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Let $A$ be a measurable Lebesgue set, with $\lambda(A)>0$($\lambda$ is the Lebesgue measure). Then, for every $b \in (0,\lambda(A))$, there exists a set $B$ measurable Lebesgue, $B\subset A$, with $\lambda(B)=b$. Can you give me a suggestion? I have no idea how to "build" the set $B$. It seems to me very likely to the Darboux property of functions..can it be used here?

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  • $\begingroup$ Is $\lambda(A)$ assumed finite? $\endgroup$ – David Feb 1 '16 at 18:52
  • $\begingroup$ The problem only specifies that $\lambda(A)>0$. $\endgroup$ – user308560 Feb 1 '16 at 18:55
  • $\begingroup$ For those interested in additional aspects about this property, see this 25 November 2005 sci.math post archived at Math Forum. Also, google the phrase "range of a measure". $\endgroup$ – Dave L. Renfro Feb 1 '16 at 19:06
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Suppose $A$ is bounded. Thus, there is an interval $[a,c]$ such that $A\subseteq [a,c]$. Moreover, $\lambda(A)<\infty$. Let $f:[a,c]\to\mathbb{R}$ given by $f(t)=\lambda([a,x]\cap A)$. Thus $f(x)<\infty$ for all $x\in[a,b]$.

On the other hand, $f(a)=\lambda([a,a]\cap A)\le\lambda([a,a])=0$ and $f(c)=\lambda([a,c]\cap B)=\lambda(B)$.

Moreover, let $x,y\in[a,c]$. Set $u=\min\{x,y\}$, $v=\max\{x,y\}$. Therefore

$\begin{eqnarray} |f(x)-f(y)|&=&|f(u)-f(v)|\\ &=&|\lambda([a,u]\cap A)-\lambda([a,v]\cap A)|\mbox{ and using $\lambda(A)<\infty$}\\ &=&\lambda([u,v]\cap A)\\ &\le&\lambda([u,v])\\ &=&v-u\\ &=&|x-y|. \end{eqnarray}$

So $f$ is uniformly continuous in $[a,c]$. Then, there is $t\in(a,c)$ such that $f(t)=b$. Set $B=[a,t]\cap A$. The last means that $\lambda([a,t]\cap B)=b$.

This shows the case $A$ bounded. Can you do for $A$ unbounded?

EDIT If $A$ is unbounded, cover it by $[-n,n]$ and apply the last.

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  • $\begingroup$ sinbadh, $f(b)$ shouldnt be $\lambda(A)$? $\endgroup$ – user308560 Feb 1 '16 at 18:25
  • $\begingroup$ I will think about the other case. $\endgroup$ – user308560 Feb 1 '16 at 18:26
  • $\begingroup$ I had a typing error. I corrected it. $\endgroup$ – sinbadh Feb 1 '16 at 18:28
  • $\begingroup$ I dont know if it is correct, but I will try an answer for $A$ unbounded.. if $A$ is unbounded, then it contains an interval of the shape $(a, \infty)$ or $[a, \infty)$ or the similary intervals with $- \infty$. suppose that $(a,\infty)\subset A$. Then, can I chose my set B to be the interval $(a,a+b)$, where $b> 0$? $\endgroup$ – user308560 Feb 1 '16 at 18:38
  • $\begingroup$ $\mathbb{Z}$ is unbounded but doesn't contain any interval like that. $\endgroup$ – David Feb 1 '16 at 18:41
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I assume $A$ is defined to be a subset of $\mathbb{R}$.

Define $f(x) = \lambda(A \cap [-x,x])$. Then $f(x)$ is continuous, $f(0) = 0$, and $\lim_{x \to +\infty} f(x) = \lambda(A)$, so by the intermediate value theorem, there is some value of $x$ for which $f(x) = b$.

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  • $\begingroup$ Why is $f$ continous? $\endgroup$ – sinbadh Feb 1 '16 at 18:43
  • $\begingroup$ You need $f(A)<\infty$ for the first equality $\endgroup$ – sinbadh Feb 1 '16 at 18:55
  • $\begingroup$ @sinbadh I've edited the question to account for the possibility that $\lambda(A)$ is infinite. $\endgroup$ – David Feb 1 '16 at 18:57
  • $\begingroup$ Yes. It is rigth. $\endgroup$ – sinbadh Feb 1 '16 at 18:58
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Hint: The Function $x \mapsto \lambda(A \cap (-\infty, x])$ is Lipschitz-continuous.

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