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Math Help - Generalizing to R^n a bound for a lebesgue measure.

  1. #1
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    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.
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  2. #2
    Senior Member Tinyboss's Avatar
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    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?
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  3. #3
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    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.
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