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If $A$ is a set of positive measure say in $\mathbb{R}^2$ then $A$ does not necessarily have a rectangle of positive measure. This is true I suppose? Because we can apply iteration in Cantor fashion but the total measure can be less than zero by choosing the thrown away ratio small say $1/4$.

Similarly I have come across a question: Given three points forming a triangle in $\mathbb{R}^2$. Show that $A$ as above having positive measure in $\mathbb{R}^2$ has vertices similar to that triangle in $\mathbb{R}^2$? I am not saying that $A$ contains that triangle but it contains the vertices forming a triangle which is a similar triangle to the given any triangle? How to show this??

Thank you!!

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    To answer the first question: yes it's true, see the [fat Cantor set](http://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set)2012-09-21

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A Cantor construction is overkill for the first part – the set of points with irrational coordinates will do.

The second part is a consequence of Lebesgue's density theorem. Pick any point $x$ of $A$ with density $1$ as one particular vertex of the triangle, and map every $y\in B_\epsilon(x)$ (an $\epsilon$-ball around $x$) to $B_\delta(x)$, with $\delta$ proportional to $\epsilon$, by placing the second vertex of the triangle at $y$ and mapping it to the resulting third vertex. If $A$ contained no similar triangle, then all of $A\cap B_\epsilon(x)$ would be mapped outside of $A$, which contradicts the fact that the density of $A$ in these balls tends to $1$ for $\epsilon\to0$.

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    How did the third vertex appear?? I really got confused in your sketch of solution Please clarify... Thank you...2012-09-23