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This is a continuation of Application of the Borel-Cantelli Lemma.

Call a real number $x$ diophantine iff $$ \exists p,c > 0 \forall q \in \mathbb Z^* \forall a \in \mathbb Z: \left | x - \frac a q \right | > \frac c {|q|^p} $$ Then almost every real number is diophantine. This is Exercise 19.2.8 from Tao Analysis II.

For fixed $p> 2$ and $c > 0$ I know that the set $$ E(p,c) := \left \{x \in [0,1] : \left | x - \frac a q \right | \leq \frac c {q^p} \text{ for infinitly many } (a,q) \in \mathbb N \times \mathbb N^* \right \} $$ has measure zero (see the link). For the question I can assume that $p > 2$ and $c,p \in \mathbb Q$. I must show that the set of non-diophantine real numbers in $[0,1]$ has measure zero. If I know that I can take a countable union of sets of measure zero, which gives a set of measure zero.

Denote $$ X := \left \{ x\in [0,1]: \forall p \in \mathbb Q_{>2} \forall c \in \mathbb Q_{>0} \exists q \in \mathbb N^* \exists a \in \mathbb N: \left |x - \frac a q \right | \leq \frac c {q^p} \right \} $$ We have to prove that $m(X) = 0$. How can I show that $m(X) = 0$ by using the fact that $m(E(p,c)) = 0$ for every $p> 2$ and $c > 0$. I see that the pairs $(p,c)$ are countable. Does this help ?

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Is there a subtlety I am missing? It seems there is nothing else to prove. If you are wondering if there is some other real $c,p$ not covered by your sets, it's not a problem, as if a particular $c,p$ works for the definition of diaphantine, then actually any larger $p$ and smaller $c$ works too (thus it won't appear in $X$). – Evan Aug 27 '13 at 14:56
What did I prove then ? :D I think it would be nice if $X \subseteq \bigcup_{p,c} E(p,c)$ but is this true ? – Epsilon Aug 27 '13 at 15:05 I was missing something... I suppose $E(p,c)$ is stronger (a subset) of the corresponding subset of $X$ for fixed $p,c$... – Evan Aug 27 '13 at 15:14
up vote 1 down vote accepted

I don't think using the $E(p,c)$ is the right approach here. Let us define

$$\begin{align} A(p,c,q,a) &:= \left\lbrace x \in [0,\,1] : \left\lvert x-\frac{a}{q}\right\rvert \leqslant \frac{c}{q^p} \right\rbrace,\\ B(p,c,q) &:= \bigcup_{a=0}^q A(p,c,q,a),\\ \Omega(p,c) &:= \bigcup_{q=1}^\infty B(p,c,q),\\ X(p) &:= \bigcap_{c > 0} \Omega(p,c). \end{align}$$

Then we have

$$\begin{align} X &:= \left \{ x\in [0,1]: \forall p \in \mathbb Q_{>2} \forall c \in \mathbb Q_{>0} \exists q \in \mathbb N^* \exists a \in \mathbb N: \left |x - \frac a q \right | \leq \frac c {q^p} \right \}\\ &= \bigcap_{p \in \mathbb Q_{>2}} \bigcap_{c \in \mathbb Q_{>0}} \bigcup_{q \in \mathbb N^*} \bigcup_{a \in \mathbb N} A(p,c,q,a)\\ &= \bigcap_{p \in \mathbb Q_{>2}} \bigcap_{c \in \mathbb Q_{>0}} \bigcup_{q \in \mathbb N^*} B(p,c,q)\\ &=\bigcap_{p \in \mathbb Q_{>2}} \bigcap_{c \in \mathbb Q_{>0}} \Omega(p,c). \end{align}$$

Now, $m(A(p,c,q,a)) \leqslant \dfrac{2c}{q^p}$, and therefore $m(B(p,c,q)) \leqslant \dfrac{2c(q+1)}{q^p} \leqslant \dfrac{4c}{q^{p-1}}$ and

$$m(\Omega(p,c)) \leqslant \sum_{q=1}^\infty m(B(p,c,q)) \leqslant \sum_{q=1}^\infty \frac{4c}{q^{p-1}} = 4c\cdot\zeta(p-1).$$


$$m(X(p)) = m\left(\bigcap_{c\in\mathbb{Q}_{>0}} \Omega(p,c)\right) \leqslant \inf_{c\in\mathbb{Q}_{>0}} m(\Omega(p,c)) = 0.$$

That also implies $m(X) = 0$.

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