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Claim. For $0<A<c/(1+c^2)$ the limit

$$f(A,c)=\lim_{N\to\infty}S(N,A,c)=\frac{12A\log c}{\pi^2}$$

exists and has this value, where $S(N,A,c)$ is the measure of the $\alpha\in(0,1)$ with $|\alpha-x/y|\le A/y^2$ for some coprime $x,y$ with $N\le y\le cN$ (the site's set of [[problems/irrationality/E1001/_index|Problem 1001]], defined with a strict inequality, differs from it by a countable set). This is Theorem III of P. Erdős, P. Szüsz and P. Turán, Remarks on the theory of diophantine approximation, Colloq. Math. 6 (1958), 119--126, received by the journal on 1957-11-16 and in revised form on 1958-06-10, and published in 1958 (the page's date is the volume's year, with the day set to its first); the card erdos_1958_remarks_theory_diophantine_approximation records the paper. In this range the approximation intervals around distinct reduced fractions with denominators in $[N,cN]$ are pairwise disjoint, so the measure is a sum over those fractions, and the value follows from the asymptotics of a weighted totient sum. The same paper proves lower bounds for $\liminf S(N,A,c)$ for all $A>0$ and $c>1$ (Theorems I and II) and an upper bound below $1$ for $A>10$ and $c>10$ (Theorem IV), and poses the existence of the limit for all parameters, and its explicit form, as its Problem I (P 241).

Covers. The existence and the value of the limit for $0<A<c/(1+c^2)$. It settles nothing for larger $A$, where the intervals overlap: existence there with closed forms for $A\le1/c$ is Kesten's, existence for all parameters is Kesten and Sós's, and the general form is given by Xiong and Zaharescu and Boca.

Acceptance. The refereed evidence is the journal publication cited above. The reviewed evidence is the documented acceptance by the catalog erdosproblems.com, whose page for the problem (the discussion link) carries the label SOLVED and whose curator, Thomas Bloom, credits Erdős, Szüsz and Turán with the value $12A\log c/\pi^2$ in this range; he is not an author of the paper. No independent check of the proof is recorded. The claim value is proved because the part it settles is proved in the affirmative, with the value found.

Depends on. Nothing in this wiki.