# Thread: show a set has measure 0

1. ## show a set has measure 0

Let $\displaystyle \Omega$ denote the set of numbers $\displaystyle x \in [0,1]$ such that there exist infinitely many
rationals $\displaystyle \frac{p}{q}$, with p and q coprime, with the property
$\displaystyle |x-\frac{p}{q}|<\frac{1}{q^3}$. Show that $\displaystyle \Omega$ has measure 0.

I think I should use the Borel Cantelli Lemma for this, however I am struggling to think of a suitable set An so that limsupAn= $\displaystyle \Omega$. Can you help me try to find a suitable set?

2. Originally Posted by ramdayal9
Let $\displaystyle \Omega$ denote the set of numbers $\displaystyle x \in [0,1]$ such that there exist infinitely many
rationals $\displaystyle \frac{p}{q}$, with p and q coprime, with the property
$\displaystyle |x-\frac{p}{q}|<\frac{1}{q^3}$. Show that $\displaystyle \Omega$ has measure 0.

I think I should use the Borel Cantelli Lemma for this, however I am struggling to think of a suitable set An so that limsupAn= $\displaystyle \Omega$. Can you help me try to find a suitable set?
It seems to work if you let $\displaystyle A_q=\{x\in[0,1] : \left|x-\frac pq\right|<\frac{1}{q^3}$ for some $\displaystyle 0\leq p<q\}$. This excludes the rationals $\displaystyle \frac{p}{q}\notin[0,1]$ but there are only finitely many that fit if $\displaystyle x\in(0,1)$, I think, so this doesn't matter (to be checked). Alternatively, you can allow $\displaystyle 0\leq p<2q$ in the definition to settle this problem more easily (if $\displaystyle |x-p/q|<1/q^3$ then $\displaystyle p/q\leq |x|+1/q^3\leq 1+1=2$)