Functions of bounded variation and a theorem of Khinchin

Jordan Bell
March 12, 2016

For q1 let

𝒜q={aq:0aq,gcd(a,q)=1}.

The sets 𝒜q are pairwise disjoint. In particular 0𝒜1 and 0𝒜q for q>1. We have

[0,1]=q1𝒜q.

Write μ for Lebesgue measure on [0,1].

The following is a version of a theorem of Khinchin about continued fractions.11 1 John J. Benedetto and Wojciech Czaja, Integration and Modern Analysis, p. 183, Theorem 4.3.3. In the literature on Diophantine approximation it is usually proved with the Borel-Cantelli lemma, rather than the machinery of bounded variation and almost everywhere differentiability.

Theorem 1 (Khinchin).

Let F:Z1R>0 and let A be the set of those x[0,1]Q such that there are infinitely many q for which there is some aqAq satisfying

|α-aq|<1qF(q).

If

q=11F(q)<,

then μ(A)=0.

Proof.

Define f:[0,1]>0 by

f(x)={1qF(q)x𝒜q0x[0,1].

Let N1. If

0=t0<t1<<tN=1,

then

j=0Nf(tj) =q=1j=0Nf(tj)1𝒜q(tj)
=q=1j=0N1qF(q)1𝒜q(tj)
q=11F(q),

and so

j=1N|f(tj)-f(tj-1)|2j=1Nf(tj)2q=11F(q).

Therefore

V(f)2q=11F(q)<

by hypothesis, where V(f) denotes the variation of f on [0,1]. The set Df of points at which f is differentiable is a Borel set,22 2 V. I. Bogachev, Measure Theory, volume 1, p. 371, Theorem 5.8.12. and because f has bounded variation, μ(Df)=1.33 3 V. I. Bogachev, Measure Theory, volume 1, p. 335, Theorem 5.2.6. Let E=Df, whose measure is μ(E)=1. Now let xE. There are xn[0,1], xnx, xnx, with which

f(x)=limnf(xn)-f(x)xn-x=limn0-0xn-x=0.

If anqnx with anqn𝒜qn, then

f(an/qn)-f(x)anqn-x=1qnF(qn)(anqn-x)f(x)=0,

so qnF(qn)|anqn-x|. There is thus some N such that if nN then

qnF(qn)|anqn-x|1,

i.e. if nN then

|x-anqn|1qnF(qn).

Assume by contradiction that xA, so there are anqn𝒜qn, anqnamqm for nm, with

|x-anqn|<1qnF(qn),

and because q=11F(q)< it holds that F(q) and thus 1qnF(qn)0. This means that x-anqn0, which implies that xE. We have shown that if xA then xE, so A[0,1]E and hence μ(A)1-μ(E)=0. ∎