Set of points $x$ in $[0,1]$ supporting a decimal expansion without $2$'s and $3$'s and find $m(E)$

Question

Let $E$ denote the set of points $x$ in $[0,1]$ supporting a decimal expansion without $2$'s and $3$'s. Show that $E$ is a measurable set, and find $m(E)$

Could you please give me a hint?

I believe, that $x=0.x_{1}x_{2}x_{3}....= \sum_{j=1}^{\infty}x_{j}10^{-j}$, with each $x_{j}\in{0,1,4,5,6,7,8,9}$

Thanks.

Answer 1

Hint: if $E_j$ is the set of points in $E$ whose first digit after the decimal point is $j$, then $m(E) = m(E_0) + m(E_1) + m(E_4) + \ldots + m(E_9)$ and $m(E_j) = m(E)/10$.

Answer 2

Well, Let $K_n$ but the set of numbers whose first $n$ decimals contain no twos or threes. There are $10^n$ possible options for the first $n$ digits to be and of those, $8^n$ have no twos or threes. So $m(K_n) = (\frac 45)^n$.

$E \subset K_n$ for all $n$ so $m(E) \le m(K_n) = (\frac 45)^n$ for all $n$. So...

Answer 3

Hint: this is just like the Cantor set except you are removing $\frac{1}{5}$ of the interval each time. As with the Cantor set, the complement is a dense open subset of $[0, 1]$.