I have this question: let $X$ be an uncountable set.
Let $\mathcal{M}=\{A \subseteq X : A $ is countable or A is co-countable$\}$, co-countable means its complement is countable. Then, define $m: \mathcal{M} \rightarrow [0, \infty]$ as
$$m(E) = \left\{ \begin{array}{rcl}
0 & \mbox{for}& E\ \ countable
\\ 1 & \mbox{for} & E\ \ co-countable
\end{array}\right.$$
Prove that $m$ is a measure.
Here is my proof:
I need to show that 1# $m(E)< \infty$ for some $E \subseteq X$, 2# $m$ is countable additive. Clearly 1 is trivial. Then, for 2 I divided it to three cases:
Case 1: $E_j$ is countable for all $j$. So that is trivial as well by the definition of $m$.
Case 2: $E_j$ is co-countable for all $j$ with the assumption that $\{E_j\}_{j\geq 1}$ is a countable collection of disjoing sets. So $X \thicksim \bigcup_{j \geq1} E_j \subseteq X \thicksim E_m$ for any $m$ positive integer which is countable, so $\bigcup_{j \geq1} E_j$ it is co-countable. Therefore, $m(\bigcup_{j \geq1} E_j )=1$. But then $\sum_{j\geq1}m(E_j)=1+1+1+…=\infty$ since for each $j$ $E_j$ assumed to be co-countable. I am getting stuck here.
Also, in
Case 3: If we have a mix of countable and co-countable sets, then $X \thicksim \bigcup_{j \geq1} E_j \subseteq X \thicksim E_m$ for some $m$ with $E_m$ is countable, so $\bigcup_{j \geq1} E_j $ is co-countable, and hence $m(\bigcup_{j \geq1} E_j )=1$. However, $\sum_{j\geq 1}E_j=0+…+0+1+1+…+1 \neq 1$. WHAT I am missing of ideas here??
Appreciate any help with that.
Thank you.
Best Answer
Let $(E_n)_{n=1}^\infty$ be a sequence of disjoint sets such that $E_i$ and $E_j$ are co-countable. Then if $E_i \cap E_j = \emptyset$ we get $$X = (E_i \cap E_j)^c = E_i^c\cup E_j^c$$ which is a countable set as a union of two countable sets. A contradiction, unless $i = j$. Hence, only $E_i$ is co-countable and $E_n$ for $n \ne i$ are countable.
$\bigcup_{n=1}^\infty E_n$ is co-countable so $m\left(\bigcup_{n=1}^\infty E_n\right) = 1$ and also $$\sum_{n=1}^\infty m(E_n) = m(E_i) + \sum_{n\ne i}m(E_n) = 1$$