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If $A\subset \mathbb{R}^d$ has positive Lebesgue measure, then for each $0\leq a<m(A)$ there is a Lebesgue measurable $E\subset A$ with $m(E)=a$.

We have that

$E\subset A \Rightarrow m(E)\leq m(A)$

and

$m(A)>a$

How can we conclude that $m(E)=a$ ??

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Define $f(t)=m(B_t(0)\cap A)$ where $B_t(0)$ is the open ball of radius $t$ centered at the origin. Show that $f$ is continuous and use the intermediate value theorem.

Define $E_t=A\cap B_t(0)$. Then the measure of $E_t$ is $f(t)$. $f(0)=0$ and $f(t)$ approaches $m(A)$ as $t$ approaches infinity, so for any $a<m(A)$ there exists a $t$ such that $m(E_t)=a$. $E_t$ therefore satisfies the condition.

To show continuity, choose $t_0>0$. We know that the volume of a ball is a continuous increasing function of the radius. Suppose the volume of a ball of radius $t$ is $V(t)$. We are given $\epsilon$ and we want to find $\delta$ such that if $|t-t_0|<\delta$ then $|f(t)-f(t_0)|<\epsilon$. By additivity of the measure, we have that if $p>q$ then $f(p)-f(q)=|f(p)-f(q)|=m(E_p-E_q)=m(A\cap (B_p(0)-B_q(0)))\leq V(p)-V(q)$, with the last inequality given by monotonicity. Choose $\delta$ so that if $|t-t_0|<\delta$, then $|V(t)-V(t_0)|<\epsilon$. Then $|f(t)-f(t_0)|<\epsilon$, so we are done.

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    $\begingroup$ Missed that. The edit above should work. $\endgroup$ – Matt Samuel Nov 10 '14 at 14:25
  • $\begingroup$ @MattS To apply the intermediate value theorem we do the following: $$0\leq a<m(A)\Rightarrow \exists E\subset A \text{ such that } m(E)=a$$ Is this correct?? At the inequality $0\leq a<m(A)$ the $0$ is the value of measure right?? $\endgroup$ – Mary Star Nov 10 '14 at 14:41
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    $\begingroup$ Perhaps using $[-t,t]^d$ instead of $B_t(0)$ would make it easier to show continuity. $\endgroup$ – Henning Makholm Nov 10 '14 at 14:53
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    $\begingroup$ @MaryStar: It looks like you're just jumping directly to the conclusion without actually understanding what the hint suggests. What does the intermediate value theorem say? What do you apply it to here? What does it give you? $\endgroup$ – Henning Makholm Nov 10 '14 at 14:54
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    $\begingroup$ It's not necessary to exactly compute the area of the annulus. You just need to know that it approaches 0. $\endgroup$ – Matt Samuel Nov 10 '14 at 15:12

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