# What is the Lebesgue measure of Luzin's non-Borel set of reals?

On Wikipedia there's a nice example of a non-Borel set due to Luzin. For completeness, I'll summarize it here. For $$x\in[0,1]$$, let \begin{align} x=a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \cfrac{1}{\ddots\,}}}} \end{align} be the continued fraction expansion of $$x$$. Let $$A$$ be the set of numbers $$x\in[0,1]$$ whose corresponding sequence $$a_0,a_1,a_2,\cdots$$ admits an infinite subsequence $$a_{k_1},a_{k_2},a_{k_3},\cdots$$ such that for each $$i$$, $$a_{k_{i+1}}$$ is divisible by $$a_{k_i}$$. Then $$A$$ is not Borel measurable.

However, $$A$$ is an analytic set, and in particular it is Lebesgue measurable. What is the Lebesgue measure of $$A$$?

My thinking is that it should be zero, since the existence of such an infinite subsequence strikes me as an improbable coincidence. However, I've forgotten whatever little I knew about continued fractions, so that's far from a proof.

I tried searching this site before asking, of course. I found only one mention, here (last sentence), where $$A$$ is again conjectured to have measure zero.

• An amazing fact! Commented Sep 11, 2023 at 20:37

I don't have a complete argument for it but I don't agree with your conjecture and believe the Lebesgue measure is 1. I might be mistaken but hope you could at least consider my position and maybe end up with a proof yourself.

Consider a much simpler system: an i.i.d sequence of integers, where each integer has a positive probability to appear. Then the probability of the Lusin event is obviously $$1$$. Indeed you just need one integer to be repeated infinitely often for the Lusin event to hold, but even if you insist on being a strict divisor, after you have seen an integer $$k$$, you are bound with probability $$1$$ to see $$2k$$ after some time, and then you could repeat the argument ensuring $$4k$$ will appear, etc.

Now what is the relevance of this toy model to the case at hand ? Surely the continued fraction expansion of a uniform random real number is not an i.i.d. sequence?

Basically what they tell you is the following: you can easily define the continued fraction expansion of $$x\in(0,1)$$ with just two functions: $$a_1(x) = \lfloor 1/x \rfloor$$ and $$h(x) = \{ 1/x\}$$. You can then set $$a_2 = a_1\circ h$$, $$a_3 = a_1 \circ h \circ h$$, etc.

You can define a probability measure on $$(0,1)$$ by setting $$\mu(dx) = \dfrac {dx}{\ln(2)(1+x)}$$. Then $$h : (0,1) \to (0,1)$$ is a measure-preserving ergodic operator called the Gauss-Kuzmin-Wirsing operator.

Consider $$X$$ with distribution $$\mu$$. The "measure-preserving" part tells you that the sequence $$X,h(X),h(h(X)), ...$$ is identically distributed. The "ergodic" part tells you that the sequence, even though it's not independent, looks independent "in the long run". That is, $$P(h^n(X) \in A \mid X \in B) \to P(X \in A)$$ as $$n$$ goes to infinity.

If you now apply $$a_1$$ to this sequence, you end up with the continued fraction expansion of $$X$$, and we have seen that it is "approximately iid". Its distribution is the image of the measure $$\mu$$ by $$a_1$$, which is the above-mentioned Gauss-Kuzmin distribution, where all integers have positive probability. As a result, one could conjecture that the Lusin event has probability $$1$$ for $$X$$.

Finally, since the measure $$\mu$$ and the Lebesgue distribution on $$(0,1)$$ are mutually absolutely continuous, then what holds with probability $$1$$ for $$\mu$$ holds also with Lebesgue measure $$1$$.

I now believe justt's conjecture that the Lebesgue measure of $$A$$ is one, and I believe I have a proof. It uses some similar ideas to justt's proof outline (in particular, Khinchin's constant), but in a more direct manner. The big idea here is that the existance (and value) of Khinchin's constant implies that for almost every $$x$$, the continued fraction expansion of $$x$$ has infinitely many coefficients which are less than $$3$$; this idea was suggested to me by Alex Mine.

As in the question, $$x$$ will be a number in $$[0,1]$$ and $$a_0,a_1,a_2,\cdots$$ will be the coefficients of its continued fraction expansion. Khinchin's result says that for almost every $$x\in[0,1]$$, the limit $$n\to\infty$$ of the geometric mean of $$a_1,\cdots,a_n$$ is some number, Khinchin's constant, which does not depend on $$x$$. In particular, per Wikipedia the constant is about $$2.7$$ or so.

Let $$B$$ denote the set of numbers whose continued fraction expansion has either infinitely many $$1$$s or infinitely many $$2$$s (or both). I claim that (i) $$B$$ is a subset of the set $$A$$ defined in the question, and (ii) $$B$$ contains every $$x$$ such that $$\lim\limits_{n\to\infty}(a_1a_2\cdots a_n)^{1/n}$$ is equal to Khinchin's constant.

For (i), notice that given $$x\in B$$, there's a subsequence of $$a_1,a_2,a_3,\cdots$$ which is either all $$1$$ or all $$2$$, and hence $$x\in A$$.

For (ii), notice that given $$x\notin B$$, the corresponding sequence $$a_1,a_2,a_3,\cdots$$ contains only finitely many numbers less than $$3$$, and hence $$\liminf\limits_{n\to\infty}(a_1a_2\cdots a_n)^{1/n}\ge 3$$. In particular, $$\lim\limits_{n\to\infty}(a_1a_2\cdots a_n)^{1/n}$$ cannot be Khinchin's constant.

Therefore, $$A$$ contains a set of full measure, and hence itself is a full measure set.

There exists a result for continued fractions similar of "normal" numbers: for almost every $$x\in [0,1]$$, the number $$k$$ occurs in the continued fraction of $$k$$ (in the limit) with probability $$\log_2 \frac{(k+1)^2}{k(k+2)}$$

So for almost all numbers $$x$$ there exists infinitely many $$1$$'s ( and $$2$$, and $$3$$ $$\ldots$$,) in the continued fraction of $$x$$.