$m(X) > 0$ Show that $Y \subset X$ where $Y$ is measurable can be such that $m(Y) = \delta$ for any $0 < \delta < m(X)$

Question

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!

Answer 1

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.