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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 –  leo Sep 21 '12 at 5:02

1 Answer 1

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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Thanks Joriki would you please answer these questions so that i could understand fully: 1) How do we know that we can pick a point x of A with density 1 is it related to A being positive measure? 2)What do you mean by with δ proportional to ϵ? Where did you create SIMILAR triangle Ok you created a triangle but why is it similar to one given? Thanks A lot!! –  Salih Ucan Sep 22 '12 at 21:46
What do you mean by map every y∈Bϵ(x) (an ϵ-ball around x) to Bδ(x) –  Salih Ucan Sep 23 '12 at 9:51
How did the third vertex appear?? I really got confused in your sketch of solution Please clarify... Thank you... –  Salih Ucan Sep 23 '12 at 9:57

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