measure of set in $\mathbb{R}$ which has only finite number of 4's in decimal expansion

Question

1. Given a set $A = \{ x | \text{ $x$ has decimal expansion which has only finite 4's } , x \in [0,1] \,\, \}$ show that $\lambda(A) =0$

2. Given a set $B = \{ x | \text{ $x$ has decimal expansion which has only finite 4's } , x \in [0,\infty) \,\, \}$ show that $\lambda(B) =0$

EDIT : I have edited answer. Can someone check now.

$C_{i} =\{x |\text{x s.t $(0..i-1)$ places can have any digits and 4 in ith place and there are no 4's after }, x \in [0,1] \} $

assume $ * \in \{0..9\} \,\, , @ \in \{0..9\} \backslash 4$

now , it has already been shown here that Measure of reals in $[0,1]$ which don't have $4$ in decimal expansion. Therefore

$C_0 = \{0.@@@... \} , \lambda(C_0) = 0$

$C_1 =\{0.4@@@...\} ,C_1 = 10^{-1} \Big[ 4+C_0 \Big] \implies \lambda(C_1) = 10^{-1} \lambda(4+C_0) = 10^{-1} \lambda(C_0) =0 $

$\displaystyle C_2 = \{0.*4@@...\},C_2 = 10^{-2}\Big[\sum_{i=0}^{9} (i*10 +4 + C_0) \Big] \implies \lambda(C_2) = 10^{-2} \lambda(C_0) = 0$

$\vdots$

Now, we can say that $A$

$1)\displaystyle A = \bigcup_{i=0}^{\infty} C_i$ (disjoint union) from which we have that $\displaystyle \lambda(A) = \sum_{i=0}^{\infty} \lambda(C_i) = 0$

$2) \displaystyle B = \bigcup_{i=0}^{\infty} (A+i)$ (disjoint union) from which we have that $\displaystyle \lambda(B) = \sum_{i=0}^{\infty} \lambda(A) = 0$ .

Is there anything wrong with above argument.

Answer 1

My favorite proof of this is with Borel-Cantelli. I'm going to show that if $$A = \{ x \in [0, 1]: x \text{ has finitely many 4's in its decimal expansion} \},$$ then $\lambda(A) = 0$, where $\lambda$ is Lebesgue measure. Then because $$B = \{ x \in [0, \infty): x \text{ has finitely many 4's in its decimal expansion} \}$$ is contained in the countable union $\bigcup_{n=0}^\infty (A + n)$ of right translations of $A$ by nonnegative integers, it follows that $\lambda(B) \leq \sum_{n=0}^\infty \lambda(A + n) = 0$ by $\sigma$-additivity.

Let $X_1, X_2, X_3, ...$ be i.i.d. discrete r.v.'s which are uniformly distributed on $\{ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 \}$. Then the r.v. $$X := \sum_{n=1}^\infty \frac{X_n}{10^n} = 0.X_1 X_2 X_3 ...,$$ which has $i$th decimal digit equal to $X_i$ for all $i \geq 1$, is uniformly distributed on $[0, 1]$. That is, $\Bbb{P}(X \in E) = \lambda(E)$ for every Lebesgue measurable subset $E \subseteq [0, 1]$.

(I'm handwaving over this part of the proof; it's not hard to show that for any interval of the form $I(a, k) := [a/10^k, (a+1)/10^k]$, where $0 \leq a \leq 10^k -1$, the probability $\Bbb{P}(X \in I(a, k)) = 10^{-k}$, and since the intervals $I(a, k)$ generate the Lebesgue $\sigma$-algebra on $[0, 1]$, we derive the claim that $\Bbb{P}(X \in E) = \lambda(E)$ for any Lebesgue measurable $E \subseteq [0, 1]$.)

Now by Borel-Cantelli, since $$\sum_{n=1}^\infty \Bbb{P}(X_n = 4) = \sum_{n=1}^\infty 0.1 = \infty,$$ it follows that $\Bbb{P}(X_n = 4 \text{ infinitely often}) = 1$. But since the $X_n$ represent the decimal digits of the random variable $X$, that's the same as saying $\Bbb{P}(X \in A^c) = \lambda(A^c) = 1,$ so $\lambda(A) = 0$.