Let $D \subset [0,1]$ be a cantor like set. Then, the outer measure, $\mu^*(D) = \varepsilon$ for $\varepsilon>0.$
Attempt: Suppose that $a^k$ is removed from the middle interval at $k$th stage, and $D_k$ be the $k$th stage of a construction of $D$. Then, we have $$\mu^*(D_k) = 1-a\sum_{n=0}^{k-1} (2a)^n = 1-a\frac{1-(2a)^k}{1-2a}. $$
Then, for $k \to \infty,$ we have $$\mu^*(D)=1-\frac{a}{1-2a}$$ for $a<1/2.$ This shows that if $a = 1/3$, then $\mu^*(D)=0.$
Of course, this does not answer my question that $\mu^*(D)=\varepsilon$, but this is what I know. So, could you give some help to tackle this question?
Thank you in advance.