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Let $X$ be a Lebesgue Measurable subset of $\mathbb{R}$ where $m(X) > 0$.

Prove that $\forall 0 < \delta < m(X)$ that $\exists Y \subset X$, measurable such that $m(Y) = \delta$

Hint: consider $f(x) = m(X \cap [-x,x]) \ \forall x > 0$

I'm having trouble figuring out how to use the hint. I am not sure what I should do with it. Maybe attempt to show it's continuous and maybe do something from there? Exactly what to do though, I'm not sure. Any advice would be appreciated!

Nolan P
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1 Answers1

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Note that $f(0)$ is also well-defined and $f(0)=0$. Also, $f$ takes value in $[0,\infty)$. By continuity of measure, we can prove that $f$ is continuous and $f(x)\rightarrow m(X)$ as $x\rightarrow \infty$. Since $\lim_{x\rightarrow \infty}f(x) = m(X)>\delta$, there exists $x_2$ such that $f(x_2)>\delta$. Now, $0= f(0) < \delta < f(x_2)$. Apply Intermediate Value Theorem, the result follows.

Danny Pak-Keung Chan
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  • Hey, can you clarify one part for me? I follow all of this and I do agree there is some $x_3$ where $f(x_3) = \delta$ from the intermediate value theorem, but how does the existence of this $x_3$ imply the existence of $Y$ where $m(Y) = \delta$? $Y$ is not mentioned in this proof, so I'm a bit unsure of how to write out the conclusion. – Nolan P Mar 04 '21 at 15:02
  • $\delta = f(x_3) = m(X \cap [-x_3,x_3])$. The set $X \cap [-x_3,x_3]$ is denoted by $Y$. – Danny Pak-Keung Chan Mar 04 '21 at 15:23
  • Perfect. That makes sense. Thank you for clarifying! Time to work on showing continuity. – Nolan P Mar 04 '21 at 15:30
  • To show that $f$ is right-continuous at a point $a$, you only need to prove the following: For any sequence $(x_n)$, if $x_1 > x_2 >\ldots > a$ and $x_n\rightarrow a$, then $f(x_n)\rightarrow f(a)$. This results is due to Heine. Note that we have invoked the Axiom of Choice. Without the Axiom of Choice, Heine's theorem is false. – Danny Pak-Keung Chan Mar 04 '21 at 15:36
  • Then, notice that the corresponding sequence of sets is decreasing, and the measure of the intersection of sets is the limit of measures of the sets (we also need the fact that the measure of some of the set is finite). This is known as the continuity of measure. – Danny Pak-Keung Chan Mar 04 '21 at 15:38
  • So I haven't learned about Heine yet. I'm trying to do it directly. I am thinking: Suppose $x > y$. Then, $|m(E \cap [-x,x]) - m(E \cap [-y,y])| = |m(E \cap ([-x,x] - [-y,y]))| \leq |m([-x,x] - [-y,y])|$. This is the part I am unsure of. Is it $|m([-x + y,x - y])| = x - y - (-y + x) = x - y - (-x + y) = x - y - y + x = 2x - 2y = 2(x - y) < 2 \delta = \epsilon$? – Nolan P Mar 05 '21 at 01:48