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

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.

• You need to clarify $C_i=10^iC$. As sets, it doesn't make sense because it would extend $C_i$ up to $[0,10^i]$. Aug 6, 2020 at 7:05
• @Chrystomath yes. I didn't think about that ! Aug 6, 2020 at 7:07
• @Chrystomath. I have edited the answer Aug 7, 2020 at 5:40
• The proof looks good to me. If you want to take a sledgehammer to crack a nut, you can prove it with the strong law of large numbers: for almost all $x$, the proportion of 4s up to the $n$th digit converges to $0.1$ when $n\to\infty$ :) Aug 7, 2020 at 6:16

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 $$\lambda(A) = 0$$.