Counting Measure on Natural Numbers is Sigma-Finite

Theorem

Let $\mu$ be the counting measure on $\struct {\N, \map \PP \N}$.

For each $n \in \N$, define:

$X_n = \N \cap \closedint 1 n = \set {1, 2, \ldots, n}$

Since $X_n \subseteq \N$ for each $n$ from Intersection is Subset, we have that:

We show that $\sequence {X_n}_{n \mathop \in \N}$ is an exhausting sequence in $\map \PP \N$ and that:

$\map \mu {X_n} < \infty$ for each $n \in \N$.

From the definition of counting measure, we also have:

$\map \mu {X_n} = n < \infty$

for each $n$.

We have that:

$\N \cap \closedint 1 n \subseteq \N \cap \closedint 1 {n + 1}$

$X_n \subseteq X_{n + 1}$

So $\sequence {X_n}_{n \mathop \in \N}$ is an increasing sequence of $\map \PP \N$-measurable sets.

It now only remains to verify:

$\ds \N = \bigcup_{n \mathop = 1}^\infty X_n$

We clearly have:

$\ds \bigcup_{n \mathop = 1}^\infty X_n \subseteq \N$

For each $k \in \N$, we have $k \in X_k$, so we also have:

$\ds \N \subseteq \bigcup_{n \mathop = 1}^\infty X_n$

so that:

$\ds \N = \bigcup_{n \mathop = 1}^\infty X_n$

So $\sequence {X_n}_{n \mathop \in \N}$ is exhausting sequence in $\map \PP \N$ and that:

$\map \mu {X_n} < \infty$ for each $n \in \N$.

So:

$\blacksquare$