Measure of the set $\{x\in [0,1]: \text{the decimal expansion of } x \text{ contains infinitely many 7.} \}$.

Question

I got this question in an old qualifier exam of Real analysis. Let $A$ be the set of all real numbers in the closed interval $[0,1]$ whose decimal expansion contains infinitely many 7. Find the Lebesgue measure of the set $A.$

(Now ownward when I say a number, it is supposed to be in the unit interval.) If I define $A_i$ to be the set of all those numbers whose decimal expansion has a $7$ at the $i$th place. It is clear that the set $A$ in the question is limit supremum of all $A_i$. And, I can show that $A_i$ are independent event (Lebesgue measure restricted to the unit interval is a probability measure). We next observe that $$\sum |A_i| = \infty.$$ An application of second Borel-Cantelli’s lemma therefore gives that the measure of $A$ is 1.

My problem is that the notion of independence is a probabilistic notion (and not introduced in most real analysis classes). So, I am interested in a more direct (say, analytical?) approach to this problem.

While writing this question this thought came to mind that I want something along the following line: Try showing that the set of all numbers whose decimal expansion has only finitely many $7$ appearing has a measure 0. In order to do so, we first obeserve that there are only countably many finite subsets of natural number. So if for a fixed finite subset $F$ of natural numbers, I can show that the set of numbers which has 7 only at those locations specified by the set $F$ has measure 0, then I will be done. Any thought on this? Would this work or not? Any alternate approach is also a welcome?

Meanwhile I will try the above idea and update accordingly.

Answer 1

Here is another approach to atone for my sins above.

Let $N$ be the set of numbers in $I$ that have no $7$s in their expansion. I claim that $mN =0$.

Let $N_{d_1,...,d_n}$ be the set of numbers in $N$ that start with digits $d_1,...,d_n$. Hopefully it is clear that $m N_{d_1,...,d_n} = {1 \over 10^n} mN$. Since $N = \cup_{d \neq 7} N_d$, we have $m N = {9 \over 10} m N$ and so $m N =0$.

It follows that $m N_{d_1,...,d_n} = 0$ and so $m (\cup_{d_1,...,d_n} N_{d_1,...,d_n}) = 0 $. Finally, $m (\cup_n \cup_{d_1,...,d_n} N_{d_1,...,d_n}) = 0$.

Hence the set of numbers that have a finite number of $7$s in their expansion is a null set.

Answer 2

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.