# Counting Measure is Measure

## Theorem

Let $\left({X, \Sigma}\right)$ be a measurable space.

Then the counting measure $\left\vert{\cdot}\right\vert$ on $\left({X, \Sigma}\right)$ is a measure.

## Proof

Let us verify the measure axioms $(1)$, $(2)$ and $(3')$ for $\left\vert{\cdot}\right\vert$.

### Proof of $(1)$

The values that $\left\vert{\cdot}\right\vert$ can take are the natural numbers $\N$ and $+\infty$.

All of these are positive, whence:

- $\forall S \in \Sigma: \left\vert{S}\right\vert \ge 0$

$\Box$

### Proof of $(2)$

It is to be shown that (for a sequence $\left({S_n}\right)_{n \in \N}$ of pairwise disjoint sets):

- $\displaystyle \sum_{n \mathop = 1}^\infty \left\vert{S_n}\right\vert = \left\vert{\bigcup_{n \mathop = 1}^\infty S_n}\right\vert$

Suppose that the cardinality of at least one $S_i$ is infinite.

Then the cardinality of:

- $\displaystyle \bigcup_{n \mathop = 1}^\infty S_n$

is infinite by Subset of Finite Set is Finite.

Hence:

- $\displaystyle \mu \left({\bigcup_{n \mathop = 1}^\infty S_n}\right) = +\infty$

Now as:

- $\mu \left( {S_i} \right) = +\infty$

it follows by the definition of extended real addition that:

- $\displaystyle \sum_{n \mathop = 1}^\infty \mu \left({S_n}\right) = +\infty$

Suppose now that all $S_i$ are finite, and $\displaystyle \sum_{n \mathop = 1}^\infty \left\vert{S_n}\right\vert$ converges.

Then by Convergent Series of Natural Numbers, it follows that for some $N \in \N$:

- $\forall m \ge N: \left\vert{S_m}\right\vert = 0$

which by Cardinality of Empty Set means $S_m = \varnothing$.

Therefore we conclude that:

- $\displaystyle \sum_{n \mathop = 1}^\infty \left\vert{S_n}\right\vert = \sum_{n \mathop = 1}^N \left\vert{S_n}\right\vert$

and by Union with Empty Set:

- $\displaystyle \bigcup_{n \mathop = 1}^\infty S_n = \bigcup_{n \mathop = 1}^N S_n$

By Cardinality of Set Union: Corollary:

- $\displaystyle \left\vert{\bigcup_{n \mathop = 1}^N S_n}\right\vert = \sum_{n \mathop = 1}^N \left\vert{S_n}\right\vert$

and combining this with the above yields the desired identity.

Suppose finally that $\displaystyle \sum_{n \mathop = 1}^\infty \left\vert{S_n}\right\vert$ diverges.

To establish the desired identity, it is to be shown that $\displaystyle \bigcup_{n \mathop = 1}^\infty S_n$ is infinite.

Suppose to the contrary that it has finite cardinality, say $k$.

By Cardinality of Set Union: Corollary, for each $N \in \N$:

- $\displaystyle \sum_{n \mathop = 1}^N \left\vert{S_n}\right\vert = \left\vert{\bigcup_{n \mathop = 1}^N S_n}\right\vert$

Now since:

- $\displaystyle \bigcup_{n \mathop = 1}^N S_n \subseteq \bigcup_{n \mathop = 1}^\infty S_n$

it follows by Cardinality of Subset of Finite Set that, for each $N \in \N$:

- $\displaystyle \left\vert{\bigcup_{n \mathop = 1}^N S_n}\right\vert \le k$

which contradicts the assumption that $\displaystyle \sum_{n \mathop = 1}^\infty \left\vert{S_n}\right\vert$ diverges.

Therefore $\displaystyle \bigcup_{n \mathop = 1}^\infty S_n$ is infinite and the identity follows.

$\Box$

### Proof of $(3')$

- $\left\vert{\varnothing}\right\vert = 0$

$\Box$

Having verified the axioms, it follows that $\left\vert{\cdot}\right\vert$ is a measure.

$\blacksquare$