Measure nonzero implies dense on a rectangle

Question

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 $\epsilon > 0$ then there exists $y \in S$: $|x-y| < \epsilon$.

Again thank you for the wealth of responses.

Answer 1

Maybe I'm misunderstanding the question, but it seems obviously wrong. A circle x^2 + y^2 <= 1 is obviously not dense in any rectangle.

Answer 2

Let $S = B(0,1)$ be the closed unit ball in $\mathbf{R}^n$. Then no such $R$ exist (draw the picture in the $n = 2$ case).

Answer 3

The other answers don't make sense, since you can just pick $R$ inside of the ball. However, I don't think your lemma is right. Here's an exercise: construct a Cantor set of positive measure (i.e. a totally disconnected, compact subset of $[0,1]$).

Answer 4

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