Counting Measure

2.1.4 - Real Analysis: Royden

A set function $c$, defined on all subsets of $\mathbb{R}$, is defined as follows. Define $c(E)$ to be $\infty$ if $E$ has infintely many members and $c(E)$ to be equal to the number of elements in $E$ if $E$ is finite; define $c(\emptyset)=0$. Show that $c$ is a countably additive and translation invariant set function. This set function is called the counting measure.

Proof:

First, we show that $c$ is countably additive. Let ${E_k}_{k=1}^{\infty}$ be a collection of nonempty (empty subsets will not affect the measure), pairwise disjoint subsets of $\mathbb{R}$. Note that

\[U = \bigcup_{k=1}^{\infty}E_k\]

has infinitely many members, so $c(U) = \infty$ regardless of the finiteness of each $E_k$. Hence,

\[\begin{aligned} \sum_{k=1}^{\infty}c(E_k) &= \sum_{k=1}^{\infty}|E_k| \\ &= \infty \\ &= c(U) \end{aligned}\]

Next, consider a finite collection of nonempty, pairiwise disjoint subsets. It suffices to consider to two, say, $E_1$ and $E_2$. If both are finite, then, we have

\[\begin{aligned} c(E_1) + c(E_2) &= |E_1| + |E_2| \\ &= |E_1 \cup E_2| \\ &= c(E_1 \cup E_2) \end{aligned}\]

Without loss of generality, suppose that $E_1$ has infinitely many members. Then,

\[\begin{aligned} c(E_1) + c(E_2) &= \infty \\ &= c(E_1 \cup E_2) \end{aligned}\]

Hence, $c$ is countably additive. Next, we show that $c$ is translation invariant. Suppose that $E \subset \mathbb{R}$ has finitely many members and let $a \in \mathbb{R}$. Since $a$ only translates each member of $E$, we have that the cardinality is unchanged:

\[c(E + a) = |E| = c(E)\]

Similarly, if $E$ has infinitely many members, then

\[c(E+a) = \infty = c(E)\]

Therefore, $c$ is also translation invariant.