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  1. Let $\epsilon>0$. Prove that the set of those $x\in [0,1]$ such that there exist infinitely many fractions $p/q$, with relatively prime integers $p$ and $q$ such that $$\bigg |x-\frac{p}{q}\bigg|\leq \frac{1}{q^{2+\epsilon}}$$ is a set of measure zero.

  2. Let $(a_n)$ be a sequence of real numbers, and let $(\alpha_n)$ be a sequence of positive numbers such that $\sum_n \sqrt {a_n}<\infty$. Prove that there exists a measurable set $A$ with $\lambda(A^c)=0$ (Lebesgue measure) such that $$\forall x\in A, \sum_n \frac{\alpha_n}{|x-a_n|}<\infty.$$

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(1) First observe that if $x\in [0,1]$ and $\left\vert x-\frac{p}q\right\vert\leq \frac{1}{q^{2+\varepsilon}}$ for some $p,q$, then $\frac pq\leq x+ \frac{1}{q^{2+\varepsilon}}\leq 2$ and hence $p\leq 2q$. Consequently, if you put $$A_q=\left\{ x\in [0,1];\; \exists p\leq 2q\;:\; \left\vert x-\frac{p}q\right\vert\leq \frac{1}{q^{2+\varepsilon}}\right\},$$ then the set you're looking at is exactly $A=\limsup_q A_q$.

Now, for each fixed $q$ the set $A_q$ is contained in the union of $2q+1$ intervals of length $\frac2{q^{2+\varepsilon}}$; so $\lambda(A_q)\leq \frac{2\times(2q+1)}{q^{2+\varepsilon}}\leq \frac{6}{q^{1+\varepsilon}}$ and hence $\sum_{q\geq 1} \lambda(A_q)<\infty$. By Borel-Cantelli, it follows that $\lambda(A)=0$.

(2) Consider the set $$ A=\{ x\in\mathbb R;\; \exists N\;\forall n\geq N\;:\; \vert x-a_n\vert\geq \sqrt{\alpha_n}\}\, .$$ Note that if $x\in A$, then $\frac{\alpha_n}{\vert x-a_n\vert}\leq\sqrt{\alpha_n}$ for all but finitely $n$ and hence $\sum_1^\infty \frac{\alpha_n}{\vert x-a_n\vert}<\infty$. So it is enough to show that $\lambda (A^c)=0$. Now, by the definition of $A$ we have

$$A^c=\limsup\, I_n \,,$$ where $I_n$ is the interval $[a_n-\sqrt{\alpha_n}, a_n+\sqrt{\alpha_n}]$. Since $\sum_1^\infty \sqrt{\alpha_n}$, one can apply Borel-Cantelli again.

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Hint for (a): if $X$ is uniform on $[0,1]$, consider the random variables $Y_q = 1$ if $|X - p/q| \le 1/q^{2+\epsilon}$ for some $p$ relatively prime to $q$, $0$ otherwise.

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