Counting measure is a measure

One has ${\mathrm{\#}}_X(\mathrm{\varnothing })=0$. For every $n$, disjointness gives ${\mathrm{\#}}_X({\bigcup }_{k<n}{E}_k)={\sum }_{k<n}{\mathrm{\#}}_X(E_k)$ whenever all those sets are finite.

If some $E_r$ is infinite, then ${\bigcup }_k{E}_k$ is infinite and both ${\mathrm{\#}}_X({\bigcup }_k{E}_k)$ and ${\sum }_k{\mathrm{\#}}_X(E_k)$ are $+\mathrm{\infty }$.

Suppose every $E_k$ is finite. If $E:={\bigcup }_k{E}_k$ is finite, only finitely many pairwise disjoint $E_k$ can be nonempty, and step 1.1 gives ${\mathrm{\#}}_X(E)={\sum }_k{\mathrm{\#}}_X(E_k)$.

Suppose every $E_k$ is finite but $E$ is infinite. For every $m\in \mathbb{N}$, the set $E$ contains more than $m$ distinct points; the finitely many indices of the $E_k$ containing those points have a strict upper bound $n$ (take one more than their maximum), so step 1.1 gives ${\sum }_{k<n}{\mathrm{\#}}_X(E_k)>m$. Hence the partial sums are unbounded and their supremum is $+\mathrm{\infty }={\mathrm{\#}}_X(E)$.

Steps 1.2, 2.1 and 2.2 cover all possibilities for the union, so countable additivity holds; with ${\mathrm{\#}}_X(\mathrm{\varnothing })=0$ from step 1.1, [L2] proves that ${\mathrm{\#}}_X$ is a measure.