# Thread: Measure of countable co-countable set

1. ## Measure of countable co-countable set

Let X be an uncountable set, let $E \subset X$
Consider the $\sigma$-algebra $\mathbb {M} = \{ E \ countable \ or \ E^c \ countable \}$

Define $\mu : \mathbb {M} \rightarrow [0, \infty ]$ by $\mu (E) = 0 \ if \ E \ countable$ and $\mu (E) = 0 \ if \ E^c \ countable$

Prove that $\mu$ is a measure.

Proof so far.

Let $E_1,E_2, ... \in \mathbb {M}$, claim: $\mu ( \bigcup _{n=1}^ \infty \mu (E_n) = \sum ^ \infty _{n=1} \mu (E_n)$

Case 1) If $E_n$ is countable for all n, then $\bigcup _{n=1}^ \infty \mu (E_n)$ is countable and $\mu ( \bigcup _{n=1}^ \infty \mu (E_n) = 0 = \sum ^ \infty _{n=1} \mu (E_n)$ since $\mu (E_n) = 0 \ \ \ \forall n$

Case 2) If $E_n ^c$ is countable. Since $( \bigcup _{n=1}^ \infty (E_n) )^c \subset \bigcup ^ \infty _{n=1} E_n^c$ and the latter is countable, I then have $\mu ( \bigcup _{n=1}^ \infty (E_n) ) = 1$

But then I also have $\sum ^ \infty _{n=1} \mu (E_n) = \mu (E_1) + \mu (E_2) = . . . = 1 + 1 + 1 + . . . = \infty$

So the two ain't equal... What did I do wrong?

Case 3) Either $E_n$ is countable or $E_n^c$ is countable. How should I proceed with this one?

Thank you!

Let X be an uncountable set
(....)
Let $E_1,E_2, ... \in \mathbb {M}$ be disjoint subsets, claim: $\mu ( \bigcup _{n=1}^ \infty E_n) = \sum ^ \infty _{n=1} \mu (E_n)$

Case 2) If $E_n ^c$ is countable. Since $( \bigcup _{n=1}^ \infty (E_n) )^c \subset \bigcup ^ \infty _{n=1} E_n^c$ and the latter is countable, I then have $\mu ( \bigcup _{n=1}^ \infty (E_n) ) = 1$

But then I also have $\sum ^ \infty _{n=1} \mu (E_n) = \mu (E_1) + \mu (E_2) = . . . = 1 + 1 + 1 + . . . = \infty$

So the two ain't equal... What did I do wrong?

Case 3) Either $E_n$ is countable or $E_n^c$ is countable. How should I proceed with this one?
You can't expect equality if the subsets aren't disjoint.

The key is given by the following question: is it possible that two sets $E_1,E_2$ are disjoint if they are such that $E_1^c$ and $E_2^c$ are countable?