Which of the following is an outer measure
Let $X$ be any set and $\mu ^* : P(X)\rightarrow [0, \infty)$ be given by
1) $$ \mu ^*(A)= \begin{cases} 0,& \text{if } A=\phi\\ 1, & \text{otherwise} \\ \end{cases}$$
2) $$ \mu ^*(A)= \begin{cases} 0,& \text{if } A & \text{is countable}\\ 1, & \text{otherwise} \\ \end{cases}$$
3) $$ \mu ^*(A)= \begin{cases} 0,& \text{if } A & \text{is finite}\\ 1, & \text{otherwise} \\ \end{cases}$$
I checked first two properties namely $\mu^*(A)\geq0$ for all $A$ and $\mu^*(\phi)=0$ and $A \subset B$ then $\mu^*(A) \leq \mu^*(B)$. My problem on showing countable subaditivity for all three case. Please help
What we want to show here is that for any countable collection $\lbrace A_{i} \rbrace$ of subsets of $X$, we have $$\mu^{*} \left( \bigcup_{i = 1}^{\infty} A_{i} \right) \leq \sum_{i=1}^{\infty} \mu^{*}(A_{i})$$
For #1, note that if $\bigcup_{i=1}^{\infty} A_{i} = \emptyset$, then each $A_{i} = \emptyset$. Clearly, $\mu^{*} \left( \bigcup_{i=1}^{\infty} A_{i} \right) = 0$ and $\sum_{i=1}^{\infty} \mu^{*}(\emptyset) = 0$. If your countable union is not empty, then the measure of the union will be 1, whereas the sum of the measures will be at least 1.
For #2, you can use a similar argument, but replace "empty" with "countable".
For #3, one can easily union a bunch of finite sets to get an infinite set, no? So, you can get that the measure of the union of these finite sets is 1, whereas the sum of the measures of these finite sets will be 0.