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

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• Nov 19th 2011, 02:53 AM
AshleyLin
Generalizing to R^n a bound for a lebesgue measure.
A standard proposition in measure theory is that if $E$ is any Lebesgue measurable set in $\mathbb{R}$ with $\lambda(E)>0$, then for any $\epsilon>0$ there is a finite, nontrivial interval $J=[a,b]$ such that $\lambda(E\cap J)>(1-\epsilon)\lambda(J)$.

To generalize this, suppose $E\subseteq\mathbb{R}^n$ with $\lambda(E)>0$. Then why for any $\epsilon>0$ is there some box $J=(a_1,a_1+\delta]\times\cdots\times(a_n,a_n+\delta]$ such that $\lambda(E\cap J)>(1-\epsilon)\lambda(J)$? Thank you.
• Nov 19th 2011, 04:44 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, 07:34 AM
AshleyLin
Re: Generalizing to R^n a bound for a lebesgue measure.
I tried adapting the argument like so: Assume $\lambda(E)<\infty$ and $\epsilon<1$. Then take finite intervals $J_m=(a_{1,m},a_{1,m}+\delta]\times\cdots\times(a_{p,m},a_{p,m}+\delta]$ such that $E\subset\bigcup_m J_m$ and $\sum_{m\geq 1}\lambda(J_m)\leq\lambda(E)/(1-\epsilon)$. Then $\lambda(E)\leq\sum_{n\geq 1}\lambda(J_m\cap E)$ and for some $m$ $(1-\epsilon)\lambda(J_m)\leq\lambda(J_m\cap E).$

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