Show that the set of all real numbers whose decimal notation includes the number 2 are borel measurable

Question

I want to show that every real number that includes the number 2 in their decimal notation is borel measurable.

I know that every singleton is borel measurable and every countable union of that is borel measurable but there is an uncountable number of real numbers with a 2 in it. I'm not sure how to show it. A set $A$ is borel measurable if $A \in \mathcal{A}$ the Borel $\sigma$ algebra. Is it possible to proof the definition of a $\sigma$ algebra that $A$ is closed under complementation and closed under countable unions?

Other than that I thought of reducing the problem to the rational numbers that are countable. Might work with the fact that $ \mathbb{Q} $ are a dense subset of $ \mathbb{R} $. Maybe that the complement of every singleton (which is closed and borel measurable) is open and in every open set with a real number is also a rational number and therefore I get a countable amount of sets?

Answer 1

I want to show that every real number that includes the number 2 in their decimal notation is borel measurable.

Real numbers can't be measurable. It's sets of real number that are measurable or not. So the correct statement would be: the set of all real numbers that include the number $2$ in their decimal notation (let's name it $A$) is Borel measurable.

A set $A$ is borel measurable if $A \in \mathcal{A}$ the Borel $\sigma$ algebra. Is it possible to proof the definition of a $\sigma$ algebra that $A$ is closed under complementation and closed under countable unions?

Absolutely not. Being closed under complementation and countable unions is a property a family of sets $\mathcal{S}$ can have; then we call that family a $\sigma$-algebra. But $A$ is a set of real numbers, so it doesn't make sense to ask whether it is closed under complementation or countable unions.

The family $\mathcal{A}$ of Borel sets is a $\sigma$-algebra and a set $S$ is Borel if $S \in \mathcal{A}$. So our goal is not to show that $A$ is a $\sigma$-algebra (which doesn't make sense), but that $A \in \mathcal{A}$. We usually do that by explicitly constructing $A$ from open sets using complements and countable unions.

Now to the question: note that $A$ can be written as

$$A = \bigcup_{k \in \mathbb{Z}} A_k$$

where $A_k$ is the set of all real numbers that have a $2$ in their decimal notation exactly on the place $k$, i.e.

$$A_k = \left\{ \pm \sum_{n=-\infty}^{m} a_n \cdot 10^n : k \leqslant m \ \& \ a_k = 2 \right\}$$

where $a_n$'s are the digits of the expansion, so $a_n \in \{ 0, 1, \ldots, 9 \}$. Try proving that each $A_k$ is Borel, which will conclude the proof, since a countable union of Borel sets is Borel.