# No $E \in \mathcal{L}(\mathbb{R})$ such that $\alpha\lambda(I) \leq \lambda(E \cap I) \leq (1-\alpha)\lambda(I)$ for all intervals $I$.

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. – nullUser Sep 20 '12 at 21:57

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.
Okay I just tried for another hour and I still don't see what you mean. So I take $A \subseteq U \subseteq (0,1)$ and $\lambda(A) \leq \lambda(U) < \lambda(A)+\epsilon$. Write $U = \cup_i J_i$ for disjoint intervals $J_i$. I can say $\lambda(U)/\lambda(A) \leq 1/k$. I can say $1 \leq \lambda(U)/\lambda(A) < 1+\epsilon/\lambda(A)$ I can say $\sum\lambda(J_i) \leq (1-k)+\epsilon$. None of these (and more) contradict anything though. – nullUser Sep 20 '12 at 19:51
Oops, I meant an upper bound on $\lambda(A)/\lambda(U)$. Specifically, $\lambda(A \cap J_i) < (1-\alpha) \lambda(J_i)$ so $\lambda(A) < (1-\alpha) \lambda(U)$. – Robert Israel Sep 20 '12 at 22:54
Thank you so much! I had it written on my page 50 times and I didn't realize that $<$ vs $\leq$ didn't matter. – nullUser Sep 20 '12 at 23:30