Denote by $\lambda$ the Lebesgue measure on $[-\infty,\infty]$, and let $E$ be a $\lambda$-measurable subset of $(0,\infty)$. Define the function $f:[0,\infty]\to[0,\infty]$ by the rule $$f(x)=\lambda\{y\in E:y\leq x\}$$ It is clear that $f$ is monotone with $f(0)=0$ and $f(\infty)=\lambda(E)$. I want to show that $f[0,\infty]=[0,\lambda(E)]$. Since the continuous image of a connected space is itself connected, it suffices to prove that $f$ is continuous. However, the proof eludes me. It seems like it should be easy, but I just can't make it work.
Perhaps it is not true after all?
Thanks!
For $x_1<x_2$, $f(x_2)-f(x_1)=\lambda\{y\in E:x_1<y\leq x_2\}\leq \lambda((x_1,x_2])=x_2-x_1.$ Hence $|f(x_1)-f(x_2)|\leq |x_1-x_2|$ for any $x_1,x_2\in\mathbb{R}^+.$ So, $f$ is continuous on any $x\in\mathbb{R}.$
Continuty at $\infty$ can be shown using continuty of measure. For any sequence $x_n\rightarrow \infty,$ $f(x_n)=\lambda(E\cap [0,x_n])\rightarrow\lambda(E),$ because $E\cap[0,x_n]\uparrow E.$
Similarly conctinuty at zero can be shown as for any sequence $x_n\downarrow0, E\cap[0,x_n]\downarrow E\cup\{0\}=\phi\Rightarrow f(x_n)=\lambda(E\cap[0,x_n])\downarrow 0.$