Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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 –  Hardy Jul 20 '12 at 21:09
1  
Is the $\delta$ dependent on the interval $(\alpha,\beta)$? –  Alex Becker Jul 20 '12 at 21:15
1  
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

1 Answer 1

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 $1$.

share|improve this answer
    
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. –  Hardy 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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.