Lebesgue measure, there exists a measurable subset $E_t\subset E$ with $m(E_t) = t$.

Question

Let $E\subset\mathbb{R}$ be a measurable set of Lebesgue measure $m(E) = 1$. Then why to any $t\in [0, 1]$, one can find a measurable subset $E_t\subset E$ with $m(E_t) = t$?

Answer 1

Consider the function $f:\mathbb R\to[0,1]$ given by $$ f(t)=m(E\cap(-\infty,t]). $$ This function is clearly monotone. And it is continuous: if $s>t$, then $$ f(s)-f(t)=m(E\cap(-\infty,s])-m(E\cap(-\infty,t])=m(E\cap(t,s])\leq m((t,s])=s-t. $$ As $\lim_{t\to-\infty} f(t)=0$ and $\lim_{t\to\infty} f(t)=1$, the Intermediate Value theorem guarantees that all values between $0$ and $1$ are achieved.