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.
 A: 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.
A: 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: 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.
A: 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.
A: We can change assumption of $A$ being subset of $[0,1]$, to be a subset of $[0,1)$. That won't change measure of but will make our calculation cleaner.
Let $A_{n}$ be partition of $[0,1)$ into $10^{n}$ left-closed intervals of equal length. Further, let $A_{n}^{\prime}$ be $A_{n}$ minus intervals which contain numbers, which fractional part of the decimal expansion up to n-th digit poses digit $4$.
For example 
$$A_{0}=A_{0}^{\prime}=\{[0,1]\}$$
$$A_{1}=\{[0,0.1), [0.1,0.2),[0.3,0.4),[0.4,0.5)...,[0.9,1)\}$$
$$A_{1}^{\prime}=\{[0,0.1), [0.1,0.2),[0.3,0.4),[0.6,0.7),[0.7,0.8),[0.9,1]\}$$
$$A_{2}=\{[0,0.01), [0.01,0.02),...,[0.99,1)\}$$
$$A_{2}^{\prime}=A_{2}- \{[0.07,0.08),[0.17,0.18),...[0.70,0.71),...,[0.79,0.80),...,[0.99,0.1)\}$$
Notice


*

*Length of each interval $X$ in $A_{n}$ (or $A_{n}^{\prime}$) is $\frac{1}{10^n}$.

*$A\subset \bigcup A_{n}^{\prime}$ for all n. As a consequence 
$$
\mu(A)\leq\mu(\bigcup A_{n}^{\prime})=\Sigma_{X\in A_{n}^{\prime}}\mu(X)=\Sigma_{X\in A_{n}^{\prime}}\frac{1}{10^n}=\frac{\#A_{n}^{\prime}}{10^n}
$$

*For $n>0$, each interval from $A_{n}$ can be uniquely paired with the sequence $a_{1}a_{2}...a_{n}$, (where $a_{k}\in\{0..9\}$) in the following way: let $X\in A_{n}$. Then X is of the form $[0.a_{1}a_{2}...a_{n}, 0.a_{1}a_{2}...a_{n}+ \frac{1}{10^n})$. On the other hand if we have sequence $a_{1}a_{2}...a_{n}$, then in $A_{n}$ there exists interval for which left bound is a number $0.a_{1}a_{2}...a_{n}$.

*If we take interval $X\in A_{n}$ that is related to the $a_{1}a_{2}...a_{n}$ sequence, and for some $k$, $a_{k}=4$ then since $0.a_{1}a_{2}...a_{n}\in X$ we deduce $X\not\in A_{n}^{\prime}$.

*On the other hand, if number $0.b_{1}b_{2}...b_{n}...$ belongs to interval $[0.a_{1}a_{2}...a_{n}, 0.a_{1}a_{2}...a_{n}+ \frac{1}{10^n})$, and for some $k\leq n$, $b_{k}=4$, then also $a_{k}=4$. So if $X\not\in A_{n}^{\prime}$ then sequence $a_{1}a_{2}...a_{n}$ related to X has some $a_{k}=4$.

*So to calculate $\#A_{n}^{\prime}$, we have to calculate how many sequences $a_{1}a_{2}...a_{n}$ are there which do not poses number 4.


For example $\#A_{0}^{\prime}=1$, $\#A_{1}^{\prime}=9$, $\#A_{2}^{\prime}=81$, and it's easy to notice that in general $\#A_{n}^{\prime}=9^{n}$.
From that 
$$
0\leq\mu(A)\leq\frac{\#A_{n}^{\prime}}{10^n}=\frac{9^{n}}{10^n}\to^{n}0
$$
so $\mu(A)=0$. Of course it works for numbers other than 4.
