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.

  • $\begingroup$ Does "$\mu^*(D)=\varepsilon$ for $\varepsilon\gt0$" mean "$\mu^*(D)=\varepsilon$ for SOME $\varepsilon\gt0$" or "$\mu^*(D)=\varepsilon$ for ALL $\varepsilon\gt0$"? If the former, it is equivalent to asking "is $\mu^*(D)\gt0,$" why make it more complicated? If the latter, the answer is no; if $\varepsilon_1\ne\varepsilon_2,$ it's impossible to have both $\mu^*(D)=\varepsilon_1$ and $\mu^*(D)=\varepsilon_2.$ $\endgroup$ – bof Aug 18 '18 at 2:05
  • $\begingroup$ In the construction of the Cantor set, at the stage $k$ you remove the middle third to each of the $2^{k-1}$ intervals obtained at the stage $k-1$. The length of each of these intervals is $\frac{1}{3^{k-1}}$, so at the stage $k$ you remove a set whose measure is $\frac{2^{k-1}}{3^k}$. The measure of the union of the sets removed up to stage $k$ is $\sum_{i=1}^{k} \frac{2^{i-1}}{3^i}$. Taking the limit, you get $\mu(D)=\mu(1)-\lim_{k\rightarrow\infty}\sum_{i=1}^{k} \frac{2^{i-1}}{3^i} = 0$. As a consequence the Cantor set is a null set and its inner measure coincides with its outer measure. $\endgroup$ – Emanuele Bottazzi Aug 18 '18 at 6:50
  • $\begingroup$ If $0<a<1/3$ then $1-a/(1-2a)$ can be any member of $(0,1)$ that you want. $\endgroup$ – DanielWainfleet Aug 18 '18 at 13:36

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