Sums and integration against counting measure: Part IIIb

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

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This post follows on from Part IIIa, where we introduced the notation and explained the setting we are working in. In particular, we are working with sets $\N=\{1,2,3,\dots\}$, $\N_0=\N\cup\{0\}=\{0,1,2,\dots\}$, $X=\N_0 \times \N_0$, and with an array of non-negative extended real numbers $a_{m,n} \jq(m,n \in \mathbb{N_0})$. Our aim is to prove that 

$$\sum_{m=0}^\infty \sum_{n=0}^\infty a_{m,n} = \sum_{n=0}^\infty \sum_{m=0}^\infty a_{m,n} = \sum_{k=0}^\infty \sum_{j=0}^{k} a_{j,k-j}\,.\tag{1}$$

We take $\nu$ to be counting measure on $X=\N_0 \times \N_0$, and define $F:X\rightarrow [0,\infty]$ by $F(m,n)=a_{m,n}\jq(m,n \in \mathbb{N_0})\,.$

We plan to prove that all of the sums in $(1)$ are equal to  $\displaystyle \int_X F\,\textrm{d}\nu$ (the integral of $F$ over $X$ with respect to counting measure).

Let $m\in\N_0$, and set $B_m=\{m\}\times \N_0$. Then $B_m$ is a countable disjoint union of a sequence of single-point sets
$$B_m=\bigcup_{n=0}^\infty\{(m,n)\}\,.$$ So

$$\int_{B_m} F\,\rd\nu = \sum_{n=0}^\infty F(m,n) =  \sum_{n=0}^\infty a_{m,n}\,.$$

But $X$ is the countable disjoint union of the sequence of sets $B_m$, and so

$$\int_X F\,\rd\nu = \sum_{m=0}^\infty \int_{B_m} F\,\rd\nu = \sum_{m=0}^\infty \sum_{n=0}^\infty a_{m,n} \,.$$

Similarly, setting $A_n=\N_0 \times \{n\} \jq (n \in \N_0)$, we obtain

$$\int_{A_n} F\,\rd\nu = \sum_{m=0}^\infty F(m,n) =  \sum_{m=0}^\infty a_{m,n}\,,$$ and
$$\int_X F\,\rd\nu = \sum_{n=0}^\infty \int_{A_n} F\,\rd\nu = \sum_{n=0}^\infty \sum_{m=0}^\infty a_{m,n} \,.$$

Finally, for each $k\in\N_0$, we set $C_k=\{(j,k-j): j=0,1,\dots,k\,\}=\{(m,n)\in\N_0\times\N_0: m+n=k\}$. Then $X$ is again the disjoint union of the $C_k$, and

$$\int_X F\,\rd\nu = \sum_{k=0}^\infty \int_{C_k} F\,\rd\nu = \sum_{k=0}^\infty \sum_{j=0}^{k} a_{j,k-j}\,,$$ as required.

This finishes the proof that all three of the sums in (1) are equal to $\int_X F\,\rd\nu$, and so they are also equal to each other!

The first equality in (1) can also be proved (in two different ways) by looking at the integral of a sum of non-negative functions over $\N_0$ with respect to counting measure $\mu$ on $\N_0$. We have sequences of functions $(f_m)_{m=0}^\infty$ and $(g_n)_{n=0}^\infty$ from $\N_0\to[0,\infty]$ defined by

$$f_m(n)=g_n(m)=F(m,n)=a_{m,n}\jq(m,n \in \mathbb{N_0})\,.$$

For each $m\in\N_0$, we have

$$\int_{\N_0}f_m\,\rd \mu = \sum_{n=0}^\infty f_m(n) = \sum_{n=0}^\infty a_{m,n}\,.$$

Set $f=\sum_{m=0}^\infty f_m$, so that, for each $n \in \N_0$, we have

$$f(n)=\sum_{m=0}^\infty a_{m,n}\,.$$

Then $$\int_{\N_0} f\,\rd\mu = \sum_{n=0}^\infty f(n) = \sum_{n=0}^\infty \sum_{m=0}^\infty a_{m,n}\,.$$

But standard theory tells us that the integral of a sum of non-negative (measurable) functions is the sum of the integrals. So

$$\int_{\N_0} f\,\rd\mu = \sum_{m=0}^\infty \int_{\N_0} f_m\,\rd\mu = \sum_{m=0}^\infty  \sum_{n=0}^\infty a_{m,n}\,.$$

The first equality in (1) now follows. We can also prove it (essentially the same way) by looking at the integral of the function $g=\sum_{n=0}^\infty g_n$.

At this point you may well wonder: are these genuine proofs, or are they circular? Do we already need facts about rearranging series in order to prove the theorems of Measure and Integration implicitly used above? Well, don't take my word for it, but I claim that the answer is no, these proofs are not circular. You do need to assume that you can rearrange the order in unions of indexed collections of sets without changing the union, but that is much more elementary. Once you have that, then countable additivity of measures automatically allows you to rearrange any sums of series coming from the measure of a disjoint union of a sequence of sets. (That gives yet another approach to all of these equalities!) However, if you haven't already done it, you should prove for yourself that counting measure really is a measure. Perhaps I'll give a short proof of that in another post.

I am not suggesting that this is a good way to teach first or second-year undergraduate students results about series of non-negative real numbers! But students of Measure and Integration may find it worth looking back at the elementary proofs they saw earlier in their studies, and comparing those proofs with the proofs above (along with the proofs of the results about integrals implicitly used above). 

In Part IV we will look at the three big convergence theorems for Lebesgue integration relating to sequences of functions. We'll see that they all fail, in general, if you work with nets of functions instead, but that they remain true for nets if you work with counting measure.