For any $x\in [0,1)$ we assign its binary representation $(x_1,x_2,\dots,x_n,\dots)$ without $1$ in period. Let $\{n_k\}_{k=1}^{\infty}$ be some increasing sequence of natural numbers, $\{a_k\}_{k=1}^{\infty}$ be some sequence of $0$ and $1$. Let $A=\{x\in [0,1): x_{n_k}=a_k \ \text{for} \ k\in \mathbb{N}\}$. Prove $\mu(A)=0$.
I came up with the following idea: For each $m\geq 1$ we define the set $$A_m=\{x\in [0,1): x_{n_k}=a_k \ \text{for} \ k=1,\dots,m\}.$$ Then we see that $A_1\supset A_2\supset \dots \supset A_n \supset A_{n+1}\supset \dots$ and $A=\cap_{m=1}^{\infty} A_m$. I was trying to show that each $A_m$ is Lebesgue measurable but I failed. Suppose hypothetically I have shown this then since $\mu(A_1)<\infty $ then $$\mu(A)=\lim _{m\to \infty}\mu (A_m).$$
Also my second hypothesis is that for each $m\geq 1$ we have $\mu(A_{m+1})=\dfrac{\mu(A_m)}{2}$.
Can anyone show to me the proof of the following moments:
1) Why each $A_m$ is Lebesgue measurable? I was not able to prove it even for $A_1$. It suffices to prove for the sets $A_1$ and the general case follows from the intersection.
2) How to prove rigorously that $\mu(A_{m+1})=\dfrac{\mu(A_m)}{2}$?
Possible answer: I guess it follows from invariance of Lebesgue measure. Denote by $A_{m+1}^0=\{x\in [0,1): x_{n_k}=a_k \ \text{for} \ k=1,\dots,m \ \text{and} \ x_{n_{m+1}}=0\}$ and $A_{m+1}^1=\{x\in [0,1): x_{n_k}=a_k \ \text{for} \ k=1,\dots,m \ \text{and} \ x_{n_{m+1}}=1\}$ then $A_m=A_{m+1}^0\cup A_{m+1}^1$ and note that this union is disjoint. And since $A_{m+1}^1=A_{m+1}^0+\dfrac{1}{2^{n_{m+1}}}$ then $\mu(A_{m+1}^1)=\mu(A_{m+1}^0)$ and $\mu(A_m)=2\mu(A_{m+1})$.
3) The question says we are considering the binary representation without $1$ in period? Am I right that in this case any number from $[0,1)$ has unique binary expansion?
Would be very grateful for answers!
Best Answer
Let's show that $A_1=\{x\in [0,1): x_{n_1}=a_1\}$ is Lebesgue measurable. Since $a_1\in \{0,1\}$ we will prove for the case when $a_1=0$. Denote by $A_1^0=\{x\in [0,1): x_{n_1}=0\}$ and $A_1^1=\{x\in [0,1): x_{n_1}=1\}$ then $A_1^0\cup A_1^1=[0,1)$ where the union is disjoint because we do not consider the representation with $1$ in the period. So if we prove that $A_1^0$ is measurable then it follows that $A_1^1$ is also measurable because $[0,1)$ is measurable.
So WLOG we assume that $A_1=\{x\in [0,1): x_{n}=0\}$ then $A_1$ is the disjoint union of the following sets $$A_1=\bigsqcup_{} A_1^{\epsilon_1,\dots,\epsilon_{n-1}},$$ where $ A_1^{\epsilon_1,\dots,\epsilon_{n-1}}=\{x\in [0,1): x_1=\epsilon_1, \dots, x_{n-1}=\epsilon_{n-1}, x_{n}=0\}$ and the union is taken over all $(n-1)$-tuples of $(\epsilon_1,\dots,\epsilon_{n-1})$ where each $\epsilon_j\in \{0,1\}$, so it means that this union contains $2^{n-1}$ sets. Also the union is disjoint because we do not consider the expansions with $1$ in the period.
We'll show that for each such $(n-1)$-tuple the set $A_1^{\epsilon_1,\dots,\epsilon_{n-1}}$ is measurable. But note that $A_1^{\epsilon_1,\dots,\epsilon_{n-1}}=B+\frac{\epsilon_1}{2}+\dots+\frac{\epsilon_{n-1}}{2^{n-1}}$, where $B=\{x\in [0,1): x_1=x_2=\dots=x_n=0\}$. And it is easy to show that $B=[0,\frac{1}{2^n})$ which is measurable and hence the $A_1^{\epsilon_1,\dots,\epsilon_{n-1}}$ is measurable and it means that $A_1$ is measurable being the union of measurable sets. Then $A_m=\{x\in [0,1): x_{n_k}=a_k \ \text{for} \ k=1,\dots,m\}$ is also measurable because $A_m=A_1^{n_1}\cap \dots\cap A_{1}^{n_m}$ where $A_1^{n_i}=\{x\in [0,1): x_{n_i}=a_i\}$.