I am trying to understand the following solution to this question. The question was
Let $\alpha \in (0,1)$. Find a Borel subset $E$ of $[-1,1]$ s.t. $$\lim_{r\to 0^{+}} \frac{m(E\cap [-r,r])}{2r}=\alpha.$$
and one of the response to this question was
You can get any density $\alpha\in (0,1)$ you want. The idea is first to consider a decreasing sequence of positive numbers $x_n \to 0$, such that $\frac{x_n}{x_{n+1}}\to 1$ ($x_n=\frac{1}{n}$ works). Then in every interval $I_n=(x_{n+1}, x_n)$ place a segment of $E$ in proportion $\alpha$ (if $\alpha=\frac{1}{3}$ say take the middle third). Now consider $E$ the union of all the pieces in the intervals $I_n$. Do this symmetrically in the negative side of $0$. You got your set $E$.
I am taking $\alpha = 1/3$.
Now, I am having trouble verifying the answer. How can I show that it is indeed the case that $\lim_{r\to 0^{+}} \dfrac{m(E\cap [-r,r])}{2r}=1/3$ for the $E$ I have above? Most importantly, I am having trouble analyzing what $m(E\cap [-r,r])$ would amount to. Any hints or help would be greatly appreciated!
I think it's easier to prove without dealing with a specific example (or without using the middle third if you do use an example, as there's no reason to choose the middle except to mimick the first-step construction of the Cantor set).
Instead, we have, for $x_{n+1} \leq r \leq x_n$, $$m(E \cap [-x_{n+1}, x_{n+1}]) \leq m(E \cap [-r,r]) \leq m(E \cap [-x_n, x_n]) \\ 2x_{n+1}\alpha \leq m(E\cap [-r,r]) \leq 2x_n\alpha.$$
Since additionally everything (i.e. $x_n, x_{n+1}, r, \alpha$) is positive, we can combine this inequality with $x_{n+1} \leq r \leq x_n$ to see $$\frac{x_{n+1}}{x_n}\alpha \leq \frac{m(E\cap [-r,r])}{2r} \leq \frac{x_n}{x_{n+1}}\alpha.$$
Taking limits as $r \to 0^+$ (and therefore $n\to \infty$) gives the result by the squeeze theorem.