Is this set borel measurable?

Question

I am having trouble on proving this.

Let $\xi $ be the set of all real numbers containing a five in their decimal representation. Why is it borel measurable?

I hope someone can help me out here. Thanks

Answer 1

Hint: Prove that

$$ A_n = \{ x\in \mathbb{R} \ : \ \lfloor 10\cdot (10^{n-1} x -\lfloor 10^{n-1} x \rfloor) \rfloor = 5 \} = \{ x\in \mathbb{R} \ : \ x \text{ has a } 5 \text{ at the nth place in the decimal expansion} \}$$

is Borel measurable. The countable union of Borel measurable sets is Borel measurable.

Added: To prove that $A_n$ is Borel. Prove that the function $$f_n:\mathbb{R}\rightarrow \mathbb{R}, \ f_n(x)= \lfloor 10\cdot (10^{n-1} x -\lfloor 10^{n-1} x \rfloor) \rfloor$$

is a Borel function (which you can either check by hand or use the fact that it is piecewise continuous). To see that $f_n$ gives the $n$th place in the decimal expansion, we consider $x= b_m \dots b_0. a_1 a_2 \dots$, then we get $$ 10^{n-1} x = b_m \dots b_0 a_1 \dots a_{n-1}. a_n a_{n+1} \dots $$ and thus $$ \lfloor 10^{n-1} x \rfloor = b_m \dots b_0 a_1 \dots a_{n-1}. \overline{0},$$ which implies $$ 10^{n-1} x - \lfloor 10^{n-1} x \rfloor = 0.a_n a_{n+1} \dots $$ and therefore $$ \lfloor 10\left(10^{n-1} x - \lfloor 10^{n-1} x \rfloor \right) \rfloor = \lfloor a_n. a_{n+1} \dots \rfloor = a_n. $$

Easier solution: Prove that $$A_n \cup \{ x \text{ has a } 6 \text{ at the } n\text{th place in the decimal expansion and only zeros afterwards.} \}$$ is closed (if every element of a sequence has a 5 at the nth place of their decimal expansion, then the limit will have a 5 at the nth place of their decimal expansion or (as pointed out by @Erdbeer0815 it converges to the number having a 6 at the nth place of their decimal expansion and only zeros afterwards) and deduce that it is a Borel measurable set.