2
$\begingroup$

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

$\endgroup$

1 Answer 1

2
$\begingroup$

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 $x<y$, 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 $h(a)<c$ and $b$ such that $h(b)>c$. Now apply the Intermediate Value Theorem on $[a,b]$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .