Prove Outer Measure of $\mathbb R$

Question

Given a function $m$ on the power set of $\mathbb R$ and how that it is an outer measure.

$m(A) = 0$ if $A$ is countable.

$m(A) = 1/2$ if $A$ and $A^c$ are uncountable.

$m(A) = 1$ if $A^c$ is countable.

To show sub additivity, I tried covering all three cases. But I got stuck with the case of $(\bigcup_{i=1}^{n}A_i)^c$ being countable. Is this the right approach?

Answer 1

An outer measure has to be $\sigma$-subadditive hence it has to hold $$m\left(\bigcup_{i=1}^\infty A_i\right) \le \sum_{i=1}^\infty m(A_i)$$ Consider that your case of a finite union is contained taking $$A_i = \emptyset$$ for $i \ge n$

Now consider 3 cases for $$B:=\bigcup_{i=1}^\infty A_i$$:

1.) $m(B) = 0 \Rightarrow B \text{ is countable } \Rightarrow \text{all } A_i \text{ are countable }$

2.) $m(B) = \frac{1}{2} \Rightarrow B$ and $B^c$ are uncountable $\Rightarrow$ at least for one $i$ it holds $A_i$ and $A_i^c$ are uncountable

3.) $m(B) = 1 \Rightarrow B^c$ countable $\Rightarrow$

1. one $A_i^c$ is countable
2. or for infinite many $A_i$ it holds $A_i$ and $A_i^c$ uncountable

Now calc $\sum_{i=1}^\infty m(A_i)$ for each of the three cases and you are done.

Answer 2

If $A_i^{c}$ is countable for some i we are in good shape. Suppose $A_i^{c}$ is uncountable for each i. Let us show that there exist $i \neq j$ such that $A_i$ and $A_j$ are both uncountable. If $A_i$ is countable for each i then so is their union and this is a contradiction because $\mathbb R$ is the union of $\cup A_i$ and its complement which makes $\mathbb R$ countable. Now suppose there exists exactly one i such that $A_i$ is uncountable, so $A_j$ is countable for each $j\neq i$. Then $\cup_{j\neq i} A_j$ is countable; call this S. Then $(A_i \cup S)^{c}$ is countable which means $A_i^{c} \setminus S$ is countable.This makes $A_i^{c}$ itself countable. But $A_i^{c}$ is uncountable for each n. is it now clear that $m(\cup A_i) \leq \sum m(A_i)$?