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In the proof of the following proposition: Let $\{E_n\}$ be a countable collection of sets of real numbers. Then $$ m^\ast\left(\bigcup E_n\right)\leq \sum m^\ast\left(E_n\right)~,$$

we suppose that $m^\ast(E_n)$ is finite for all $n$. Then for each $E_n$, there is a countable collection $\{I_{k}^{n}:k\geq 1\}$ such that $E_n\subset \bigcup_{k}I_{k}^{n}$ $~~$ and $$\sum l\left(I^{n}_{k}\right) \leq m^\ast(E_n)+\frac{\epsilon}{2^n},~~~\epsilon >0.$$

Here is my question. I don't understand why the last inequality true. Explanations will be very much appreciated.
Thanks.

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The exterior measure of $E_n$ is the infimum of the sum of the lengths of all possible coverings of $E_n$ by open intervals. In particular, this infimum is smaller than what's on the left side of this inequality. This means you can find an $\epsilon$ such that what you have written holds. –  ae0709 Oct 11 '11 at 2:47
    
I get that the left hand side $\leq m^\ast(E_n)+\epsilon$. Where is the $2^n$ coming from? –  Nana Oct 11 '11 at 2:51
    
It's just a trick so that when you take the sum it vanishes and you're left with $\epsilon$. Your collection $\{E_k\}$ is a countable collection, so you can associate a natural number to each $E_k$. This natural number is what we're using in our choice for $\dfrac{\epsilon}{2^k}$ –  ae0709 Oct 11 '11 at 2:56
    
oh ok. If you would convert you comments as answer, I'll gladly accept it. –  Nana Oct 11 '11 at 3:13
    
Done. Thank you, and I hope it helped.. –  ae0709 Oct 11 '11 at 3:20
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1 Answer

up vote 8 down vote accepted

$m^*(E_n)=\inf{\left\{\displaystyle\sum_{k=0}^\infty \ell(I_k)\mid E_n\subseteq\bigcup_{i=0}^\infty I_k \right\}}$, where $I_k$ is an open bounded interval. Since what is on the left hand side of your inequality is one of these possible coverings, the infimum, $m^*(E_n)$ is less than that. This means you can find an $\epsilon>0$ such that $m^*(E_n) + \epsilon > \displaystyle\sum_{k=0}^\infty \ell(I_k)$, where $\{I_k\}_{k=0}^\infty$ is one possible colleciton of bounded open sets containing $E_n$. Since we have a countable collection of these sets, $E_n$, we can associate to each set a natural number $n$ so that your inequality holds for the choice of $\dfrac{\epsilon}{2^n}$ for the $n$th set.

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Thanks. It was certainly helpful –  Nana Oct 11 '11 at 3:57
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