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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?

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

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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 $\displaystyle \sum_{n \in \mathbb N} f(n)$.

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  • $\begingroup$ Ok, when f is defined on R, what would I get instead? $\endgroup$
    – Steven33
    Oct 8, 2019 at 20:22
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    $\begingroup$ 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, $\int \vert f\vert =\infty$. $\endgroup$ Oct 8, 2019 at 20:26
  • $\begingroup$ Thank you very much:) $\endgroup$
    – Steven33
    Oct 8, 2019 at 20:29

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