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

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For other posts in this series, see

https://explaining-maths.blogspot.com/search?q=Sums+and+integration+against+counting+measure

In case the MathJax mathematics below does not display properly in your browser, I will make a PDF of the main article available via my WordPress blog

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,∞].

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 ∑x∈Xf(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 

∑x∈Xf(x)=sup{∑x∈Ef(x):E is a finite subset of X}.

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⊆F, then 

∑x∈Ef(x)≤∑x∈Ff(x).

We note that (1) is then valid for all sets X and all functions f:X→[0,∞], and that (2) holds for all (not necessarily finite) sets E, F such that E⊆F⊆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→[0,∞], and set 

M=sup{∑x∈Ef(x):E is a finite subset of N}.

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

∞∑n=1f(n)=M.

For each N∈N, set EN={1,2,…,N} (so that EN is a finite subset of N) and set 

SN=∑x∈ENf(x)=N∑n=1f(n).

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

∞∑n=1f(n)=limN→∞SN=sup{SN:N∈N}.

By definition of M, we have SN≤M for all N∈N, and so (3) gives us

∞∑n=1f(n)≤M.

For the reverse inequality, let E be a finite subset of N. Then there exists an N∈N such that E⊆EN, and then (by  (2) and (3))

∑x∈Ef(x)≤∑x∈ENf(x)=SN≤∞∑n=1f(n).

Thus, for all finite sets E⊆N, we have

∑x∈Ef(x)≤∞∑n=1f(n).

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

M≤∞∑n=1f(n),

as required. So equality holds.◻

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 lim inf that were claimed in Part IV.

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