# Counting measure and Lebesgue intregal

Consider $$f: \mathbb {N} \rightarrow \mathbb{R}$$

I want to compute the Lebesgue integral with counting measure.

$$\int_N f d \mu$$

Why does it make sense to integrate over N and not for example R?

• Go through definition
– MANI
Oct 8, 2019 at 19:33
• Because your measure space is $\mathbb N.$
– zhw.
Oct 8, 2019 at 19:46

The counting measure can be defined on any $$(A, \mathcal P(A))$$. See Wikipedia article for more details.
Now, if your function $$f$$ is only defined on $$\mathbb N$$, the integral will only make sense on $$\mathbb N$$ also. And its value will be $$\displaystyle \sum_{n \in \mathbb N} f(n)$$.
• You would get $\int_{\mathbb R} f$... but knowing that $\int$ isn’t in that case the Lebesgue integral but $\int_{\mathbb R} f =\sum_{x \in \mathbb R} f(x)$ with the meaning of $\sum$ provided in the Wikipedia article I referenced. In particular, as soon as $f$ has an uncountable number of non vanishing values, $\int \vert f\vert =\infty$. Oct 8, 2019 at 20:26