How to construct a compact totally disconnected subset of the unit interval with specific measure

Question

How to construct a compact totally disconnected subset $E \subset [0,1]$ with Lebesgue measure $m(E) = \epsilon$ for any given $\epsilon \in (0,1)$, without removing any intervals. This question is from Rudin Real and Complex Analyis Chapter 2 Execise 7, but I have added the "without removing any intervals" as the solution to Execise 8 seems to require compact totally disconnected sets with positive measure that are constructed without removing any subsets I believe. I can see how a construction of such sets without the restriction of non-removal of intervals is possible, similar to the Cantor set construction.

Answer 1

Since $\mathbb{Q} \cap I$ is a countable subset of $I=[0,1]$. Write $\mathbb{Q}\cap (0,1)=\{r_n:n \in \mathbb{N}\}$. Now for $0<\epsilon<1$ consider the set $A=\bigcup_{n\in \mathbb{N}}(r_n- \frac{\epsilon}{2^{(n+1)}},r_n+\frac{\epsilon}{2^{(n+1)}})$. $A$ is an open dense subset of $\mathbb{R}$ having measure $\epsilon$. If I assume $A\subset I$ then $I-A$ will work as your set just you have to replace $1-\epsilon$ with $\epsilon$.

But getting $A\subset I$ is a tough task we have to rearrange $r_n$'s value.

Consider $J_1=\{ n\in \mathbb{N}: r_n- \frac{\epsilon}{2^{(n+1)}}<0\}$. $J_1$ is a subsequence say $\{{n_k}\}_{k=1}^{\infty}$ For each $r_{n_k}$ we want to find a suitable member say $r_{n_k}'$ to replace with it. Let's see on the interval $ (\frac{\epsilon}{2^{(n+1)}},1-\frac{\epsilon}{2^{(n+1)}})$ because it is safe for $r_{n_k}$, any choice from this interval will not make any trouble for $r_{n_k}$. Now since $ (\frac{\epsilon}{2^{(n+1)}},1-\frac{\epsilon}{2^{(n+1)}})$ has infinitely many rational numbers I will take a rational $r_m$ with a very large index m such that $\frac{\epsilon}{2^{m}}<r_{n_k}$. So now if I replaced them then $|r_n'-\frac{\epsilon}{2^{m}}|=|r_n-\frac{\epsilon}{2^{m}}|>0$. Now whatever $J_1$ is finite or infinite we can do this process which will short out our problem. And we will get a subset $A\subset I$ which is open dense having measure $\epsilon$. Take the complement of it.

There may be overlapping of the intervals so length will be less than $\epsilon$ I am planning to edit this.