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 $\mu$.

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:


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

Proof of $(2)$
It is to be shown that (for a sequence $\left({S_n}\right)_{n \in \N}$ of pairwise disjoint sets):


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

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

Then the cardinality of:


 * $\left({\bigcup_{n \mathop = 1}^\infty S_n}\right$

is infinite.

Hence:


 * $\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:


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

It remains to be shown that the case holds where no $S_i$ is infinite.

Proof of $(3')$
Note that $\varnothing \in \Sigma$ as $\Sigma$ is a $\sigma$-algebra.

By definition the empty set has precisely zero elements:


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

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