Let us have a fixed interval $I_0=[a,b]$ and let $A$ be a subset of $I_0.$ Compute $$\mu^* \left( \left\{\left( 1+ \frac{1}{n}\right)^n | n \in \mathbb{N} \right\} \right)$$
I've been thinking that if this somehow is related to the fact that if $A$ is countable the the outer measure $\mu^* =0$. The inside of the brackets seems countable and therefore I expect the result to be $0$(?)
Please help! Thank you!
On
You're right. You can prove this the following way (this proof holds for any countable set $A=\{a_1,a_2,...,a_n,...\}$):
Take some $\epsilon>0$. Then for any $a_i\in A$, cover $a_i$ with the interval $[a_i-\frac{\epsilon}{2^{i+1}},a_i+\frac{\epsilon}{2^{i+1}}]$. Then: $A\subseteq\bigcup_{i\in\mathbb{N}}[a_i-\frac{\epsilon}{2^{i+1}},a_i+\frac{\epsilon}{2^{i+1}}]$ and so:
$$\mu^*(A)\leq\sum_{i\in\mathbb{N}}\mu^*([a_i-\frac{\epsilon}{2^{i+1}},a_i+\frac{\epsilon}{2^{i+1}}])= \sum_{i\in\mathbb{N}}\frac{\epsilon}{2^i}=\epsilon$$ for all $\epsilon>0$, and so $\mu^*(A)=0$. Therefore the outer lebesgue measure of any countable set is zero and so is the lebesgue outer measure of yours.
Hint: Just use the definition of the outer measure.