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Let $s>2$ and let $A_s$: The set of $x\in [0,1]$ with the property that there is an infinity of positive integers $n$ such that for some integer $m\in[0,n],\ |x-\dfrac{m}{n}|<\dfrac{1}{n^s}$ Show that $\mu(A_s)=0$, with $\mu$ Lebesgue's measure

I do not know how to start with this problem.

Thanks in advance

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    $\begingroup$ hint: Borel Cantelli $\endgroup$ – Phicar May 7 '18 at 5:25
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Let $$ B_{m,n}=\left\{x\in[0,1]:\ \left|x-\frac mn\right|<\frac1{n^s}\right\}=\left(\frac mn-\frac1{n^s},\frac mn+\frac1{n^s}\right).$$ Because each $x$ belongs to infinitely many $B_{m,n}$, we have for each $n_0\in\mathbb N$ $$ A_s\subset\bigcup_{n\geq n_0}\bigcup_{m=1}^nB_{m,n}. $$ Then $$ \mu(A_s)\leq\sum_{n\geq n_0}\sum_{m=1}^n\mu(B_{m,n})=\sum_{n\geq n_0}\sum_{m=1}^n\frac2{n^s}=\sum_{n\geq n_0}\frac2{n^{s-1}}. $$ As $s>2$, the series on the right becomes arbitrarily small if $n_0$ is large. So $\mu(A_s)=0$.

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