Measure of a Cantor like set It is a well known fact how to construct Cantor sets of any measure $\alpha\in[0,1)$ by removing middle third intervals. Here is a different kind of procedure of constructing a nowhere dense set of positive length I found in Bogachev's Measure theory, Volume 1:

*

*Starting with $I=[0,1]$, for any $n\in\mathbb{N} (=\{1,2,\ldots\})$, divide $I$ in $2^n$ subintervals of equal length.

*For each of the $2^n$ subintervals $I_{n,j}$ remove the open subinterval of length $2^{-2^{n+1}}$ whose right endpoint is that of $I_{n,j}$.

*Let $A_n$ be the union of the remaining closed subintervals, and define
$$ A=\bigcap_nA_n$$
The set $A$ has positive Lebesgue measure.

Edit: What I am trying to check is whether I got the correct estimate for the measure of $|A_n|$:
$$\begin{align*}
\alpha_n&:=|A_n|=4\cdot 2^{-2^{n+1}}\prod^n_{j=1}(2^{2^j}-1)\\
&=4\cdot 2^{-2^{n+1}}2^{2^1+2^2+\ldots +2^n}\prod^n_{j=1}(1-2^{-2^j})\\
&=\prod^n_{j=1}(1-2^{-2^j})
\end{align*}$$
The limit of this product is indeed positive, as the explanations by David Ulrich and Oliver Diaz show. If anyone could help me corroborate the expression I got for $\alpha_n$ I would appreciate it.
 A: This is more a comment than a solution expanding on David C. Ulrich's

*

*Recall that a sequence $q_n=\prod^n_{k=1}(1+a_k)$, with $a_n\neq-1$ for all $n\in\mathbb{N}$ converges to $p\neq0$ iff for any $\varepsilon>0$ there is $N_\varepsilon>0$ such that
$$\Big|\prod^k_{j=1}(1+a_{n+k}) -1\Big|<\varepsilon,\qquad n\geq N, \, k\in\mathbb{N}$$
See for example, T. Apostol's Mathematical Analysis, p.p. 207.


*Also, recall that a product $q_n=\prod^n_{k=1}(1+a_n)$ is said to converge absolutely (again $a_n\neq-1$ for all $n\in\mathbb{N}$) iff $Q_n=\prod^n_{k=1}(1+|a_k|)$ converges. Based on the criteria in (1), it is not difficult to check that if $q_n$ converges absolutely, then $p=\lim_nq_n$ exists and $p\neq0$.


*Finally, recall that $q_n=\prod^n_{k=1}(1+a_n)$  converges absolutely (again $a_n\neq-1$ for all $n\in\mathbb{N}$) absolutely iff $\sum_n|a_n|<\infty$. This follows from the inequality
$$ \sum^n_{k=1}|a_k|\leq \prod^n_{k=1}(1+|a_k|)\leq \exp\Big({\sum^n_{k=1}|a_n|}\Big)$$
I did not checked whether the expression you give is indeed the length of the intervals that make up $A_n$, but the expression you gave does converge (absolutely as in (2)) to something positive on account that (1) $2^{-2^k}\neq1$ for all $k\in\mathbb{N}$ and (2) $\sum^\infty_{k=1}\frac{1}{2^{2^k}}<\infty$.
I don't know if there is a known expression for $\prod^\infty_{n=1}(1-2^{-2^n})$ though.
