Integration on ${\displaystyle \mathbb {N} }$ with counting measure

Take the measure space ${\displaystyle (\mathbb {N} ,2^{\mathbb {N} },\mu )}$, where ${\displaystyle 2^{\mathbb {N} }}$ is the set of all subsets of the naturals and ${\displaystyle \mu }$ the counting measure. Take any measurable ${\displaystyle f:\mathbb {N} \to [0,\infty ]}$. As it is defined on ${\displaystyle \mathbb {N} }$, ${\displaystyle f}$ can be represented pointwise as

$${\displaystyle f(x)=\sum _{n=1}^{\infty }f(n)1_{\{n\}}(x)=\lim _{M\to \infty }\underbrace {\ \sum _{n=1}^{M}f(n)1_{\{n\}}(x)\ } _{\phi _{M}(x)}=\lim _{M\to \infty }\phi _{M}(x)}$$

Each ${\displaystyle \phi _{M}}$ is measurable. Moreover ${\displaystyle \phi _{M+1}(x)=\phi _{M}(x)+f(M+1)\cdot 1_{\{M+1\}}(x)\geq \phi _{M}(x)}$. Still further, as each ${\displaystyle \phi _{M}}$ is a simple function

$${\displaystyle \int _{\mathbb {N} }\phi _{M}d\mu =\int _{\mathbb {N} }\left(\sum _{n=1}^{M}f(n)1_{\{n\}}(x)\right)d\mu =\sum _{n=1}^{M}f(n)\mu (\{n\})=\sum _{n=1}^{M}f(n)\cdot 1=\sum _{n=1}^{M}f(n)}$$

$${\displaystyle \int _{\mathbb {N} }fd\mu =\lim _{M\to \infty }\int _{\mathbb {N} }\phi _{M}d\mu =\lim _{M\to \infty }\sum _{n=1}^{M}f(n)=\sum _{n=1}^{\infty }f(n)}$$

Discussion

The counting measure is a special case of a more general construction. With the notation as above, any function ${\displaystyle f:X\to [0,\infty )}$ defines a measure ${\displaystyle \mu }$ on ${\displaystyle (X,\Sigma )}$ via

$${\displaystyle \mu (A):=\sum _{a\in A}f(a)\quad {\text{ for all }}A\subseteq X,}$$

$${\displaystyle \sum _{y\,\in \,Y\!\ \subseteq \,\mathbb {R} }y\ :=\ \sup _{F\subseteq Y,\,|F|<\infty }\left\{\sum _{y\in F}y\right\}.}$$

${\displaystyle f(x)=1}$

${\displaystyle x\in X}$

See also