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

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 $${\displaystyle \int _{\mathbb {N} }\phi _{M}d\mu =\int _{\mathbb {N} }\left(\sum _{n=1}^{M}f(n)1_{\{n\}}(x)\right)d\mu =\sum _{n=1}^{M}f(n)\mu (\{n\})=\sum _{n=1}^{M}f(n)\cdot 1=\sum _{n=1}^{M}f(n)}$$Hence by the monotone convergence theorem $${\displaystyle \int _{\mathbb {N} }fd\mu =\lim _{M\to \infty }\int _{\mathbb {N} }\phi _{M}d\mu =\lim _{M\to \infty }\sum _{n=1}^{M}f(n)=\sum _{n=1}^{\infty }f(n)}$$