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 $\lambda(A)$?
[Math] Lebesgue measure of numbers whose decimal representation contains at least one digit equal $9$
decimal-expansionlebesgue-measuremeasure-theory
Related Solutions
See, study the complement of the set. That is, look at integers which contain $5$ in their decimal expansion. Caveat : note that $0.6 = 0.5\overline{9}$ also counts as a decimal which is expressed with a $5$ as one of the digits, so belongs in the set. In particular, any terminating decimal which terminates with $6$ can be considered to belong to the set.
The first such category we can think of is: Those that have $5$ as the first digit following the decimal point. This is the set of numbers $[0.5,0.6]$. This has measure $0.1$, so the left over measure is $0.9$.
Now, from the remaining set, remove the set of all numbers with second digit $5$. This consists of $[0.05,0.06[, [0.15,0.16] \ldots [0.95,0.96]$ without $[0.55,0.56]$, since that was already removed earlier. Now, each of these has measure $0.01$, so we have removed $0.09$ more from the system. Hence, the left over is $0.81$.
By induction, prove the following : at the $n$th step, the set left over has measure $\frac{9^n}{10^n}$. Now, as $n \to \infty$, we see that the given set has measure zero (I leave you to rigorously show this, you can use the Borel-Cantelli lemma). This also incorporates the fact that the given set is measurable, since it's measure is computable(and is $0$).
This feels very similar to the Cantor set, and copper.hat's statement can be amended and used. This I think is similar to the method you are proposing. We wish to show the measure of the set of numbers with finitely many 7's in it's decimal expansion (call this set $C$) is 0.
Let $C_n$ be the set of numbers in $[0, 1]$ with no 7's past the first n digits of it's decimal expansion. Then $C = \cup_n C_n$ thus to show $\mu(C) = 0$ it suffices to show $\mu(C_n) = 0$ for all $n$ (where $\mu$ is Lebesgue measure).
But $C_n$ has a close relation to the Cantor set. For each $(x_1, \ldots, x_n) \in \{0, 1, \ldots, 9\}^n$ let $C_{x_1, \ldots, x_n}$ be the set of numbers in $[0, 1]$ with decimal expansion $0.x_1 x_2 \cdots x_n \cdots$ where there are no $7$'s past the $n$th digit $x_n$. The Cantor set is the set of numbers in $[0, 1]$ which have no 2's in their ternary expansion. We show the Cantor set has measure 0 by showing it is the intersection of sets with measure going to 0. The same technique can be used to show that $C_{x_1, \ldots, x_n}$ has measure 0.
Thus while $C_n$ is uncountable, $C_n = \bigcup_{(x_1, \ldots, x_n) \in \{0, 1, \ldots, 9\}^n} C_{x_1, \ldots, x_n}$ is a countable union of sets of measure 0 and hence has measure 0 as required.
Best Answer
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: $$\lambda(A)=\sum_{n=1}^{\infty}\frac{9^{n-1}}{10^n}=\frac{1}{9}\sum_{n=0}^{\infty}\Big(\frac{9}{10}\Big)^n-\frac{1}{9}=\frac{1}{9}\cdot\frac{1}{1-\frac{9}{10}}-\frac{1}{9}=1$$