My question is related to the following (EDIT: I know the solution and I am not asking for one.):
Let $A=[0.5,0,6)$, $f:[0,1)\rightarrow [0,1)$, $f(x)= 10x \mod{1}$, $$A^\star = \{x\in [0,1): f^n(x)\not\in E, \forall n\geq 0 \}.$$ Calculate the Lebesgue measure of $A^\star$.
I know the proof that since $f$ is ergodic w.r.t. the Lebesgue measure we have that $\mu (A^\star)=0$.
The following attempt for calculating $\mu (A^\star)$ clearly has a mistake. Even if it is not correct, does the calculated value corresponds to the measure of some other specific set?
$$ f^n(x)=10^n x\mod{1}\in A\\ \Leftrightarrow x\in \Big[\frac{5}{10^{n+1}},\frac{6}{10^{n+1}} \Big)$$
$$ \mu((A^\star)^c)=\sum_{n=0}^\infty \mu\Big( \Big[\frac{5}{10^{n+1}},\frac{6}{10^{n+1}} \Big)\Big) = \sum_{n=0}^\infty \frac{1}{10^{n+1}}=\frac{1}{9}\\ \Leftrightarrow \mu (A^\star)=\frac{8}{9} $$
$\mu(A^\star)=0$.
Hint. Clearly, $A^\star=\bigcap_{n=1}^\infty A_n$, where $$ A_n=\big\{x\in[0,1] : \text{the $n-$th digit of $x$ is not $5$}\big\}. $$ We have $$ \mu(A_1)=\frac{9}{10},\quad \mu(A_1\cap A_2)=\frac{9^2}{10^2}, \ldots, \mu(A_1\cap \cdots \cap A_n)=\frac{9^n}{10^n}. $$ The construction of $A^\star$ as the intersection of the $A_n$'s is identical to the definition of Cantor set.