# Generalizing to R^n a bound for a lebesgue measure.

• Nov 19th 2011, 01:53 AM
AshleyLin
Generalizing to R^n a bound for a lebesgue measure.
A standard proposition in measure theory is that if $\displaystyle E$ is any Lebesgue measurable set in $\displaystyle \mathbb{R}$ with $\displaystyle \lambda(E)>0$, then for any $\displaystyle \epsilon>0$ there is a finite, nontrivial interval $\displaystyle J=[a,b]$ such that $\displaystyle \lambda(E\cap J)>(1-\epsilon)\lambda(J)$.

To generalize this, suppose $\displaystyle E\subseteq\mathbb{R}^n$ with $\displaystyle \lambda(E)>0$. Then why for any $\displaystyle \epsilon>0$ is there some box $\displaystyle J=(a_1,a_1+\delta]\times\cdots\times(a_n,a_n+\delta]$ such that $\displaystyle \lambda(E\cap J)>(1-\epsilon)\lambda(J)$? Thank you.
• Nov 19th 2011, 03:44 AM
Tinyboss
Re: Generalizing to R^n a bound for a lebesgue measure.
Do you understand the proof of the R^1 case? Have you tried to generalize the argument to R^n? If not, that's a good place to start. If so, where did you get stuck?
• Nov 19th 2011, 06:34 AM
AshleyLin
Re: Generalizing to R^n a bound for a lebesgue measure.
I tried adapting the argument like so: Assume $\displaystyle \lambda(E)<\infty$ and $\displaystyle \epsilon<1$. Then take finite intervals $\displaystyle J_m=(a_{1,m},a_{1,m}+\delta]\times\cdots\times(a_{p,m},a_{p,m}+\delta]$ such that $\displaystyle E\subset\bigcup_m J_m$ and $\displaystyle \sum_{m\geq 1}\lambda(J_m)\leq\lambda(E)/(1-\epsilon)$. Then $\displaystyle \lambda(E)\leq\sum_{n\geq 1}\lambda(J_m\cap E)$ and for some $\displaystyle m$ $\displaystyle (1-\epsilon)\lambda(J_m)\leq\lambda(J_m\cap E).$

Is this the correct idea? I'm essentially following the approach in the $\displaystyle \mathbb{R}^1$ case that I've read, but I'm unsure of some steps. For instance, why is it possible to find such $\displaystyle J_m$ such that $\displaystyle \sum_{m\geq 1}\lambda(J_m)\leq\lambda(E)/(1-\epsilon)$? It seems to be asking a lot from such sets, and I don't see why they should exist.