With the help I got in an earlier post, (Proving that $\mathbb Q\cap [0,1]$ is a null subset of $\mathbb R$), I figured out the proof that every countable set is a null set w.r.t. Lebesgue-measure and was able to apply it to examples from different dimensions. But now at this example: $$A =\Bigl\{\left(\begin{array}{c}x\\\frac{1}{x}\\\end{array}\right)|x \in [1,2]\Bigr\} \subset \mathbb{R}^2$$, I'm stuck again since there is no way to put decreasing intervals around every point since there is infinitely many. So how would I provide a proof with the definition being:
$A \subset\mathbb Q$ is called a null set, if $\forall \epsilon > 0$, there exists a countable number of cuboids $\{Q_k\}_{k=1}^{\infty}$ with volume $\sum _{k=1}^{\infty} \operatorname{vol}(Q_k) < \epsilon $ with $A \subset \bigcup_{k=1}^{\infty} Q_k $.
The uniform continuous function $\varphi(x)=1/x$ on $[1,2]$ is such that $A=\{(x,\varphi(x)):x\in[1,2]\}$. Given $\epsilon>0$, there is an $n_{0}$ such that $|x-y|<1/n_{0}$ implies $|\varphi(x)-\varphi(y)|<\epsilon$ for all $x,y\in[1,2]$. Then \begin{align*} A\subseteq\bigcup_{i=1}^{n}\left(I_{i}\times\left[\min_{I_{i}}f,\max_{I_{i}}f\right]\right), \end{align*} and hence $|A|\leq\displaystyle\sum_{i=1}^{n}(\max_{I_{i}}f-\min_{I_{i}}f)|I_{i}|$, where $I_{i}=[1+(i-1)/n_{0},1+i/n_{0}]$, and hence $|A|<\epsilon\displaystyle\sum_{i=1}^{n}|I_{i}|=\epsilon$.