Measure nonzero implies dense on a rectangle This would be a very handy lemma for me but I have been unable to prove it thus far. 


  If $S \in \mathbb{R}^n$ is bounded and is not of measure zero, then there exists a rectangle $R$ such that $S$ is dense in $R$.


Does anyone have any ideas on how to prove it? Can someone provide a counterexample?

Thank you!
Brian

Edit: I should have clarified what I meant (moreover this may not be (or even worse, it could be completely different from) the standard definition of dense): $S$ is dense on $R$ if for every $x \in R$, if $\epsilon &gt; 0$ then there exists $y \in S$: $|x-y| &lt; \epsilon$.

Again thank you for the wealth of responses.
Any fat Cantor set in $\mathbb R$ has positive measure yet is not dense in any segment. See http://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set
Maybe I'm misunderstanding the question, but it seems obviously wrong.  A circle x^2 + y^2 &lt;= 1 is obviously not dense in any rectangle.