We know that $[0,1]\cap \mathbb{Q}$ is a dense subset of $[0,1]$ and has measure zero, but what about $([0,1]\cap \mathbb{Q})\times([0,1]\cap \mathbb{Q})$? Is it also a dense subset of $[0,1]\times[0,1]$ and has measure zero too?

Besides, what about its complement? Is it dense in $[0,1]\times[0,1]$ and has measure zero?

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    $\begingroup$ If $[0,1]\cap\mathbb{Q}$ is dense and has measure 0, then the complement (which is also dense) must have measure 1. So yes, the product set has measure 0, and the complement (of the product set) would have measure 1 - and is certainly dense in the unit square. $\endgroup$ – Old John Aug 22 '12 at 12:12
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    $\begingroup$ There is a good mnemonic formula for this case: $0 \cdot \infty = 0$. That is, if you have a (measurable) set $A \times B$ and if $A$ is measure zero, then the measure of $A \times B$ is zero regardless of $B$. $\endgroup$ – Sangchul Lee Aug 22 '12 at 12:20
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    $\begingroup$ @sos440: The relevant formula here is $0\cdot 0=0$. $\endgroup$ – Chris Eagle Aug 22 '12 at 12:23
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    $\begingroup$ Another way to see it is to notice that this a countable set, so it has zero Lebesgue measure. $\endgroup$ – Ahriman Aug 22 '12 at 12:42
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    $\begingroup$ With this many great comments, one should write an answer, even if rather short. $\endgroup$ – Sasha Aug 22 '12 at 13:20

I'll post a slick proof here.

Let $d(r)$ denote the denominator of the rational number $r$. In other words, if $r=p/q$ holds for positive integer $q$ and integer $p$, where $p,q$ are co-prime, we have $d(r)=q$. Then $d(r)\ge1$ for all $r\in\Bbb Q$.

Suppose that $D(m)=\left\{\,x\,\big|\,0\le x\le1\land d(x)=m\,\right\}$, we have $D(1)=\{0,1\}$, and $0<|D(m)|<m$ for $m>1$, therefore $$\sum_{x\in D(m)}\frac1{d(x)^4}=\sum_{x\in D(m)}m^{-4}<m^{-3}$$ Thus $$\sum_{x\in [0\mathinner{..}1]}\frac1{d(x)^4}=2+\sum_{m>1}\sum_{x\in D(m)}\frac1{d(x)^4}<2+\sum_{m>1}m^{-3}=C$$

For all $0\le x,y\le1$ and $x,y\in\Bbb Q$, we draw a circle whose center is $(x,y)$ and radius is $(d(x))^{-2}(d(y))^{-2}\epsilon$. The total area is \begin{align} \sum_{\substack{0\le x,y\le1\\x,y\in\Bbb Q}}\frac{\pi\epsilon^2}{d(x)^4d(y)^4} &=\pi\epsilon^2\sum_{\substack{0\le x,y\le1\\x,y\in\Bbb Q}}\frac1{d(x)^4d(y)^4}\\ &=\pi\epsilon^2\sum_{\substack{0\le x\le1\\x\in\Bbb Q}}\frac1{d(x)^2}\sum_{\substack{0\le y\le1\\y\in\Bbb Q}}\frac1{d(y)^2}\\ &\le\pi\epsilon^2C^2 \end{align} Therefore the measure is not greater than $\pi\epsilon^2C^2$. Let $\epsilon\to0$, we get the answer.

Note The summation-interchanging works well because the terms are all nonnegative.


Any countable set is of zero Lebesgue measure, and the product of (finitely many) countable sets is countable. Also, a product of finitely many dense sets is dense. Thus, the answer is yes.


To give a somewhat comprehensive answer:

  • the set in question is countable (as a product of countable sets), so it is of measure zero (because any countable set is zero with respect to any continuous measure, such as Lebesgue measure).
  • it is also dense, because it is a product of dense sets.
  • it has measure zero, so its complement has full measure.
  • its complement has full measure with respect to Lebesgue measure, so it's dense in $[0,1]^2$
  • $\begingroup$ can I understand the last item of your comments as this:If a subset E of A has full measure,with respect to Lebesgue measure,then it is dense in A? $\endgroup$ – user38448-po Aug 23 '12 at 3:01
  • $\begingroup$ @user38448: Notice that every open set in $[0,1]^2$ has positive measure, so any set of full measure must be dense. $\endgroup$ – tomasz Aug 23 '12 at 4:24

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