Binary expansion has a zero in each even position is Lebesgue measurable

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Exercise: Show that the subset of points $A$ of the interval $[0,1]$ whose binary expansion has a zero in each even position of its expansion ($\frac{2}{3}=0,101010...\in A$ for example) is Lebesgue measurable and zero Lebesgue measure.

I know the following definition of measurability: If $\mu^*:\mathscr{P}(\Omega)\to\mathbb{R}^+$ is a outer measure. Then $E_i$ is measurable with respect to $\mu^{*}$ iff $\forall A\in\mathscr{P}(\Omega)\:\:\mu^*(A)=\mu(A\cap E)+\mu(A\setminus E)$.

However despite knowing this definition I do not know if it supposed to be used.

I have no idea on how to solve this exercise.

Question:

Can someone provide an hint about how to prove measurability and that the measure is zero?

Thanks in advance!

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Let $L_0=[0, 1[$ ($\{1\}$ has null measure) for every $k\in\mathbb N$ we have $$ L_0=\bigcup_{i=0}^{2^k-1}\left[\frac{i}{2^k}, \frac{i+1}{2^k}\right[ $$ if $n\in\left[\frac{i}{2^k}, \frac{i+1}{2^k}\right]$ then $n$ has a $0$ at position $k$ in its binary expansion if and only if $i$ is even. Then let $$ L_k=\bigcup_{i=0}^{2^{2k-1}-1}\left[\frac{2i}{2^{2k}}, \frac{2i+1}{2^{2k}}\right[\\ L=\bigcap_{k\in\mathbb N}L_k $$ clearly $L$ is Lebesgue measurable

Prove that $$ \lvert L_n\cap L_{n-1}\cap\dotsb\cap L_1\rvert =\frac 12\lvert L_{n-1}\cap\dotsb\cap L_1\rvert $$ then you can easily continue the proof.