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I am trying to show that, a set $E$ in $\left( 0,1\right) $ is such that, if $\left( \alpha,\beta\right) $ is any interval, then $$\mu\left(E \cap \left( \alpha ,\beta \right) \right) \ge \delta \left( \beta -\alpha \right) $$ where $\delta > 0 $ then the $\mu\left(E\right)=1$.

What have i tried. I could not help notice the case by case breakdown that $E$ might be completely contained in $\left( \alpha,\beta\right) $ in which case $$\mu\left(E\right) = \mu\left(E \cap \left( \alpha ,\beta \right) \right) \ge \delta \left( \beta -\alpha \right)$$

Similarly $E$ might have 2 parts so to calculate it's measure we'll need $$\mu\left(E\right) = \mu\left(E \cap \left( \alpha ,\beta \right) \right) + \mu\left(E \backslash \left(E \cap \left( \alpha ,\beta \right) \right)\right)$$

I am assuming a case of $\left(E \cap \left( \alpha ,\beta \right) \right) = \emptyset$ can not occur as that would imply $\alpha = \beta$ given the other conditions.

I was hoping if some one would be kind to give me a hint or a clue, so i could make progress.

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If you suppose that $\delta >1,$ then $\mu(E)=\mu(E\cap (0,1))\ge \delta\cdot 1 >1,$ which means that $\mu(E)\neq 1$. – Andrew Jul 20 '12 at 20:58
@Andrew If $0 < \delta \leq 1$, I think this is still a meaningful question? – William Jul 20 '12 at 21:04
@Andrew Thanks i am inferring from your post that you were highlighting that both $\delta > 1$ and the other set $(0,1)$ could not happen together but no reason why $\delta > 1 $ but $(1/4, 3/4) $ could not be the values – Comic Book Guy Jul 20 '12 at 21:09
Is the $\delta$ dependent on the interval $(\alpha,\beta)$? – Alex Becker Jul 20 '12 at 21:15
If the question is $\exists E~ \exists \delta~ \forall (\alpha,\beta) \ldots,$ then $0<\delta\leq 1.$ But if the question should read $\exists E~ \forall (\alpha,\beta)~ \exists \delta\ldots,$ then you should ignore my remark :-) – Andrew Jul 20 '12 at 21:28
up vote 2 down vote accepted

The original argument is flawed.

I am assuming that $E$ is a measurable set. By Andrew's comment, $0 < \delta \leq 1$.

Pick any $x \in (0,1)$.

$\frac{\mu(E \cap (x - \epsilon, x + \epsilon))}{2\epsilon}\geq \delta$

where $\epsilon \leq \text{min}(|x|, |1 - x|)$. Taking limit as $\epsilon \rightarrow 0$, this represent the density of $x$ in $E$. By the Lebesgue Density Theorem, for almost all $x$ in $E$, the density is $1$. This implies (since the complement is measurable), that for almost all points in $(0,1) - E$, the density in $E$ is $0$. Hence for every measurable set, for almost all points, the density is either $1$ or $0$. Since $\delta > 0$ in the above which holds for all $x$, one has that for almost all points the density is $1$. So $E$ has measure $1$.

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I was going to use LDT too, but it seemed to me this should be proveable with something much weaker. – Alex Becker Jul 20 '12 at 21:22
@William I could not follow how we can infer $\delta \ge 1$, any help would be much appreciated. – Comic Book Guy Jul 20 '12 at 21:33
For those like myself who move at a more pedestrian pace, the conclusion of the density theorem is that $1_E(x) \geq \delta > 0$ a.e. $x \in (0,1)$. Since $1_E$ is integer valued, then this implies that $1_E(x) =1$ a.e. $x \in (0,1)$. It doesn't really imply that $\delta \geq 1$, but we get the desired conclusion. – copper.hat Jul 20 '12 at 21:35
@William: It doesn't matter here, but you can't conclude that if $1 \geq \delta$ then $\delta \geq 1$. – copper.hat Jul 20 '12 at 21:41

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