Measure of reals in $[0,1]$ which don't have $4$ in decimal expansion

It's an exercise in E. M. Stein's "Real Analysis."

Let $A$ be the subset of $[0,1]$ which consists of all numbers which do not have the digit $4$ appearing in their decimal expansion. What is the measure of $A$?

I would be grateful if someone can give me some hints.

Thank you.

You can construct the set $A$ as a limit of nested sequence, so you prove measurability of $A$ and find its measure at the same time. With $n$-th digit of a number we refer to the $n$-th digit after the delimiter in the decimal expansion of the number, e.g. $2$ is the $4$-th digit of $0.434256$

The answer is $\mu(A) =0$. The informal proof is simple: each time you restrict the $n$-th digit, you truncate the measure by multiplying it with $9/10$. So, $\mu(A) = \lim\limits_{n\to\infty}\frac{9^n}{10^n} = 0$.

About the formal proof: we elaborate the idea by Chandrasekhar. Let us denote let $A_n = \{x\in [0,1]:\text{ first n digits of }x\neq 4\}$. Clearly, $$ A_{n+1}\subseteq A_n, \quad A = \lim\limits_{n\to\infty}A_n = \bigcap\limits_{n=1}^\infty A_n,\quad \mu(A) = \lim\limits_{n\to\infty}\mu(A_n). $$ E.g. $A_1 = [0,0.4)\cup [0.5,1]$ with $\mu(A_1) = 0.9$. To calculate $A_2$ we first notice that it is a subset of $A_1$ such that $2$-th digit of any number in $A_2$ is any digit but $4$.

That gives an idea that each time it's just sufficient to consider first-step truncation. Let us denote $$ K(B) = \{x\in B:\text{ first digit of }x\neq 4\} $$ and $10^kB = \{10^kx:x\in B\}$. Clearly, we have $A_1 = K([0,1])$ and $A_{n+1} = 10^{-n}K(10^nA_n)$.

Note that each time $10^n A_n$ is a union of intervals with integer bounds, so $$ \mu(K(10^nA_n)) = 10^{n}\frac9{10}\mu(A_n) = 9\cdot 10^{n-1}\mu(A_n) $$ so $$ \mu(A_{n+1}) = \frac{9}{10}\mu(A_n) $$ and we come to the finish line: $$ \mu(A) = \lim\limits_{n\to\infty}\mu(A_n) = 0. $$

Notice that equality $\mu(10^k B) = 10^k \mu(B)$ we just need for the finite unions of intervals, so you can easily prove it.

A quick way to see the solution is to consider a random (uniformly distributed) number in $[0,1]$. By the infinite monkey principle, the decimal expansion of such a random number must contain a $4$ almost surely. But the probability measure of the uniform distribution is just the Lebesgue measure on $[0,1]$, so we're excluding a set of measure $1$. Therefore $A$, consisting of the numbers that are left, must have measure $0$.

Making this rigorous probably entails doing something like Chandrasekhar's comment.

Let $B = A \setminus \{1\}$.

The first digit of any element $x$ of $B$ is not 4. The fractional part of $10x$ is also in $B$. We thus have a disjoint union

$$B = \bigcup_{i \in \{0,1,2,3,5,6,7,8,9\}} \left( \frac{i}{10} + \frac{1}{10} B \right) $$

so $\mu(B) = 9 \cdot \frac{1}{10} \mu(B)$. Solving gives $\mu(B) = 0$ or $\mu(B) = +\infty$, and the latter is clearly impossible.

Here's another solution which is related to Henning's. This solution uses methods from algorithmic randomness.

Any Martin-Löf random (indeed, any Kurtz random) must have at least one (indeed, infinitely many) 4's in its decimal expansion. To see this, let $r$ be a real number with no 4's in its decimal expansion. The computable betting strategy (i.e. martingale) that spreads all current capital evenly over all digits except 4 will then succeed (with a computable rate of success) on $r$. In other words, having seen some initial segment of the decimal expansion of $r$, this strategy bets that then next bit isn't a 4. Since the house (i.e. the Lebesgue measure) gives uniform odds for each digit, we're guaranteed a payback factor of $10/9$. Since $(10/9)^{n} \to \infty$ (computably), we'll win arbitrarily much.

Thus your set $A$ is contained in the complement of the collection of Martin-Löf (or Kurtz) randoms. Since the collection of Martin-Löf (or Kurtz) randoms has measure 1, your set $A$ must have measure 0.