Let $A\subseteq \mathbb{R}$ be the set $A=\{x\in (0,1):$ the decimal expansion of $x$ contains infinitely many 7's}. Show that $A$ is a Borel set.
My thoughts: The collection of rational numbers $\in (0,1)$ whose decimal exp. contains $\infty$-many 7's is clearly Borel because the rational numbers are countable and any subset of a countable set is itself countable and hence Borel. Now the tricky part is to deal with the irrational numbers between 0 and 1. I know that the collection of irrationals $\in (0,1)$ is Borel but why should a subset of a Borel set be Borel? I thought about using the Theorem "Every continuous function is Borel measurable", to prove this but I am a bit lost on how to do this. If the function,
$
f(x) =
\begin{cases}
0 & \text{if x doesn't have a decimal exp w/infinitely many 7's} \\
1 & \text{if x has a decimal expansion with infinitely many 7's} \\
\end{cases}
$
were continuous then the issue would be resolved, but it is clearly not when the domain is irrational numbers and the range is the reals. Is there some way to adjust the function so that it is continuous? Thanks for reading.
Best Answer
There is another way to attack your problem: write down a conditional that defines your set.
Consider the following set: $$A_n := \{x\in(0,1)\mid \text{the $n$th digit of the decimal expression of $x$ is 7}\}.$$ You can see that $A_n$ is Borel, as it is a finite union of half-intervals. Then your set can be described as follows: $$x\in A \iff \forall n\in\mathbb{N}\exists m>n [x\in A_m].$$ Hence $A=\bigcap_{n=1}^\infty\bigcup_{m>n} A_m$, which is also Borel.
I want to remark that my strategy is a common way to determine the complexity of a given set. With some trick (by replacing $\epsilon$ or $\delta$ to $1/n$ for natural numbers $n$), you can see that the set $$\{x\in \mathbb{R} \mid \text{$f$ is continuous at $x$}\}$$ is also Borel regardless of what $f$ is.