Notation for counting measure in the case of two infinite sums, one being converted to an integral over $\mathbb{N}$?

Question

Using the properties of Lebesgue integration and the counting measure $\mu:\mathbb{N}\to [0,\infty]$ it can be shown that $$ \sum_{i=1}^\infty\bigg(\sum_{j=1}^\infty g_{ij}\bigg) = \sum_{j=1}^\infty\bigg(\sum_{i=1}^\infty g_{ij}\bigg), $$ for every map $\mathbb{N}\times\mathbb{N}\to [0,\infty]:(i,j) \to g_{ij}$,

Let $g_{ij}:\mathbb{N}\times\mathbb{N}\to [0,\infty]:(i,j) \to g_{ij}$ be a map. Then we have, \begin{align*} \sum_{i=1}^\infty\bigg(\sum_{j=1}^\infty a_{ij}\bigg) & = \int_\mathbb{N} \sum_{j=1}^\infty a_{ij} d\mu \\ & = \sum_{j=1}^\infty \int_\mathbb{N} a_{ij} d\mu \\ & = \sum_{j=1}^\infty\bigg(\sum_{i=1}^\infty a_{ij}\bigg). \end{align*} Is my notation correct for using the counting measure on the natural numbers? In particular, how do we know that the integral over $\mathbb{N}$ is signifying that the elements of $\mathbb{N}$ should be fed into the $i$ component of the map $a_{ij}$?

Answer 1

You could write the counting measure as $\ \#: \>{\cal P}({\mathbb N})\to[0,\infty]$ and then write the integral as $$s_i=\int_{\mathbb N} a_{ij}\>d\#(j)\ .$$

Answer 2

It is implicit. For instance, when you integrate $f$ with respect to some measure $\mu$ as $\int f d\mu$, it is assumed that you are feeding the measure into the integrand.

So you can write this as $$\int_{\mathbb N} \sum_{j=1}^\infty a_j d\mu$$

or $$\int_{\mathbb N} \sum_{j=1}^\infty a_{ij} d\mu(i).$$