Asking about a hint: constructing a cantor-like set This problem is a part of an exercise in Stein's Real Analysis. It reads:
Construct a measurable set $E \subset [0, 1]$ such that for any non-empty open sub-interval $I$ in $[0, 1]$, both sets $E \cap I$ and $E^c \cap I$ have positive measure. 
The problem gives me a hint: consider a Cantor-like set of positive measure, and add in
each of the intervals that are omitted in the first step of its construction, another
Cantor-like set. Continue this procedure indefinitely.
I'm not asking for an answer here, but I just wanted to clarify what the hint means: Let's say that I have a cantor-like set $\hat{C}$ such that it has a positive measure. What do I do next? In the $k$th step of its construction, I would remove $2^k$ intervals. Therefore, in the first step, I would remove just one interval. I'm not sure what this hint means.
 A: The hint means the following:
$[0,1] \setminus \hat C$ consists of countably many open intervals. Fill each of these intervals (in fact, its closure) with another Cantor-like set with positive measure (with the same diameter as the length of the interval just being filled). Now you get again a Cantor-like set, it's just bigger. You repeat this construction and the sought-after set $E$ is the union of this increasing sequence of Cantor-like sets (which is not a Cantor-like set anymore because it's not closed). You just have to be careful about the measures of the sets added in each step so the final set doesn't have full measure in any open interval.
An equivalent approach which might be easier to check:
Take a sequence of positive numbers strictly decreasing to $0$, say e.g. $(1/n)$. Do the same construction as above but in $n$th step only for those contiguous intervals which are longer than $1/n$ (of course, in each step we consider the Cantor-like set obtained in the previous step). Advantage of this approach is that in each step we add only a finite number of new Cantor-like sets thus e.g. closedness of the resulting sets in each step is obvious. And again, the set $E$ is obtained as the union of all these sets.
