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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.

A. Howells
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3 Answers3

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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?

Well I just found these wikipedia pages:

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 $1$.

justt
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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.

A. Howells
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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 $$\log_2 \frac{(k+1)^2}{k(k+2)}$$

see Gauss-Kuzmin theorem

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

orangeskid
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