Lebesgue measure of numbers whose decimal representation contains at least one digit equal $9$ Let $A$ be the set of numbers on $[0, 1]$ whose decimal representation contains at least one digit equal $9$. What is its Lebesgue measure $\lambda(A)$?
 A: Let $0.a_1a_2a_3a_4\ldots$ be a number in $[0,1]$. If the $n^{th}$ digit is the first 9 in the expansion, there are $9^{n-1}$ possible assignments of digits ($0,1,\ldots,8$) to $a_1, a_2, \ldots, a_{n-1}$. For each of these assignments, we have an interval of length $\frac{1}{10^n}$ of numbers having a nine in their expansions.
Summing up over $n$, we get our answer: $$\lambda(A)=\sum_{n=1}^{\infty}\frac{9^{n-1}}{10^n}=\frac{1}{9}\sum_{n=0}^{\infty}\Big(\frac{9}{10}\Big)^n-\frac{1}{9}=\frac{1}{9}\cdot\frac{1}{1-\frac{9}{10}}-\frac{1}{9}=1$$
A: We don't care about the countable set of $x\in[0,1]$ possessing two different decimal expansions. Denote by $A_k$ $(0\leq k\leq9)$ the set of numbers in $[0,1]$ having $k$ as first digit after the decimal point. The sets $A_k$ with $k<9$ can be written as  $$A_k={k\over10}+{1\over 10}A\qquad(0\leq k<9)\ .$$ The scaling property of Lebesgue measure $\lambda$ then implies that $\lambda(A_k)={1\over10}\lambda(A)$ for these $k$. In this way we obtain
$$\lambda(A)=9\cdot{1\over10}\lambda(A)+{1\over 10}\ ,$$
whereby the last term measures the set $A_9$.
It follows that $\lambda(A)=1$.
A: Recall that a real number is said to be normal (in base $10$) if every sequence of $n$ consecutive digits appears in its decimal expansion with frequency $1/10^n$.
In 1909 Borel proved that almost every real number (wrt the Lebesgue measure) is normal. In particular, this means that $\lambda(A) = \lambda([0,1]) = 1$.
