Sums and integration against counting measure: Appendix 3: Defining sums over sets without using Measure and Integration

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 this appendix, I'll look at defining sums over general sets without referring to Measure and Integration. Our starting point will be the following.

Let $X$ be a set and let $f$ be a function from $X$ to $[0,\infty]$.

Although we are using non-negative extended real numbers, nevertheless the arguments here will have more of a first-year flavour than some of the other posts in this particular series, while continuing with the theme of bootstrapping from results for finite sets to results for general sets.

If $X$ is a finite set, we define $\sum_{x \in X} f(x)$ in the usual way, and we take it as standard that we obtain the same value whichever order we add the values. (By convention, the empty sum is $0$.) 

If $X$ is an infinite set, for the purposes of this post we define 

\[\sum_{x \in X} f(x) = \sup \left\{ \sum_{x \in E} f(x): E \textrm{ is a finite subset of } X \right\}.\tag{$1$}\]

We take it as standard that, because the values of $f$ are non-negative, we have the following monotonicity result: if $E$ and $F$ are finite subsets of $X$ with $E\subseteq F$, then 

\[\sum_{x \in E} f(x) \leq \sum_{x \in F} f(x)\,.\tag{$2$}\]

We note that ($1$) is then valid for all sets $X$ and all functions $f:X\to[0,\infty]$, and that ($2$) holds for all (not necessarily finite) sets $E$, $F$ such that $E\subseteq F \subseteq X$.

(If you want to check those last claims, you may may want to be careful not to overuse $E$ in your proof! In ($1$) $E$ is a 'dummy variable', so you can use a different symbol when you need to.)

See earlier posts for a discussion of why this version of the definition of sum agrees with the definitions we used earlier. But here is a quick standalone argument for the case of series, when $X=\N$. (We did something similar in an earlier post.)

Let $f :\N \to [0,\infty]$, and set 

\[M=\sup \left\{ \sum_{x \in E} f(x): E \textrm{ is a finite subset of } \N \right\}\,.\]

Our aim is to use results for finite sets to prove that

\[\sum_{n=1}^\infty f(n) = M\,.\]

For each $N\in\N$, set $E_N=\{1,2,\dots,N\}$ (so that $E_N$ is a finite subset of $\N$) and set 

\[S_N=\sum_{x\in E_N} f(x) = \sum_{n=1}^N f(n)\,.\]

We see that this sequence of partial sums $S_N$ is monotone increasing, and (by definition of the sum of a series, and the fact that the $S_N$ are monotone increasing)

\[\sum_{n=1}^\infty f(n) = \lim_{N\to \infty} S_N = \sup\{S_N:N\in\N\,\}\,.\tag{$3$}\]

By definition of $M$, we have $S_N\leq M$ for all $N \in\N$, and so ($3$) gives us

\[\sum_{n=1}^\infty f(n) \leq M\,.\]

For the reverse inequality, let $E$ be a finite subset of $\N$. Then there exists an $N \in \N$ such that $E\subseteq E_N$, and then (by  ($2$) and ($3$))

\[\sum_{x \in E} f(x) \leq \sum_{x \in E_N} f(x) = S_N \leq \sum_{n=1}^\infty f(n)\,.\]

Thus, for all finite sets $E\subseteq \N$, we have

\[\sum_{x \in E} f(x) \leq \sum_{n=1}^\infty f(n)\,.\]

It follows (by the definition of $M$ as a supremum) that

\[M \leq \sum_{n=1}^\infty f(n)\,,\]

as required. So equality holds.$\quad\square$

That is a special case of bootstrapping from results for finite sets to results for infinite sets.

Defining sums over general sets in this way allows us to give a version of Fatou's Lemma for sums without any mention of Measure and Integration. We'll do that in Appendix 5. But first, in Appendix 4, we'll fill in some details about $\liminf$ that were claimed in Part IV.