Sums and integration against counting measure: Part IIIa

For other posts in this series, see
https://explaining-maths.blogspot.com/search?q=Sums+and+integration+against+counting+measure

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In the last post we saw that, when we sum a series of non-negative extended real numbers, whichever order we choose to add the terms in we still obtain the same result, namely the integral of the relevant function against (or with respect to) counting measure.

In this post we will look at adding up an infinite array of non-negative extended real numbers. 

Recall that, for us, $\N=\{1,2,3,\dots\}$. For convenience we denote by $\N_0$ the set of non-negative integers, so that  $\N_0=\N\cup\{0\}=\{0,1,2,\dots\}$.

Below we will want to use counting measure on various sets simultaneously, so we use a variety of symbols. However the sigma algebras we use will always be the whole of the power set of each set involved, so that all sets (and all functions) will be measurable below.

Set $X=\N_0 \times \N_0$. Throughout this post, we assume that we have an array of non-negative extended real numbers $a_{m,n} \in [0,\infty]\jq(m,n \in \mathbb{N_0})$. Associated with this array (or rather essentially the same thing as this array) is a function $F:X\rightarrow [0,\infty]$ defined by
$$F(m,n)=a_{m,n}\jq(m,n \in \mathbb{N_0})\,.$$ In theory we should probably write $F((m,n))$ here, but it is standard to omit the extra parentheses.

We also have sequences of functions $(f_m)_{m=0}^\infty$ and $(g_n)_{n=0}^\infty$ from $\N_0\to[0,\infty]$ defined by
$$f_m(n)=g_n(m)=F(m,n)=a_{m,n}\jq(m,n \in \mathbb{N_0})\,.$$

The results most often used are probably identities such as
$$\sum_{m=0}^\infty \sum_{n=0}^\infty a_{m,n} = \sum_{n=0}^\infty \sum_{m=0}^\infty a_{m,n} = \sum_{k=0}^\infty \sum_{j=0}^{k} a_{j,k-j}\,,\tag{1}$$

and variants of these. Again you can think of this type of result as a combination of being able to add up our non-negative extended real numbers in any order, and also to group them however we want.

So, how do these results fit with counting measure and the standard theorems of Measure and Integration? Well, the first equality in $(1)$ above looks like some form of Fubini's Theorem, and that is certainly one way to approach it. We will take a slightly more elementary approach. But still, let's start with two copies of counting measure, say $\mu_1$ and $\mu_2$, on $\N_0$, and let's also take counting measure $\nu$ on $\N_0 \times \N_0 = X$. If we planned to use Fubini's Theorem, we would note that the product measure of $\mu_1$ with $\mu_2$ is just $\nu$.

We can decompose $X$ as a countable disjoint union of sets in various ways. For each $n\in\N_0$, set $A_n=\N_0 \times \{n\}$, and for each $m \in \N_0$ set $B_m=\{m\}\times \N_0$. Also, for each $k\in\N_0$, set $C_k=\{(j,k-j): j=0,1,\dots,k\,\}$. Then
$$X=\N_0\times\N_0=\bigcup_{n=0}^\infty{A_n}=\bigcup_{m=0}^\infty{B_m}=\bigcup_{k=0}^\infty{C_k}\,,$$
and in each of these three unions the sets involved are pairwise disjoint. For the last of these, note that $C_k$ can also be written as $\{(m,n)\in\N_0\times\N_0: m+n=k\}$. 

The equalities in $(1)$ can all be obtained from the fact that each of these double sums is equal to the integral of $F$ over $X$ with respect to counting measure,  $\displaystyle \int_X F\,\textrm{d}\nu$. However the first equality can also be obtained by considering the integral with respect to counting measure of either of the functions $\sum_{m=0}^\infty f_m$ or $\sum_{n=0}^\infty g_n$.

To be continued in part IIIb