# Functions of bounded variation and a theorem of Khinchin

Jordan Bell
March 12, 2016

For $q\geq 1$ let

 $\mathscr{A}_{q}=\left\{\frac{a}{q}:0\leq a\leq q,\gcd(a,q)=1\right\}.$

The sets $\mathscr{A}_{q}$ are pairwise disjoint. In particular $0\in\mathscr{A}_{1}$ and $0\not\in\mathscr{A}_{q}$ for $q>1$. We have

 $[0,1]\cap\mathbb{Q}=\bigcup_{q\geq 1}\mathscr{A}_{q}.$

Write $\mu$ 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 $F:\mathbb{Z}_{\geq 1}\to\mathbb{R}_{>0}$ and let $A$ be the set of those $x\in[0,1]\setminus\mathbb{Q}$ such that there are infinitely many $q$ for which there is some $\frac{a}{q}\in\mathscr{A}_{q}$ satisfying

 $\left|\alpha-\frac{a}{q}\right|<\frac{1}{qF(q)}.$

If

 $\sum_{q=1}^{\infty}\frac{1}{F(q)}<\infty,$

then $\mu(A)=0$.

###### Proof.

Define $f:[0,1]\to\mathbb{R}_{>0}$ by

 $f(x)=\begin{cases}\frac{1}{qF(q)}&x\in\mathscr{A}_{q}\\ 0&x\in[0,1]\setminus\mathbb{Q}.\end{cases}$

Let $N\geq 1$. If

 $0=t_{0}

then

 $\displaystyle\sum_{j=0}^{N}f(t_{j})$ $\displaystyle=\sum_{q=1}^{\infty}\sum_{j=0}^{N}f(t_{j})\cdot 1_{\mathscr{A}_{q% }}(t_{j})$ $\displaystyle=\sum_{q=1}^{\infty}\sum_{j=0}^{N}\frac{1}{qF(q)}\cdot 1_{% \mathscr{A}_{q}}(t_{j})$ $\displaystyle\leq\sum_{q=1}^{\infty}\frac{1}{F(q)},$

and so

 $\sum_{j=1}^{N}|f(t_{j})-f(t_{j-1})|\leq 2\sum_{j=1}^{N}f(t_{j})\leq 2\sum_{q=1% }^{\infty}\frac{1}{F(q)}.$

Therefore

 $V(f)\leq 2\sum_{q=1}^{\infty}\frac{1}{F(q)}<\infty$

by hypothesis, where $V(f)$ denotes the variation of $f$ on $[0,1]$. The set $D_{f}$ 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, $\mu(D_{f})=1$.33 3 V. I. Bogachev, Measure Theory, volume 1, p. 335, Theorem 5.2.6. Let $E=D_{f}\setminus\mathbb{Q}$, whose measure is $\mu(E)=1$. Now let $x\in E$. There are $x_{n}\in[0,1]\setminus\mathbb{Q}$, $x_{n}\neq x$, $x_{n}\to x$, with which

 $f^{\prime}(x)=\lim_{n\to\infty}\frac{f(x_{n})-f(x)}{x_{n}-x}=\lim_{n\to\infty}% \frac{0-0}{x_{n}-x}=0.$

If $\frac{a_{n}}{q_{n}}\to x$ with $\frac{a_{n}}{q_{n}}\in\mathscr{A}_{q_{n}}$, then

 $\frac{f(a_{n}/q_{n})-f(x)}{\frac{a_{n}}{q_{n}}-x}=\frac{1}{q_{n}F(q_{n})\left(% \frac{a_{n}}{q_{n}}-x\right)}\to f^{\prime}(x)=0,$

so $q_{n}F(q_{n})\left|\frac{a_{n}}{q_{n}}-x\right|\to\infty$. There is thus some $N$ such that if $n\geq N$ then

 $q_{n}F(q_{n})\left|\frac{a_{n}}{q_{n}}-x\right|\geq 1,$

i.e. if $n\geq N$ then

 $\left|x-\frac{a_{n}}{q_{n}}\right|\geq\frac{1}{q_{n}F(q_{n})}.$

Assume by contradiction that $x\in A$, so there are $\frac{a_{n}}{q_{n}}\in\mathscr{A}_{q_{n}}$, $\frac{a_{n}}{q_{n}}\neq\frac{a_{m}}{q_{m}}$ for $n\neq m$, with

 $\left|x-\frac{a_{n}}{q_{n}}\right|<\frac{1}{q_{n}F(q_{n})},$

and because $\sum_{q=1}^{\infty}\frac{1}{F(q)}<\infty$ it holds that $F(q)\to\infty$ and thus $\frac{1}{q_{n}F(q_{n})}\to 0$. This means that $x-\frac{a_{n}}{q_{n}}\to 0$, which implies that $x\notin E$. We have shown that if $x\in A$ then $x\notin E$, so $A\subset[0,1]\setminus E$ and hence $\mu(A)\leq 1-\mu(E)=0$. ∎