# Proving that $\mathbb Q\cap [0,1]$ is a null subset of $\mathbb R$

I'm trying to proof that the set $$A = \mathbb Q \cap [0,1] \subset \mathbb{R}$$ is a null set.

My definition of the null set is that $$A \subset\mathbb Q$$ 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 $$A \subset \bigcup_{k=1}^{\infty} Q_k$$.

Since my lecturer didn't provide much more than just the definition, I'm stuck at getting an intuition and a way to tackle the problem. Could I maybe argue, that the $$x$$-axis in $$\mathbb{R}^2$$ is a null set and therefore A as a subset must also be a null set? Any help or tips are highly appreciated.

• By your logic, then $\Bbb R$ would be a null set (set of measure zero) since it's the $x$-axis in $\Bbb R^2$. But that would be silly. That is a deliberate confusion between one-dimensional and two-dimensional Lebesgue measure. Nov 5 '19 at 6:30
• Every countable set is a null set. The standard proof is to put the $n$-th point in an interval of length $\epsilon/2^n$. Nov 5 '19 at 6:32

The question has noting to do with $$\mathbb R^{2}$$. You are asked to show that a certain subset of $$\mathbb R$$ has measure $$0$$ w.r.t. Lebesgue measure on $$\mathbb R$$.
$$A$$ is a countable set so you can write it as $$\{a_1,a_2,...\}$$. Consider the intervals $$(a_i-\frac {\epsilon} {2^{i+1}},a_i+\frac {\epsilon} {2^{i+1}})$$. These cover $$A$$ and their total length is less than $$\epsilon$$.