Let \(E\subset[0,\infty)\) be a measurable set with \(p:=m(E)>0\). Show that for any number \(c\in[0,p)\), there is a number \(x\in[0,\infty)\) such that \(m([0,x]\cap E)=c\).
Solution
1. Define the cumulative measure function.
Set \[ F(x):=m(E\cap[0,x]),\qquad x\ge0. \] If \(0\le x<y\), then \[ 0\le F(y)-F(x) =m(E\cap(x,y]) \le y-x. \] Hence \(F\) is nondecreasing and \(1\)-Lipschitz, therefore continuous on \([0,\infty)\). Also \[ F(0)=m(E\cap\{0\})=0. \]
2. Show that \(F(x)\) tends to \(p\).
The sets \[ E\cap[0,n] \] increase to \(E\) as \(n\to\infty\). By continuity of measure from below, \[ F(n)=m(E\cap[0,n])\longrightarrow m(E)=p. \] Thus for every \(c<p\) there exists \(N\) such that \(F(N)>c\).
3. Apply the intermediate value theorem.
Fix \(c\in[0,p)\). Since \(F(0)=0\le c<F(N)\) for some \(N\) and \(F\) is continuous on \([0,N]\), the intermediate value theorem gives an \(x\in[0,N]\) such that \[ F(x)=c. \] Equivalently, \[ \boxed{m([0,x]\cap E)=c.} \]