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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 $$\int_{\mathbb{N}} f\,\text{d}\mu = \sum_{n=1}^\infty f(n)$$ $\endgroup$ Apr 22, 2014 at 4:29

2 Answers 2

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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, $$ \int_{\mathbb{N}} f_n\,d\mu\to\int_{\mathbb{N}} f\,d\mu\text{ as }n\to\infty. $$ 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, $$ \begin{align*} \int_{\mathbb{N}} f_n\,d\mu&=\int_{\{1\}}f_n\,d\mu+\cdots+\int_{\{n\}}f_n\,d\mu+\int_{\{n+1,n+2,\ldots\}}f_n\,d\mu\\ &=\int_{\{1\}}f_n(1)\,d\mu+\cdots+\int_{\{n\}}f_n(n)\,d\mu+\int_{\{n+1,n+2,\ldots\}}0\,d\mu, \end{align*} $$ where we have used that $f_n$ is constant on each of these sets, by definition. So, we see that $$ \int_{\mathbb{N}} f_n\,d\mu=1\cdot f_n(1)+1\cdot f_n(2)+\cdots+1\cdot f_n(n)+0=f(1)+\cdots+f(n). $$ So, we have that $$ \int_{\mathbb{N}} f\,d\mu=\lim_{n\to\infty}\int_{\mathbb{N}} f_n\,d\mu=\lim_{n\to\infty}(f(1)+\cdots+f(n))=\sum_{k=1}^{\infty}f(k). $$ 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, 2014 at 4:49
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    $\begingroup$ @Algebra The same. $\int f\,\mathrm d\mu=\sum_{x\in X}f(x)$ $\endgroup$ Apr 22, 2014 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. $\endgroup$ Apr 22, 2014 at 23:54
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    $\begingroup$ @NickPeterson Can we do it like this instead: $\int_{\mathbb{N}} f \mathrm{d} \mu = \int_{\cup_{n \in \mathbb{N}} \{n\} } f \mathrm{d} \mu = \sum_{n \in \mathbb{N}} \int_{ \{n\} } f \mathrm{d} \mu = \sum_{n \in \mathbb{N}} \int_{ \{n\} } f(n) \mathrm{d} \mu = \sum_{n \in \mathbb{N}} f(n) \mu(\{n\}) = \sum_{n \in \mathbb{N}} f(n)$ ? $\endgroup$
    – goblinb
    May 20, 2020 at 3:00
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    $\begingroup$ @strawberry-sunshine Yes, $\int_E 1\,d\mu=\mu(E)$ for any measurable set. And more generally, for any fixed $\alpha$, $\int_E\alpha\,d\mu=\alpha\int_E1\,d\mu=\alpha\mu(E)$. $\endgroup$ Mar 5, 2021 at 21:20
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$\color{red}{\mathbf{\text{general case}:}}$

if $X$ be arbitrary set and if $(X,P(X),\mu)$ be counting measure space on $P(X)$ such that

$\underset{\{x\}\xrightarrow{\hspace{1.2cm}}1}{\mu:P(x)\rightarrow [0,\infty]}$

and if

$f:X\rightarrow [0,\infty]$

be an arbitrary function on $X$, then integral of $f$ is equal to

$\int_Xfd\mu\color{blue}=\underset{\underset{\text{$\phi$ is simple}}{\phi\leq f}}{\sup}\left(\int_{X}\phi\; d\mu\right)=\underset{\underset{\underset{\text{F is finite}}{F\subseteq X}}{\phi\leq f}}{\sup}\left(\underset{x\in F}{\sum} \phi(x)\right) =\underset{\underset{\text{F is finite}}{F\subseteq X}}{\sup}\left(\underset{x\in F}{\sum} f(x)\right) \color{red}=\underset{x\in X}{\sum}f(x)$

in fact equality in blue, is definition of integral and equality in red is definition of sum of arbitrary positive function on arbitrary set $X$.

now, let $f:X\rightarrow \mathbb{C}$. if $$\int_X |f|d\mu<\infty$$ then integral of $f$ is defined by:

\begin{align*} &\int_X fd\mu\\ &=\Bigg[\int_X f^+_{Re}d\mu-\int_X f^-_{Re}d\mu\Bigg]+i\Bigg[\int_X f^+_{Im}d\mu-\int_X f^-_{Im}d\mu \Bigg]\\ &=\Bigg[\underset{x\in X}{\sum}f^+_{Re}(x)+ \underset{x\in X}{\sum}f^-_{Re}(x)\Bigg]+i\Bigg[ \underset{x\in X}{\sum}f^+_{Im}(x) - \underset{x\in X}{\sum}f^-_{Im}(x) \Bigg] \end{align*}

$\color{red}{\textbf{the case $X=\mathbb{N}$}}$

now, if $X=\mathbb{N}$, then for function $f\geq 0$, we have

$\int_{\mathbb{N}}fd\mu=\underset{n\in \mathbb{N}}{\sum}f(n)$

and integral for complex function, is defined as above.

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