I have some problems with the question below, although I think that it's not that difficult to solve.
Let $A\subset\mathbb{R}$ be a measurable and bounded. Show that for all $x\in\left[0,l_{1}(A)\right]$ there exists measurable subset $A_{x}\subset A$ that satisfies $l_{1}\left(A_{x}\right)=x$.
Because A is bounded there exists such a and b ($a<b$) that $A\subset\left[a,b\right]$.
I tried starting with the function $f\left(x\right)= m\left(\left[a,\,x\right]\cap\,A\right)$
It's increasing and I have to show that it's also continuous which I can't. Then, using property of Darboux, we will have the thesis.
Can you help me with that, please?
Let $x <y$ Then $|f(x)-f(y)| =m((x,y]\cap A)$ because $[a,y]$ is the disjoint union of $[a,x]$ and $(x,y]$. Hence $|f(x)-f(y)| \leq m((x,y])=y-x$. hence $f$ is continuous.