Sums and integration against counting measure: Appendix 2: Proof of a claim about integrals from Part I

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

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Let $\mu$ be counting measure on (the power set of) a set $X$, and let $f$ be a function from $X$ to $[0,\infty]$.

In Part I we defined \[\displaystyle \sum_{x \in X} f(x) = \int_X f\,\mathrm{d}\mu\,,\] and we noted that this agreed with our usual notions if the set $X$ is countable. We then claimed that (even if $X$ is uncountable), we have

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

The aim of this post is to prove ($1$). In terms of integrals, we can rewrite $(1)$ as

\[\int_X f\,\rd\mu= \sup \left\{\int_E f\,\rd\mu: E \textrm{ is a finite subset of } X \right\}\,.\tag{$2$}\]

This is the form we used in Part IV of this series.

Set

\[S= \sup \left\{ \sum_{x \in E} f(x): E \textrm{ is a finite subset of } X \right\}\,,\] the right hand side of ($1$). It is clear that $S$ is equal to the right hand side of ($2$), and that this is less than or equal to the left hand side of $(2)$ (or ($1)$). So we need the reverse inequality,
\[\int_X f\,\rd\mu \leq S\,,\tag{$3$}\] or equivalently 

\[\sum_{x \in X} f(x) \leq S\,,\tag{$3'$}\]

and we focus on proving these below.

I'm not quite sure what the most efficient route is here. One approach is to consider non-negative, simple functions which are less than or equal to $f$, and that is what I first intended. But another approach is to bootstrap from the finite case, through the countably infinite case to the general case. Let's look at that approach instead, as it somehow fits well with some of the other things we have been doing.

The inequalities ($3)$ and ($3'$) are clear if $X$ is finite.

Suppose that $X$ is countably infinite, say $X=\{x_1,x_2,\dots\}$ (with the $x_i$ distinct). For each $n\in\N$, set $E_n=\{x_1,x_2,\dots,x_n\}$. Then (since $E_n$ is a finite subset of $X$)

\[\sum_{i=1}^n f(x_i)=\sum_{x \in E_n} f(x) \leq S\,.\]

Letting $n\to \infty$ gives us \[\sum_{i=1}^\infty f(x_i) \leq S\,,\] and so ($3'$) holds,

\[\sum_{x \in X} f(x) \leq S\,.\]

The remaining case is when $X$ is uncountably infinite.

Set $Y=\{x\in X: f(x)>0\}$, and, for each $n \in \N$, set $Y_n=\{x \in X: f(x)\geq 1/n\,\}$. Then we have

\[Y=\bigcup_{n \in \N} Y_n\,,\]

and

\[\sum_{x \in X} f(x)=\int_X f\,\rd\mu = \int_Y f\,\rd\mu = \sum_{x \in Y} f(x)\,.\tag{$4$}\]

(Note that $f(x)=0$ for all $x$ in $X\setminus Y$.)

We consider two cases.

• If all of the sets $Y_n$ are finite, then $Y$ is countable, and so we can use what we already know about countable sums. Since $Y$ is countable, we have
  \[\sum_{x \in Y} f(x) \leq \sup \left\{ \sum_{x \in E} f(x): E \textrm{ is a finite subset of } Y \right\} \leq S.\]
  Combining this with ($4$) then gives us ($3$) and ($3'$) for $X$.

• Otherwise, there exists an $n$ such that $Y_n$ is infinite. In this case, we claim that $S=\infty$, so that ($3$) and ($3'$) are trivially true.
  Let $m \in \N$. Then there is a finite set $E \subseteq Y_n$ such that $E$ has $m$ elements. But then (since $f(x) \geq 1/n$ for all $x \in E$)
  \[\int_E  f\,\rd\mu \geq m/n\,,\] and so $S \geq m/n$.
  This holds for all $m \in \N$, and so $S=\infty$, as claimed.

The result follows. $\quad\square$

For the last bullet point, an alternative is to take a countably infinite subset $A$ of $Y_n$, and use what we know about countable sums.

Perhaps another approach is to give $X$ the discrete metric, so that $X$ is a locally compact metric space, and then the finite sets are the same as the compact subsets. (But if $X$ is uncountable then it won't be $\sigma$-compact here.) Our fact can then be interpreted in terms of (what is sometimes called) the inner regularity of the measure $f\rd\mu$. Whether that provides a good alternative route to a proof is unclear to me at the moment, but at least it is another way to think about it. 

You may have noticed that in our sums/integrals with respect to counting measure, it looks to be rather trivial if the function $f$ actually takes the value $\infty$ at some point of the domain of integration. Generally the results are more useful when we apply them to non-negative real-valued functions. Nevertheless, when we look at double sums of arrays of non-negative extended real numbers $a_{m,n}$, even if none of the numbers themselves are $\infty$, it is possible that one or more of the inner sums could be infinity. So it is probably just as well to allow the value infinity throughout the discussion.