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Normal numbers have a 'random' expansion. For example, in base 10 it means that all digits $0,1,\dots,9$ occur 'equally often' in its decimal expansion. A longstanding open problem is: is $\pi$ a normal number?

I am interested in the set of normal numbers. This set is not open and also not closed. It is everywhere dense in $\mathbb R$, but its complement is also everywhere dense. A basic question to ask about a subset of $\mathbb R$ is: is it a Borel set?

To be precise, one has to define what normal number is. For simplicity, we consider only elements of the real interval $[0,1]$. A simply normal number is defined for a specific base $b\geq2$. Then, a normal number is a number that is simply normal in every base $b$. I think that it would be easiest to start with the most simple case: simply normal numbers in base $2$. The set then looks as follows:

$$S=\left\{\sum_{n=1}^{\infty} a_n\cdot 2^{-n}\;\bigg|\; a_n\in \{0,1\} \bigg|\; \lim_{n\rightarrow\infty}\frac{a_1+\ldots+a_n}{n}=\frac12\right\}.$$ Is this a Borel set?

Daniele Tampieri
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Riemann
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  • Thank you for pointing this out! There was an error in my question. I have now corrected it – Riemann Dec 10 '23 at 22:49
  • Please leave a comment to explain why you downvoted my question. – Riemann Dec 10 '23 at 22:51
  • Which community guideline does this question violate? – Riemann Dec 10 '23 at 22:53
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    Hi! That happens a lot (uncommented down-votes), but I think in this case it's not too difficult to guess: You posted a bare question, without any context as to where this question comes from (e.g., is it from a course?), no thoughts about how you've tried to proceed, or what you know that might be helpful. It would help to take the time to edit your post to add these details in! – Brian Tung Dec 10 '23 at 22:55
  • I also suspect someone is holding the fact that you're not new here against you; that is, they're probably expecting you to know that such a question (more or less a PSQ) is not up to scratch. ETA: The down-vote may also have come before you fixed the question. – Brian Tung Dec 10 '23 at 22:57
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    Ok I have now added a lot of context! – Riemann Dec 10 '23 at 23:03
  • Great! It might also be helpful to give an ordinary English-language characterization of $S$, both as a guide for readers and as a check that you're correctly understood. "Real values in $[0, 1]$ whose binary representation is half zeros and half ones in the limit," or something like that (if I've read it correctly myself). – Brian Tung Dec 10 '23 at 23:30
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    Yes, re-framing in terms of normal numbers is good! Incidentally, not only are simply normal numbers Borel, but absolutely normal numbers are also Borel. There's a paper by Becher on this, I believe. – Brian Tung Dec 11 '23 at 00:48

1 Answers1

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Yes, this is a Borel set.

It is often quite difficult to prove something is a Borel set by means of “mapping in” to the set. Your set is defined as the image of a certain measurable set. The trick is to instead define it as the inverse image of a certain measurable set.

More formally, define, for all $x \in [0, 1]$ and $n \in \mathbb{N}_+$, $x_n \in \{0, 1\}$ to be the $n$th bit of $x$’s binary expansion after the radix point. That is, we have $x = \sum\limits_{n = 1}^\infty x_n 2^{-n}$. It turns out not to matter how you define the expansion of a dyadic number (one with a terminating and a nonterminating expansion),so just pick one definition and stick with it.

It is fairly easy to show that each function $x \mapsto x_i$ is Borel measurable. Now define $f_i(x) = \frac{x_1 + \cdots + x_i}{i}$, which is also clearly measurable. Then $S = \{x \in [0, 1] \mid \lim\limits_{n \to \infty} f_n(x) = 1/2\}$ (check this for dyadic numbers). Note that we redefine $S$ by mapping out of $[0, 1]$, not by mapping in.

By expanding the definition of $\lim\limits_{n \to \infty}$, we see that $S = \bigcap\limits_{q \in \mathbb{Q}_+} \bigcup\limits_{n \in \mathbb{N}_+} \bigcap\limits_{m \in \mathbb{N}, m \geq n} f_m^{-1}(B_q(1/2))$, where $B_q$ is a ball of radius $q$ as usual. Using measurability and the properties of a $\sigma$-algebra, $S$ is Borel.

Mark Saving
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    This proof can probably be generalized to other bases $b$. Now taking the intersection of all those, we get theset of all normal numbers. Hence this is also a Borel set. – Riemann Dec 10 '23 at 23:42
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    @Riemann Indeed. I also imagine that the measure of each such set is $1$, so the measure of their intersection will also be $1$. – Mark Saving Dec 10 '23 at 23:53
  • @ Mark Saving, that is an interesting thing to consider. How about all the sets $S_t:={\sum a_n2^{-n}\mid \lim_{n\rightarrow\infty}\frac{a_1+\dots+a_n}{n}=t}$? Would they then all have measure $0$, except for $t=\frac12$? – Riemann Dec 10 '23 at 23:56
  • I will actually make another question about this. – Riemann Dec 10 '23 at 23:58
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    @Riemann I just looked it up on Wikipedia and it appears that the non-normal numbers have measure $0$. – Mark Saving Dec 11 '23 at 00:00
  • Here is my new question: https://math.stackexchange.com/questions/4824191/do-the-normal-numbers-have-lebesgue-measure-1 – Riemann Dec 11 '23 at 00:04