Consider the dyadic expansion for numbers on the unit interval.
Let $A_n=\left\{x\colon d_{n+1}(x)=\dots=d_{2n}(x)=0\right\}$, where $d_k(x)$ is the $k$th dyadic digit.
I want to find $A=\lim_n A_n$ and to prove that it has Lebesgue measure $0$.
What I've tried so far is to consider the dyadic expansion for $x_a\in A_n$. It will be of the form
\begin{align} x_{a_n}=&\sum_{i=1}^{n}\frac{d_i}{2^i}+\sum_{i=2n+1}^{\infty}\frac{d_i}{2^i} \end{align}
from there I argued that when $n\to \infty$ the last term vanishes, and I thought the limit set are all numbers with a terminating expansion, which are therefore a subset of $\lbrack0,1\rbrack\cap\mathbb{Q}$ and hence are countable and have Lebesgue measure 0. But I'm not so sure anymore.
Is this correct?
I think I found what I was looking for.
A little thought should be enough to realize that the $k$-th digit of the dyadic expansion is zero on the set
\begin{equation} B_k=\bigcup_{i=1}^{2^{k-1}}\left\lbrack \frac{2(i-1)}{2^k},\frac{2\left(i-1\right)+1}{2^k} \right). \end{equation}
Now, what I need is the intersection, namely
\begin{align} A_n=& B_{n+1}\cap \dots \cap B_{2n}\\ =& \bigcup_{i=1}^{2^{n}}\left\lbrack \frac{2^n(i-1)}{2^{2n}},\frac{2^n\left(i-1\right)+1}{2^{2n}} \right). \end{align} I figured this out with help of a drawing. Since every subinterval has length $1/2^{2n}$, we get
\begin{align} \lambda \left(A_n\right)=& 2^n\frac{1}{2^{2n}}\\ =& \frac{1}{2^n} \end{align}
and $\lambda(A_n)\to 0$ as $n\to \infty$.