# Lebesgue measure and Intermediate Value Theorem

Question

Let $$(\mathbb{R}, M_{Leb}, m)$$ be the Lebesgue measure space, and let $$E\in M_{Leb}$$ be such that $$0 < m(E) < \infty$$. Show that the function defined so that $$h(x) = m(E \cap (−\infty, x])$$ for $$x\in\mathbb{R}$$ is uniformly continuous on $$\mathbb{R}$$, and show that for every $$c \in (0,m(E))$$ there is a Lebesgue measurable subset $$A \subseteq E$$ such that $$m(A) = c$$.

Attempts

I'm really unsure for the first part. It seems we could take $$\delta=\epsilon$$ in the usual definition for uniform continuity but I could be wrong.

For the second part, I noticed that $$h(x)\to 0$$ as $$x\to -\infty$$ and $$h(x)\to m(E)$$ as $$x\to \infty$$, so we could maybe apply the Intermediate Value Theorem but is that justified on a closed interval $$(0,m(E))$$ where the endpoints are limits? I'm uncertain.

Thank you

I assume you mean $$h(x)=m(E\cap(-\infty,x])$$.
Note that if $$A\subset B$$, then $$m(B)-m(A)=m(B\setminus A)$$. If $$x, then $$h(y)-h(x)=m(E\cap(-\infty,y])\setminus (E\cap(-\infty,x]))=m(E\cap(x,y])\leq m((x,y])=y-x.$$ So yes, as you say, given $$\varepsilon>0$$ you can take $$\delta=\varepsilon$$. If $$|y-x|<\delta$$, then $$|h(y)-h(x)|<\varepsilon$$.
The second part is a calculus exercise. You have $$h:\mathbb R\to[0,m]$$, continuous, and such that $$\lim_{x\to-\infty}h(x)=0,\qquad\qquad \lim_{x\to\infty}h(x)=m.$$ Given $$c\in(0,m)$$, by definition of limit there exists $$a$$ such that $$h(a) and $$b$$ such that $$h(b)>c$$. Now apply the Intermediate Value Theorem on $$[a,b]$$.