The Duffin-Schaeffer conjecture is an old conjecture in metric number theory which has withstood attempts to solve it for about 70 years. The statement can be found here: http://en.wikipedia.org/wiki/Duffin%E2%80%93Schaeffer_conjecture

My question concerns a special case of the conjecture. It is an immediate corollary to the Borel-Cantelli Lemma that if there exists a function $f : \mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that the $E_q$'s are independent (where $\displaystyle E_q = \left[0, \frac{f(q)}{q}\right] \cup \left[1 - \frac{f(q)}{q}, 1\right] \cup \bigcup_{\substack{1 \leq p \leq q \\ \gcd(p,q) = 1}} \left[\frac{p}{q} - \frac{f(q)}{q}, \frac{p}{q} + \frac{f(q)}{q}\right]$. interpreted as the event that $\alpha$ is in the right hand side), then the Duffin-Schaeffer conjecture is true; namely that $\displaystyle \sum_{q=1}^\infty \frac{f(q)}{q} \phi(q) = \infty$ implies that $\mu(\limsup_{q \rightarrow \infty} E_q) = 1$.

So my questions are: 1) Are there any known examples of $f$ such that the $E_q$'s are independent, and 2) is there any hope that a notion of 'almost independent' or 'weakly correlated' can be applied to solve the conjecture? An example of something like this would be Janson's inequality generalizing Chernoff's inequality.

  • $\begingroup$ About 2) to apply Borel-Cantelli (in that direction) it is enough to have pairwise independence (every pair or $E_q$ are independent). More generally, it is a consequence of the second moment method, so having good bounds on $P(E_q \cap E_r)$ is also enough. $\endgroup$ May 16, 2011 at 22:15
  • $\begingroup$ To apply the second moment method, one would need a random variable... what would be the random variable in this case? $\endgroup$ May 19, 2011 at 3:08
  • $\begingroup$ The random variable $X_q$ would be the indicator function of the event $E_q$. Then $\limsup E_q$ is the event that $\sum X_q$ diverges. (By the way, Ori probably did not get notified of your comment since it does not begin by @Ori.) $\endgroup$
    – Did
    May 22, 2011 at 16:27

1 Answer 1


First of all what Ori said about pairwise independence being sufficient for the divergence part of the Borel-Cantelli Lemma is correct- I believe this observation is originally due to Erdos and Renyi. Now I will try to answer both of your questions, in reverse order:

To prove the Duffin-Schaeffer Conjecture it is enough to show that the sets $E_q$ are quasi-independent on average, whenever the sum of their probabilities diverges. Quasi-independence on average is the condition that $$\sum_{1\le q,r\le Q} P(E_q\cap E_r)\ll \left(\sum_{1\le q\le Q}P(E_q)\right)^2,$$ where $\ll$ is Vinogradov notation. Actually it is sufficient just to show that this inequality holds over a subsequence of $Q$ tending to $\infty$. Then by a standard variance argument (for example see Lemma 2.3 of Glyn Harman's book Metric Number Theory) you get that $$P(\limsup E_q)>0.$$ Finally there is a zero-one law due to Gallagher which then allows you to conclude that the limsup set has measure $1$.

For the first question, I don't have an example where the sets $E_q$ are actually bona-fide independent, or even pairwise independent. However there is a trivial estimate for the measure of the intersection of $E_q$ and $E_r$ (for $q\not= r$), which is that $$P(E_q\cap E_r)\le 8f(q)f(r).$$ Now let $f:\mathbb{N}\rightarrow [0,\infty)$ be the function defined by $f(q)=1/2q$ if $q$ is prime, and $f(q)=0$ otherwise. Then it is easy to check that when $q$ is prime, $$P(E_q)=\frac{\varphi(q)}{q^2}=\frac{1}{q}\left(1-\frac{1}{q}\right)\ge f(q)/2.$$ Therefore in this case we have that the sum of the measures of the sets $E_q$ diverges, and that all of the sets are pairwise quasi-independent (i.e. not just quasi-independent on average).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .