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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 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 :-)2012-07-20

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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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    @William: It doesn't matter here, but you can't conclude that if $1 \geq \delta$ then $\delta \geq 1$.2012-07-20