# Co-Countable Measure is Measure

## Theorem

Let $X$ be an uncountable set.

Let $\Sigma$ be the $\sigma$-algebra of countable sets on $X$.

Then the co-countable measure $\mu$ on $X$ is a measure.

## Proof

Let us verify the measure axioms $(1)$, $(2)$ and $(3')$ for $\mu$.

### Proof of $(1)$

For all $S \in \Sigma$, $\mu \left({S}\right)$ is $0$ or $1$.

In either case, $\mu \left({S}\right) \ge 0$.

$\Box$

### Proof of $(2)$

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

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

Suppose that at least one $S_n$ is co-countable, say $S_N$.

Since the $S_n$ are pairwise disjoint, it follows that, for all $n \in \N$, $n \ne N$:

$S_n \subseteq X \setminus S_N$

From Subset of Countably Infinite Set is Countable, the $S_n$ with $n \ne N$ are all countable, as $S_N$ is co-countable.

Therefore:

$\mu \left({S_n}\right) = \begin{cases}1 & \text{if$n = N$}\\ 0 & \text{if$n \ne N$}\end{cases}$

and subsequently:

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

From Superset of Co-Countable Set, $\displaystyle \bigcup_{n \mathop = 1}^\infty S_n$ is co-countable.

Hence:

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

verifying $(2)$ for $\mu$ in this case.

If on the other hand, all $S_n$ are countable, then for all $n \in \N$:

$\mu \left({S_n}\right) = 0$

and so:

$\displaystyle \sum_{n \mathop = 1}^\infty \mu \left({S_n}\right) = 0$
$\displaystyle \bigcup_{n \mathop = 1}^\infty S_n$

is also countable, hence:

$\displaystyle \mu \left({\bigcup_{n \mathop = 1}^\infty S_n}\right) = 0$

verifying $(2)$ for $\mu$ in this case as well.

Hence $(2)$ holds for $\mu$, from Proof by Cases.

$\Box$

### Proof of $(3')$

Note that $\varnothing \in \Sigma$ as $\Sigma$ is a $\sigma$-algebra.

By Empty Set is Countable, $\varnothing$ is countable, whence:

$\mu \left({\varnothing}\right) = 0$

$\Box$

Having verified the axioms, it follows that $\mu$ is a measure.

$\blacksquare$