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Let $A\subseteq \mathbb{R}$ be a Lebesgue measurable set of measure $m(A)=p>0$. Then for all $0<q<p$, show that there is a subset $B\subseteq A$ with $m(B)=q$.

Which Theorem do i have to use here, regularity or density of Lebesgue measure or something else ?

derivative
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1 Answers1

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Consider $f(x) = \int_A 1_{(-\infty, x]}$. Then $\lim_{x \to -\infty} f(x) = 0$, $\lim_{x \to +\infty} f(x) = p$ and $f$ is continuous.

Then, use the intermediate value theorem.

ViktorStein
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copper.hat
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  • Why is f continuous ? I think we can't prove it with $\epsilon-\delta$ criterion ( $m(A)<\delta\Rightarrow\int_{A} \mathbb{1}_{(-\infty,x]}<\epsilon $) because this is precisely the exercise, to show that such a $\delta$ exists. – derivative Nov 15 '13 at 14:01
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    No, this is not precisely the exercise. Suppose $x>y$, then $|f(x)-f(y)| = \int_A 1_{(y,x]} \le \int 1_{(y,x]} = |x-y|$. Similarly for $x>y$. Hence $f$ is Lipschitz continuous with rank 1. – copper.hat Nov 15 '13 at 16:32