Understanding the usage of lemma 16 on pg.47 (in Royden "Real analysis") in solving a problem. Here is the I want to answer:
Define $E = \{x \in [0,1] :  |x - \frac{p}{q}| < q^{-3} \text{  for infinitely many } p,q \in \mathbb{N}  \}.$ Prove that $m(E) = 0.$
My thoughts are:
1-I was thinking about using the fact that: Any countable set is measurable. But I do not know how to give a rigorous proof that the given set $E$ is countable. Could anyone help me in doing so?
2- Also I was having an idea of using lemma 16 on pg.47 (in Royden "Real analysis") which states that:
Let $E$ be a bounded measurable set of real numbers. Suppose there is a bounded, countably infinite set of real numbers $\Lambda$ for which the collection of translates of $E, \{\lambda + E \}_{\lambda \in \Lambda},$ is disjoint. Then $m(E) = 0.$
But I do not know how to use this. And I am guessing that we can not use this lemma here in our question as my set $E$ is unbounded (I am not sure from this. Am I correct?). Also, I do not know if there is a bounded, countably infinite set of real numbers $\Lambda$ for which the collection of translates of $E, \{\lambda + E \}_{\lambda \in \Lambda},$ is disjoint.
Also, in the statement of the lemma 16, I am not sure why we need  $\Lambda$ to be bounded. Could anyone explain this to me, please?
Could anyone help me in refining my thoughts and telling me what is correct and what is wrong and also, help me prove my question?
 A: Let $A_{n}=\bigcup_{k=0}^{n}(\frac{k}{n}-\frac{1}{n^{3}},\frac{k}{n}+\frac{1}{n^{3}})$.
Firstly, observe that $E\subseteq\limsup_{n}A_{n}:=\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_{k}$.
For, let $x\in E$. Choose sequences of non-negative integers $(p_{k})$,
$(q_{k})$ such that $0<q_{1}<q_{2}<\ldots$ and $|x-\frac{p_{k}}{q_{k}}|<\frac{1}{q_{k}^{3}}.$
If $p_{k}>q_{k}$, then $|x-\frac{p_{k}}{q_{k}}|=\frac{p_{k}}{q_{k}}-x\geq\frac{q_{k}+1}{q_{k}}-x\geq\frac{1}{q_{k}}$,
contradicting to $|x-\frac{p_{k}}{q_{k}}|<\frac{1}{q_{k}^{3}}$. It
follows that $x\in A_{q_{k}}$ because $x\in(\frac{p_{k}}{q_{k}}-\frac{1}{q_{k}^{3}},\frac{p_{k}}{q_{k}}+\frac{1}{q_{k}^{3}})$. Hence, $x$ appears infinitely often in $(A_n)$, i.e., $x\in\limsup_n A_n$.
Next, we show that $\lambda(\limsup_{n}A_{n})=0$, where $\lambda$
denotes the usual Lebesgue measure. By direct calculation, $\lambda(A_{n})\leq(n+1)\frac{2}{n^{3}}$.
Therefore $\lambda(\cup_{k\geq n}A_{k})\leq\sum_{k=n}^{\infty}\frac{2(k+1)}{k^{3}}$.
By continuity of measure (and the fact that $\lambda(\bigcup_{n=1}^{\infty}A_{n})<\infty$),
\begin{eqnarray*}
\lambda(\limsup_{n}A_{n}) & = & \lim_{n\rightarrow\infty}\lambda\big(\bigcup_{k\geq n}A_{k}\big)\\
 & \leq & \lim_{n\rightarrow\infty}\sum_{k=n}^{\infty}\frac{2(k+1)}{k^{3}}\\
 & = & 0.
\end{eqnarray*}
This shows that $E$ is a subset of the Lebesgue null set $\limsup_n A_n$.
Since Lebesgue measure is complete, $E$ is also Lebesgue measurable and $\lambda(E)=0$.
A: Here is a solution that involves a nice application of the Borel-Cantelli lemma.
For integers $p$ and $q$, $q\geq1$ define $I_{p,q}=\big[\frac{p}{q}-\frac{1}{q^3},\frac{p}{q}+\frac{1}{q^3}\big]$. It is easy to check that
$$
x\in I_{p,q} \quad\text{iff}\quad qx-\frac{1}{q^2}\leq p\leq qx+\frac{1}{q^2}
\tag{1}\label{one}$$
As any interval of length $\frac{2}{q^2}$ can only contain at most one integer if $q\geq2$ or three if $q=1$,   $x\in E$ iff $x$ belongs to infinitely many sets of the form
$$ A_q =\bigcup^\infty_{p=0}\big(I_{p,q}\cap[0,1]\big)$$
That is $E\subset\{A_q\quad\text{i.o}\}:=\bigcap^\infty_{r=1}\bigcup^\infty_{q=r}A_q$.
If $I_{p,q}\cap[0,1]\neq\emptyset$, then from $\eqref{one}$
$$
-\frac{1}{q^2}\leq p\leq q+\frac{1}{q^2}.
$$
which is equivalent to $0\leq p\leq q$ when $q\geq2$. Putting things together,
$\lambda(A_q)\leq (q+1)\lambda(I_{p,q})=\frac{2(q+1)}{q^3}$ and so,
$$\lambda\Big(\bigcup^\infty_{q=1}A_q\Big)\leq\sum^\infty_{q=1}\lambda(A_q)<\infty$$
The conclusion follows by the Borel Cantelli's lemma (direct part).
