Prove that the Lebesgue measure of a particular set is zero. I am doing revision and got extremely stuck with the following exercise, which appeared in an exam from the previous year.
Consider the measure space $(\mathbb R, \mathbb B, \lambda)$ where $\mathbb B$ is the Borel sigma algebra and $\lambda$ the Lebesgue measure. Let $\delta > 0$ and 
\begin{align*}
V := \left\{x \in \mathbb R: \exists \text{ infinitely many } p \in \mathbb Z, \,  q \in \mathbb N \text{ where } \left|x - \frac{p}{q}\right| \le \frac{1}{q^{2+\delta}}\right\}.
\end{align*}
Show that $\lambda(V) = 0$.
I wish I had an attempt for this, but unfortunately I don't know how to start, I'm afraid. 
Please help me.

For any finite $K > 0$, consider the set
$$
V(K) := \left\{x \in \mathbb R: (\lvert x\rvert < K)\land \left( \exists \text{ infinitely many } p \in \mathbb Z, \,  q \in \mathbb N \text{ where } \left|x - \frac{p}{q}\right| \le \frac{1}{q^{2+\delta}}\right)\right\}.
$$
Proving that each $V(K)$ has measure $0$ suffices, because $V = \bigcup\limits_{k=1}^\infty V(k)$.
For $p \in \mathbb{Z}, q\in \mathbb{N}$, let
$$A(p,q)= \left\lbrace x \in \mathbb{R} : \left\lvert x-\frac{p}{q}\right\rvert \leqslant \frac{1}{q^{2+\delta}} \right\rbrace.$$
Then $\lambda(A(p,q)) = 2q^{-(2+\delta)}$. Further, let
$$B(q,K) = \bigcup_{\lvert p\rvert < 2Kq} A(p,q).$$
Then $\lambda(B(q,K)) \leqslant 5Kq\cdot 2q^{-(2+\delta)} = 10K\cdot q^{-(1+\delta)}$, and
$$\sum_{q=1}^\infty \lambda (B(q,K)) \leqslant 10 K \sum_{q=1}^\infty \frac{1}{q^{1+\delta}} < \infty.$$
By the Borel-Cantelli lemma, the set
$$W(K) = \bigcap_{n=1}^\infty \bigcup_{q=n}^\infty B(q,K)$$
is a null set. But $V(K) \subset W(K)$ for $K \geqslant 1$.
