Let $X$ be an uncountable set, let $\mathfrak{M}$ be the collection of all sets $E \subset X$ such that either $E$ or $E^c$ is at most countable ( i.e, finite or countable), and define $\mu(E)=0$ in the first case, $\mu(E)=1$ in the second. Prove that $\mu$ is measure on $\mathfrak{M}?$
My attempt : Here to prove $\mu$ is measure mean that $\mu$ is countably additive on $\mathfrak{M}$
we have two cases
$1.$when $E_n$ is countable
$2.$ when $E_n^c$ is countable
Case $1.$ : let $E_1, E_2…$ be a countable sequence of pairwise disjoint element of $\mathfrak{M}$
then their union is also countable
so we have the following inequality
$\mu(\bigcup E_n)=\sum_{n=1}^{\infty}\mu(E_n)=0$
My doubt : How to show $\mu$ is measure when $E_n^c$ is countable ?
Best Answer
Let $F$ be a countable subset of $M.$
If every $f\in F$ is countable then $\cup F$ is countable so $$\cup F\in M \;\land\;\mu(\cup F)=0=\sum_{f\in F}0=\sum_{f\in F}\mu(f).$$ If there exists an uncountable $f_0\in F$ then
(i) since $f_0\in M,$ we have $\cup F\in M$ because $(\cup F)^c$ is a subset of the countable set $(f_0)^c,$ and
(ii) since $(\cup F)^c$ is countable by (i), we have $$\mu(\cup F)=\infty=\mu(f_0)\le\sum_{f\in F}\mu(f)\le \infty.$$