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I am having trouble computing integration w.r.t. counting measure. Let $(\mathbb{N},\scr{P}(\mathbb{N}),\mu)$ be a measure space where $\mu$ is counting measure. Let $f:\mathbb{N}\rightarrow{\mathbb{R}}$ be a non-negative bounded measurable function. Then, what is $\int_{\mathbb{N}}fd\mu$? What's gonna happen if we remove the boundedness in $f$, i.e. just let $f$ be an arbitrary non-negative measurable function?

What happens if we relace $\mathbb{N}$ by a general set $X$?

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    $\begingroup$ If $\mu$ is the counting measure on $\mathbb{N}$, then it would seem that $$\int_{\mathbb{N}} f\,\text{d}\mu = \sum_{n=1}^\infty f(n)$$ $\endgroup$ – Nicholas Stull Apr 22 '14 at 4:29
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Counting measure is just summation!

To see this, you can approach from a few different angles; how about we consider the Monotone Convergence Theorem. To that end, for $n\in\mathbb{N}$, define $f_n:\mathbb{N}\to\mathbb{R}$ by $$ f_n(k)=\begin{cases}f(k) & \text{if }1\leq k\leq n\\ 0 & \text{else}\end{cases}. $$ Then clearly, as $n\to\infty$, $f_n\to f$ pointwise; it is also monotone increasing, because $f(k)\geq0$ for all $k\in\mathbb{N}$. So, by the MCT, $$ \int f_n\,d\mu\to\int f\,d\mu\text{ as }n\to\infty. $$ Now, consider these $f_n$. Note that we can write $$ \mathbb{N}=\{1\}\cup\{2\}\cup\cdots\cup\{n\}\cup\{n+1,n+2,\ldots\}, $$ and that these sets are all measurable. So, $$ \begin{align*} \int f_n\,d\mu&=\int_{\{1\}}f_n\,d\mu+\cdots+\int_{\{n\}}f_n\,d\mu+\int_{\{n+1,n+2,\ldots\}}f_n\,d\mu\\ &=\int_{\{1\}}f_n(1)\,d\mu+\cdots+\int_{\{n\}}f_n(n)\,d\mu+\int_{\{n+1,n+2,\ldots\}}0\,d\mu, \end{align*} $$ where we have used that $f_n$ is constant on each of these sets, by definition. So, we see that $$ \int f_n\,d\mu=1\cdot f_n(1)+1\cdots f_n(2)+\cdots+1\cdot f_n(n)+0=f(1)+\cdots+f(n). $$ So, we have that $$ \int f\,d\mu=\lim_{n\to\infty}\int f_n\,d\mu=\lim_{n\to\infty}(f(1)+\cdots+f(n))=\sum_{k=1}^{\infty}f(k). $$ The boundedness is of no consequence here -- since our terms are non-negative, the series either converges or diverges to $\infty$; in either case, the integral is exactly the sum.

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    $\begingroup$ @ Nicholas: Your arguments really help me to grasp the concept of counting measure. Thanks a lot! What's gonna happen if we replace $\mathbb{N}$ by a general set $X$? $\endgroup$ – user54992 Apr 22 '14 at 4:49
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    $\begingroup$ @Algebra The same. $\int f\,\mathrm d\mu=\sum_{x\in X}f(x)$ $\endgroup$ – Hagen von Eitzen Apr 22 '14 at 5:05
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    $\begingroup$ @Algebra I'm glad that you found it helpful! Also: Hagen von Eitzen is correct regarding general counting measures. $\endgroup$ – Nick Peterson Apr 22 '14 at 23:54
  • $\begingroup$ Can we do the same for $f:\mathbb{R}\rightarrow\mathbb{R}$? But mass is defined on points $supp(f)\subset\mathbb{N}\subset\mathbb{R}$. $\endgroup$ – ZHU Feb 17 '18 at 23:10
  • $\begingroup$ Yes, you can. The same argument applies. $\endgroup$ – Nick Peterson Feb 19 '18 at 2:15
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When X is arbitrary, the counting measure is not countably additive and more care will be needed in applying limit theorems.

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