Q:

Lebesgue Measure, distribution function

Let $\mu$ be the Lebesgue measure on $\mathbb{R}$ and let A be a Borel set such that $0 < \mu(A) < \infty$. Prove that for any $0 < \alpha < \mu(A)$ there exists a Borel set $A_{\alpha} \subset A$ such that $\mu(A_\alpha)=\alpha$.
Hint: Consider the sets $B_x = A \cap (-\infty,x]$ and the function $f(x)= \mu(B_x)$.
I follow the hint and set $f(x)= \mu(B_x)$. By the definition, I get $0 \le f(x) \le \mu(A)$. Then, I don't know what I should do for the next step.

A:

Set up $f$ as in your hint. Fix $x\in\mathbb R$. Then for each $\varepsilon>0$, there exists some $x'\in\mathbb R$ such that $|x'-x|<\varepsilon$. Then 
$$|f(x')-f(x)|=|\mu(B_x)-\mu(B_{x'})|\le|x'-x|<\varepsilon,$$
showing that $f$ is continuous on $\mathbb R$ (this is because we have either $B_x\subseteq B_{x'}$ or $B_{x'}\subseteq B_x$. If, without loss of generality, the former holds, then $\mu(B_{x'})=\mu(B_x)+\mu(B_{x'}\backslash B_x)\le \mu(B_x)+|x'-x|$). Then the desired property follows from continuity, as $f(-\infty)=0$ and $f(+\infty)=\mu(A)>\alpha$.  

