I have a question about an exercise, any help will be thankful.
Let $X$ be a set and $\mu: 2^{X} \to [ 0,+ \infty]$
Which satisfies:
-$\mu ( \emptyset) =0$
-$\mu (A)=1$ if $A$ is a pointwise subset of $X$
-$\mu (A) = \sqrt{3} $ if $A$ has at least two different elements
Show that that $\phi$ is an outer measure.\
I've reach the firts two conditions of outer measuer with the 3 initial conditions of the hypothesis, the last one is the countable subadditivity property
I was thinking something like this:
Take $(A_n)_{n\in \mathbb{N}}$ as $(\lbrace a_1 \rbrace ,\lbrace \lbrace a_1 \rbrace, \lbrace a_2 \rbrace \rbrace ,...)$ such that
$\mu ( \bigcup_{n\in \mathbb{N}} A_n)=\sqrt 3 \leq \mu(A_1) + \mu(A_2) +...+ =1+ \sqrt 3 +\sqrt 3+...=\sum_{n\in \mathbb{N}} \mu (A_n)$
and therefore
$\mu ( \bigcup_{n\in \mathbb{N}} A_n)\leq \sum_{n\in \mathbb{N}} \mu (A_n)$
I do not know if it is the easiest way to do it but I can not find a way to prove the last point of the definition, any help will be very grateful.
By the way, English is not my mother lenguage, forgiveness for grammatical errors
No, you cannot assume what $(A_{n})$ is, rather, we know that if $\displaystyle\bigcup_{n}A_{n}$ contains two or more elements, then $\mu\left(\displaystyle\bigcup_{n}A_{n}\right)=\sqrt{3}$, in this case, we can consider if there are two sets $A_{N}$ and $A_{M}$ that contains only one element each, then $\mu(A_{N})+\mu(A_{M})=2>\sqrt{3}$, so it is definitely true that $\displaystyle\sum_{n}\mu(A_{n})\geq\mu(A_{N})+\mu(A_{M})\geq\mu\left(\displaystyle\bigcup_{n}A_{n}\right)$, or we can consider there is some $A_{K}$ that contains two or more elements, then $\displaystyle\sum_{n}\mu(A_{n})\geq\mu(A_{K})=\sqrt{3}=\mu\left(\displaystyle\bigcup_{n}A_{n}\right)$.
Since $\mu\left(\displaystyle\bigcup_{n}A_{n}\right)$ cannot be strictly greater than $\sqrt{3}$, by considering the case that $\displaystyle\bigcup_{n}A_{n}$ has two or more elements suffices for the proof.