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 $X$?

**Answer**

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.

**Attribution***Source : Link , Question Author : Community , Answer Author : Nick Peterson*