Let $\alpha > 2$ be a real number. Define
$$E = \{x \in [0,1] \ s.t. \ |x - \frac{p}{q}| < \frac{1}{q^{\alpha}} \ \forall p,q\}$$ Prove $m(E) = 0$
Hint: fix $p,q$ and calculate the measure of that set. Apply Borel-Cantelli.
Well, this hint has not gotten me too far. If I let $p = q = 1$, I'd get
$|x - 1| < 1$
$x - 1 < 1 \to x < 2$
$-(x - 1) < 1 \to 1 - x < 1 \to 0 < x$
$m((0,2)) = 2 - 0 = 2$ which is not equal to $0$. Did I make a mistake here?
For the application of Borel Cantelli, here is my attempt:
Let $E_i = \{x \ s.t. \ |x - \frac{p_i}{q_i}| < \frac{1}{q^{\alpha}}\}$
This is countable because $\frac{p_i}{q_i}$ is rational and the rationals are countable
Then, $x \in [\frac{p_i}{q_i} - \frac{1}{q^{\alpha}}, \frac{p_i}{q_i} + \frac{1}{q^{\alpha}}]$
So, $\sum_{i = 1}^{\infty} m(E_i) = \sum_{i = 1}^{\infty} 2 * \frac{1}{q^{\alpha}} < \infty$
Therefore, Borel-Cantelli applies, and the set is of measure $0$.
I must have done something terribly wrong here because my instructor was angry when I turned this part in (I didn't submit the (0,2) error because that one was obviously a mistake). This part alone made him noticeably upset, so I must be making a terrible mistake. From looking at this though, I really don't see where I went wrong. Can anyone clear up my confusion here?
Edit: I am still stuck. Is anyone able to help?
Edit 2: Still have not been able to solve this. The class is well over now, but I have still been trying to solve some questions on the side. Can anyone solve this one so I can take note of it?
Let $F$ be the set of those $x\in [0,1]$ for which there exist infinitely many $(p,q)\in \Bbb N^2$ satisfying $|x-p/q|<1/q^{\alpha}.$ Clearly $E\subseteq F$. We show $m(F)=0$.
For $n\in \Bbb N$ let $S_n=\cup \{(j/n-1/n^{\alpha},\,j/n+1/n^{\alpha}): n\ge j\in\Bbb N_0\}.$
For $n'\in \Bbb N$ let $T_{n'}=\cup \{S_n:n\ge n'\}.$
Then $x\in F\implies x\in\cap \{T_{n'}:n'\in \Bbb N\}.$
We have $m(S_n)\le 2(n+1)/n^{\alpha}\le 4/n^{\alpha - 1}.$
Therefore $m(T_{n'})\le \sum_{n=n'}^{\infty}4/n^{\alpha-1}=O(1/n'^{(\alpha-2}) =o(1)$ as $n'\to\infty.$
So $m(F)=0$ because $F\subseteq T_{n'}$ for every $n'.$