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I have a typical problem concerning the measure theory. I would like to know if any one has a good general strategy to solve this kind of problem because I always don't have idea where have I to begin to prove this kind of things.

The problem which I have to solve is:

If $C\subset \mathbb{R}$ is a lebesgue meas. set of measure $\lambda (C)=l>0$. To show is: $\forall 0<r<l \exists B\subset C$ with $\lambda (B)=r$.

Thanks for any answer

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  • $\begingroup$ Perhaps one could try the following, in case $l<\infty$: consider the set $C_x = C\cap (-\infty,x)$ for $x\in\mathbb R$. Then $x\mapsto\lambda(C_x)$ is decreasing and should be continuous. Furthermore, $\lim_{x\to-\infty} \lambda(C_x) = 0$. Now apply the mean value theorem. $\endgroup$ – Rasmus Nov 14 '13 at 8:58
  • $\begingroup$ Indeed the function $x\mapsto\lambda(C\cap(-\infty,x))$ is nondecreasing and continuous because 1-Lipschitz. $\endgroup$ – Did Nov 14 '13 at 9:13
  • $\begingroup$ So that's mean that I only have to check that the function you defined above it is continuous, apply the mean value theorem and conclude saying that there exists an $y\in \mathbb{R}$ such that $\lambda (C\bigcap (-\infty ,y))=r$ Am I forgetting something? $\endgroup$ – sky90 Nov 15 '13 at 8:55
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This is so called Serpinski's theorem. For the sketch of the proof for general non-atomic measures see this wikipedia article.

For your particular case, consider function $f(x)=\lambda(C\cap(-\infty,x))$. It is non-decreasing, so to prove continuity it is enough to prove continuity on the right and on the left of each point. But the latter is nothing more than so called continuity of measure. Since $f$ is continuous, the desired result follows from itermediate value theorem.

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  • $\begingroup$ Ok! Then I understood it. Thanks a lot $\endgroup$ – sky90 Nov 16 '13 at 14:31

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