Borel-Cantelli (proof and application)

Question

Hi I was reading the second volume of the Tao's Analysis book and in one exercise he's asking for a proof of Borel-Cantelli

If we have a sequence $s_n\in \Omega$ of measurable sets s.t. $\sum_{n\in \mathbb{N}} m(s_n)<\infty$. Then $M=\{x\in \Omega: x\in s_n \text{ infinitely often }\}$ has measure zero.

Here is my approach: Define $f=\sum_{n\in \mathbb{N}}1_{s_n}$, then

$$\int_\Omega f=\int_\Omega\sum_n1_{s_n}=\sum_n\int_\Omega1_{s_n}=\sum_nm(s_n)<\infty \tag1$$

Using the Lebesgue monotone convergence theorem in $\sum_{n\le N} s_n=S_N\le S_{N+1}\le \ldots$. So $\{x\in \Omega:f(x)=\infty\}$ has measure zero. Since this set is the same as $M$ the claim follows.

(2) Now in other point of the problem where I'm stuck says. Let $p>2$ and $c>0$. Let $$\{x\in [0,1]\, : \, \rvert x-a/q \lvert\le c/q^p \text{ for infinitely many positive integers } a, q \}$$ has measure zero.

My first idea was something very naive:

$$A(a,q)=\left\{x\in [0,1]: \left|x-\frac{a}{q}\right|<\frac{c}{q^{p}} \ \text{ for } a\le q \right\}$$ So the set $A(a,q)$ consists of about $q$ intervals of size at most $2c/q^{p}$ from here since $c>0$ is a positive real we have $\sum_{q\in\mathbb N}|A(a,q)|<\infty$

Any ideas for the second point? I can't figure out how to use the Borel-Cantelli. I believe the sets are similar as $A(a,q)$ I really appreciate some extra-hint.

Thanks in advance.

Answer 1

Your argument for (1) sounds right.

For (2), you want to apply Borel-Cantelli, so what should the set $s_n$ be? Key estimate: there are at most $q-1$ reduced fractions with denominator $q$, and $\sum_q q/q^p < \infty$ since $p >2$.

Answer 2

I think your proof of (1) is correct. Here is a proof of (1) without using integration nor any theorem. Write $\{s_{n_k}\}$ for the subsequence with $M\subset s_{n_k}$. Then, for any $k\in\mathbb N$, $$ k\,m(M)\leq\sum_1^km(s_{n_j})\leq\sum_nm(s_n), $$ so $$ m(M)\leq\frac1k\,\sum_nm(s_n). $$ As $k$ is arbitrary, $m(M)=0$.