Following up on this post Find an interval such that the intersection has measure $\epsilon$ I was trying to show that given $E$ a Lebesgue-measurable set with finite measure, then for every $0\leq \epsilon < m(E)$, there exists $x \geq 0$ such that $m(E\cap [-x,x])=\epsilon$.
Based on a suggestion from that post I've been able to prove that $g(x) = m(E \cap [-x,x])$ is Lipschitz, hence continuous. I would like now to conclude that $Im(g) \supseteq [0, m(E)]$. If I am able to show this, as $g$ is continuous then for every $\epsilon \in [0,m(E)], \ \exists x \geq 0: g(x) = \epsilon$.
If $x = 0$, then $g(x) = m(E \cap \{0\}) = 0$. Now If $x$ goes to infinity then $[-x,x] \to \mathbb{R}$. If $\lim_{x\to\infty}m(E \cap [-x,x]) = m(lim_{x\to\infty}E \cap [-x,x]) = m(E \cap \mathbb{R}) = m(E)$ then I think we can conclude that $Im(g) \supseteq [0,m(E)]$. But I don't know if I can take the limit from outside the measure into it. Can I do this? If so, why?
$[-x,x]\to \Bbb{R}$ sure does sound intuitive, but what you really need is a basic form of “monotone convergence”: for any measure space $(X,\mathfrak{M},\mu)$, if $A_1\subset A_2\subset\dots$ is a sequence of measurable sets and $A=\bigcup A_n$ their union, then \begin{align} \mu(A)=\lim\limits_{n\to\infty}\mu(A_n)=\sup\limits_{n\geq 1}\mu(A_n). \end{align} This is called continuity from below. To prove this, let $B_1=A_1$, and for $n\geq 2$, $B_n=A_n\setminus A_{n-1}$. Then, the $B$’s are pairwise disjoint, and $A_n=\bigcup_{k=1}^nB_k$ and $A=\bigcup_{k=1}^{\infty}B_k$. Use countable additivity of measures and the definition of a series.
Edit:
Now that I think about the question further, the question is false as stated. It should be $0\leq \epsilon< m(E)$. The strict inequality is important in order for you to apply the intermediate-value theorem. We have $\lim\limits_{x\to \infty}g(x)=m(E)$ and $g$ is an (weakly-)increasing function on $[0,\infty)$, so given $0\leq\epsilon<m(E)$, we can find some $R>0$ large enough such that for all $x\geq R$, we have $g(x)\in (\epsilon,m(E))$. Now we can apply the intermediate-value theorem on $[0,R]$.
To see that we have to exclude $\epsilon=m(E)$, consider a set $E$ of finite measure but “going off to infinity while carrying positive mass”. For example, let $E$ be the union of sets $E_n=\left(n,n+\frac{1}{2^n}\right)$ so that $E_n$ has measure $\frac{1}{2^n}$ and they are all disjoint. Then, $E=\bigcup_{n=1}^{\infty}E_n$ has finite measure, but for every $x\geq 0$, $E\cap [-x,x]$ has strictly smaller measure than $E$.
If on the other hand, you restrict $E$ to be bounded, then you can go all the way up to $\epsilon=m(E)$.
And finally a fun variant for you to think about: given $\epsilon\in (0,1)$, construct a set $E\subset [0,1]$ such that $E$ is dense and $m(E)=\epsilon$.