Prove, that the set of those $x\in \mathbb{R}$ such that there exists infinitely many fractions $\frac{p}{q}$, with relatively prime integers $p$ and $q$ such that $$\left|x-\frac{p}{q}\right|\leq \frac{1}{q^3}$$ is a set of measure zero.
My solution: Let's fix $p\in \mathbb{Z}$ and consider the following set for each $q\in \mathbb{N}$: $$E_{p,q}=\left( \frac{p}{q}-\frac{1}{q^3},\frac{p}{q}+\frac{1}{q^3}\right).$$ We see that $E_{p,q}$ is measurable and $m(E_{p,q})=\dfrac{2}{q^3}$, where $m$ - Lebesgue measure. Also note that $\sum \limits_{q=1}^{\infty}m(E_{p,q})=\sum \limits_{q=1}^{\infty}\frac{2}{q^3}<\infty$. Let's use the Borel-Cantelli lemma.
Then $E_p:=\limsup \limits_{q\to \infty} E_{p,q}=\{x\in \mathbb{R}: x \ \text{belongs to infinitely many} \ E_{p,q}\}$ has zero measure, i.e. $m(E_p)=0$. Then $E=\bigcup \limits_{p=1}^{\infty}E_p$ has alzo zero measure.
I know that this question appeared many times in MSE but please do not close it since many solutions I have seen is a bit different so that's why I've created topic in order to know is my solution correct or not?
Would be very grateful for feedback.