Sums and integration against counting measure: Appendix 1: Proof that counting measure really is a measure

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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

Counting measure is a very standard example of a measure, but not everyone checks the details. I'm going to attempt to give a fairly efficient proof! But here are a few comments first about what we will be assuming before we start.

  1. By convention the empty set, $\emptyset$, is a finite set, and the number of points in $\emptyset$ is $0$.
  2. For a finite set $A$, we denote the cardinality of $A$ (number of points in $A$) by $|A|$. Here $|A|$ is a non-negative integer, and $|A|=0$ if and only if $A=\emptyset$.
  3. A finite union of finite sets is finite.
  4. If we have finitely many pairwise disjoint finite sets $A_1, A_2, \dots, A_m$, then
    \[\left|\bigcup_{k=1}^m A_k \right| = \sum_{k=1}^m |A_k|\,.\]
It is, of course, possible to start further back, but we do have to start somewhere!

Now let $X$ be a set, and denote the power set of $X$ by $\Pset(X)$. Define counting measure $\mu$ from $\Pset(X)$ to $[0,\infty]$ as follows. For $A\in \Pset(X)$, we define
\[\mu(A)=\begin{cases}|A|\quad&\textrm{if $A$ is a finite set},\\ \infty&\textrm{otherwise}.\end{cases}\]

Let's prove that $\mu$ really is a measure.
First note that (by our first comment above), $\mu(\emptyset)=|\emptyset|=0$. So we just need to check countable additivity.
Let $(A_k)_{k=1}^\infty$ be a sequence of pairwise disjoint subsets of $X$, and set
\[A=\bigcup_{k=1}^\infty A_k\,.\]
We have to show that
\[\mu(A)=\sum_{k=1}^\infty \mu(A_k)\,.\tag{$1$}\]
  • First note that if any of the sets $A_k$ are infinite, then so is $A$, and both sides of ($1$) are clearly equal to $\infty$.
  • Similarly, if infinitely many of the (pairwise disjoint) sets $A_k$ are non-empty, then the set $A$ is infinite and both sides of ($1$) are infinity.
  • The only remaining case is when all of the sets $A_k$ are finite, and only finitely many of the $A_k$ are non-empty. In this case there is a natural number $m$ such that $A_k=\emptyset$ for all $k>m$, and then we have
    \[A=\bigcup_{k=1}^m A_k\,.\]
    But then, by our 3rd and 4th comments above, $A$ is a finite set and
    \[|A|=\left|\bigcup_{k=1}^m A_k \right| = \sum_{k=1}^m |A_k|\,,\] and so
    \[\mu(A) = \sum_{k=1}^m \mu(A_k) = \sum_{k=1}^\infty \mu(A_k)\,,\] as required.
For that last equality, note that $\mu(A_k)=\mu(\emptyset)=0$ for all $k>m$.

That does it!

Note that what we have done, in some sense, is to bootstrap up from results for finitely many finite sets to the general case. This is, in some sense, the flavour of the whole topic (as seen in the earlier posts).

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