# Is the set obtained by removing intervals of $1/4^n$ a borel set?

I have a set defined by: $D_0 = [0, 1]$.

$D_1$ is obtained from D0 by removing an open interval of length $1/4$ from the middle, so $D_1 = [0, 3/8] \cup [5/8, 1]$.

$D_2$ is obtained from $D_1$ by removing an open interval of length $1/16$ from the middle of each of the $2$ intervals, so $$D_2 = \left[0, \frac{5}{32}\right] \cup \left[\frac{7}{32}, \frac 3 8\right] \cup \left[\frac 5 8, \frac{25}{32}\right] \cup \left[\frac{27}{32}, 1\right]$$

$D_n$ is obtained from $D_{n−1}$ by removing an open interval of length $1/4^n$ from the middle of each of the $2^{n−1}$ intervals. Set $$D = \bigcap_{n=0}^\infty D_n$$

Is $D$ countable? Is $D$ Borel? And is $D$ measurable? And justify your answers.

I think D is uncountable as $D$ becomes the union of closed intervals that is in total length $1/2$. And each interval would have a bijection with the real numbers $[0,1]$ which is uncountable. I think it is Borel as it can be constructed from countable unions and countable intersections of closed sets. I also think it's measurable as I think the Lebesgue measure is $1/2$ but I'm really not sure.

Any help is appreciated, thanks in advance.

• Intuitive answer: Did you, at any time during the construction of this set, say the words "axiom of choice"? No? Then it's Borel. Commented Dec 10, 2015 at 16:31

You are correct that $D$ is uncountable (and has positive Lebesgue measure), although your justification is wrong. The set $D$ does not contain any non-trivial closed intervals: One way to see this is to use the fact that the largest interval at stage $n$ of the construction has length tending to zero (even exponentially quickly).
However, you can mimic the answers to each of these questions for the usual $1/3$ Cantor set. For example, the question of countability can be resolved by viewing $D$ as the set of numbers in a certain base lacking certain digits, just as the $1/3$ Cantor set is the set of numbers in $[0, 1]$ whose ternary representations have no ones.
To see that $D$ is Borel, just notice that it's a countable intersection of Borel sets. $D_n$ is Borel because it's a finite union of intervals. This also implies $D$ is measurable.
Finally, to show that the set has positive measure, you can compute the measure of the complement, which is a countable union of open sets. At stage $1$, we remove an interval of length $1/4$; at stage $2$, we remove two intervals of length $1/16$ which are not only disjoint, but disjoint from that of the previous stage. This process continuous, and the measure of what's eventually removed is
$$\frac 1 4 + 2 \cdot \frac 1 {4^2} + 2^2 \cdot \frac{1}{4^3} + \cdots = \frac 1 2$$ Taking complements gives that $m(D) = \frac 1 2$. By the way, this gives an independent proof that $D$ is uncountable.