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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.

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  • $\begingroup$ 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. $\endgroup$ Nov 5 '19 at 6:30
  • $\begingroup$ Every countable set is a null set. The standard proof is to put the $n$-th point in an interval of length $\epsilon/2^n$. $\endgroup$ Nov 5 '19 at 6:32
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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 $\epsilon$.

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