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.

  • $\begingroup$ 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

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 $\epsilon$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.