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For $\alpha \in (0,\frac{1}{2}]$, I am trying to show that there is no $E \in \mathcal{L}(\mathbb{R})$ such that for every interval $I$ we have $ \alpha\lambda(I) \leq \lambda(E \cap I) \leq (1-\alpha)\lambda(I). $

I know of the Lebesgue Density Theorem which immediately blows this question out of the water, but we are far away from that point in class, and I am sure that proving this theorem is not the intent of the problem. I have already shown that there is a set $E \in \mathcal{L}(\mathbb{R})$ such that $ 0< \lambda(E \cap I) < \lambda(I). $

I have stared at it all day now and I think I need some help. There is supposedly a very short proof using only basic properties about $\lambda$. If you have any hints I would appreciate them.

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    I'm still really stuck on this one. Does anyone have any more help? I don't see how what @Robert said helps.2012-09-20

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I presume that $\cal L$ is the Lebesgue measurable sets, and $\lambda$ is Lebesgue measure.

Hint: Suppose such a set $E$ existed. Let $A = E \cap (0,1)$. Get an open set $U$ such that $A \subseteq U \subseteq (0,1)$ and $\lambda(U)$ is just a little bigger than $\lambda(A)$. Write $U$ as the union of countably many disjoint open intervals, use your assumption to get an upper bound on $\lambda(U)/\lambda(A)$, and get a contradiction.

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    Thank you so much! I had it written on my page 50 times and I didn't realize that < vs $\leq$ didn't matter.2012-09-20