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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?

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    $\begingroup$ Hint: Consider the set of real numbers whose decimal expansions do not contain a $2$ (and recall facts about the Cantor set). Incidentally, the fact that some real numbers have two decimal expansions needs to be handled at some point, unless you were explicitly told to only consider non-terminating decimal expansions. $\endgroup$ – Dave L. Renfro Oct 29 '17 at 11:37
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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.

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  • $\begingroup$ Thanks a lot for the detailed explanation. I have an Idea how to show that $A_{k}$ is borel measurable. For example for all real numbers that start with the number 2. After the comma I still have all real numbers and they are as the complement of the empty set borel measurable and therefore all numbers that start with a 2 are borel measurable? Not sure if this conclusion is right but if so you could keep doing this for all $A_{k}$ $\endgroup$ – Erdbeer0815 Oct 29 '17 at 20:53
  • $\begingroup$ Sorry, it doesn't make any sense. What do you mean by "after the comma I still have all real numbers"? Are you claiming that $A_0$ (which is the set of all numbers that have a $2$ right before the comma) equals $\mathbb{R}$? Definitely not, since $5 \notin A_0$. I suggest that you try drawing the set $A_0$ on the real line to see what it looks like. $\endgroup$ – Adayah Oct 30 '17 at 6:48
  • $\begingroup$ No 5 would be 2.5 and 14 would be 2.14 ... you can find every real number after the comma. Not sure if you can make it out of that. But I came up with a better idea. Every limit of a sequence in $A_{k}$ has a 2 at the kth position and is therefore element of $A_{k}$ and therefore closed. $\endgroup$ – Erdbeer0815 Oct 30 '17 at 8:24
  • $\begingroup$ You correctly identified that $A_k$ is closed (which suffices to complete the proof), but that would require some better justification. $\endgroup$ – Adayah Oct 30 '17 at 8:38
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    $\begingroup$ The limit of $2.9$, $2.99$, $2.999$, $\ldots$ is $2.9999\ldots = 3$ so it can be written such that it contains a $2$, hence it is in $A_0$. The next idea is good too, but needs to be worked on a little more. The number $527$ has a $2$ on the $1$st place ($7$ is on $0$th), but $527 \notin [20, 30]$. $\endgroup$ – Adayah Oct 30 '17 at 21:15

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