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Show that if $E$ has finite measure and $\epsilon>0$, then $E$ is the disjoint union of a finite number of measurable sets, each of which has measure at most $\epsilon$.

Any hints or ideas would be very appreciated. Thank you.

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You can prove that there exists $M>0$ such that $m(E\setminus[-M,M])<\varepsilon$, and then break up $E\cap[-M,M]$ into small pieces.

Jonas Meyer
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  • How to break up a set into small pieces? That's what the problem is about, so I don't know if I can do that with set $E \cap [-M,M]$. –  Sep 10 '13 at 04:59
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    @antihappiness: E.g., $E\cap[a,a+\varepsilon)$ has measure at most $\varepsilon$. For bounded sets you will have no problem breaking up the set in this way. – Jonas Meyer Sep 10 '13 at 05:00
  • Ok, now, to find $M$, if $E=\Bbb Q$ (for example) then any $M$ works, because $\Bbb Q$ has measure zero. If $E$ is bounded, we can take $M=\max{|\sup E|, |\inf E|}-\epsilon/2$ (right?) but I don't know what will be $M$ in the general case :S is there one or I have to do it by cases? –  Sep 10 '13 at 05:24
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    @anti-happiness: The bounded monotone sequence $(m(E\cap[-n,n])){n=1}^\infty$ has a limit which you can show to be $m(E)$, hence for sufficiently large $n$, $m(E\setminus[-n,n])=m(E)-m(E\cap[-n,n])<\varepsilon$. Or use $m(E)=\sum\limits{n=1}^{\infty}m(E\cap{x:n-1\leq |x|\leq n})$, and by convergence of the series, $\lim\limits_{k\to\infty}m(E\setminus[-k,k])=\lim\limits_{k\to\infty}\sum\limits_{n=k+1}^\infty m(E\cap{x:n-1\leq|x|\leq n})=0$. – Jonas Meyer Sep 10 '13 at 05:42