The Counting Measure

# The Counting Measure

Recall from the General Measurable Spaces and Measure Spaces page that if $X$ is a set and $\mathcal A$ is a $\sigma$-algebra on $X$ then the pair $(X, \mathcal A)$ is called a measurable space. We said that the sets $E \in \mathcal A$ are called measurable sets.

We said that a function $\mu : \mathcal A \to [0, \infty]$ is called a measure on $\mathcal A$ if $\mu (\emptyset) = 0$ and if for every countable collection of disjoint measurable sets $(E_n)$ we have that $\displaystyle{\mu \left ( \bigcup_{n=1}^{\infty} E_n \right ) = \sum_{n=1}^{\infty} \mu (E_n)}$, and we defined the triple $(X, \mathcal A, \mu)$ to be a measure space.

We now look a special measure called the counting measure.

 Definition: Let $X$ be any set and let $\mathcal P(X)$ denote the power set of $X$. The Counting Measure $c : \mathcal P(X) \to [0, \infty]$ on $\mathcal P(X)$ is defined for all $E \in \mathcal P(X)$ by $c(E) = |E|$, and the triple $(X, \mathcal P(X), c)$ is called a Counting Measure Space.

Let $X$ be any set. We will verify that $(X, \mathcal P(X), c)$ is a measure space.

First we have that:

(1)
\begin{align} \quad c(\emptyset) = |\emptyset| = 0 \end{align}

Now let $(E_n)_{n=1}^{\infty}$ be any countable collection of mutually disjoint sets in $\mathcal P(X)$. There are three cases to consider.

Case 1: Suppose that for some $k \in \mathbb{N}$ we have that $|E_k| = \infty$. Then clearly:

(2)
\begin{align} \quad c \left ( \bigcup_{n=1}^{\infty} E_n \right ) = \biggr \lvert \bigcup_{n=1}^{\infty} E_n \biggr \rvert = \infty = |E_k| = \sum_{n=1}^{\infty} |E_n| = \sum_{n=1}^{\infty} c(E_n) \end{align}

Case 2: Suppose that for all $n \in \mathbb{N}$ we have that $|E_n| < \infty$ and that there are infinitely many $n$ such that $|E_n| > 0$. $\displaystyle{\bigcup_{n=1}^{\infty} E_n}$ must contain infinitely many elements, and $\displaystyle{\sum_{n=1}^{\infty} |E_n| = \infty}$ as it is a divergent series of positive numbers. Therefore:

(3)
\begin{align} \quad c \left ( \bigcup_{n=1}^{\infty} E_n \right ) = \sum_{n=1}^{\infty} c(E_n) \end{align}

Case 3: Suppose that for all $n \in \mathbb{N}$ we have that $|E_n| < \infty$ and that there are finitely many $n$ such that $|E_n| > 0$, say these sets are $\{ E_1, E_2, ..., E_N \}$ without loss of generality, and that $|E_k| = t_k$ for each $k \in \{ 1, 2, ..., N \}$. Since $(E_n)_{n=1}^{\infty}$ is a collection of mutually disjoint sets, we have that:

(4)
\begin{align} \quad c \left ( \bigcup_{n=1}^{\infty} E_n \right ) = \biggr \lvert \bigcup_{n=1}^{\infty} E_n \biggr \rvert = \biggr \lvert \bigcup_{n=1}^N E_n \biggr \rvert = \sum_{n=1}^{N} t_n = \sum_{n=1}^{N} |E_n| = \sum_{n=1}^{\infty} |E_n| = \sum_{n=1}^{\infty} c(E_n) \end{align}

Therefore $\mu$ is a measure on $\mathcal P(X)$ and $(X, \mathcal P(X), c)$ is a measure space.