Measures of subsets of a measurable set

Question

Suppose $S$ is a Lebesgue measurable set on the real line with positive measure $s$. It seems natural to guess the following: for any $m \in [0,s]$, there is a subset $S' \subset S$ such that the Lebesgue measure of $S'$ is $m$. I'm failing to come up with a proof though.

Answer 1

If $m=0$ or $m=s$ then this is obvious. Now consider $m\in (0,s)$. Consider $F(x)=\int_{-\infty}^x 1_{S} d\mu=\mu(S\cap (-\infty, x))$. If you can prove that $F$ is continuous, your claim would follow from the intermediate value theorem. (Note that $\lim_{x\to -\infty}F(x)=0$ and $\lim_{x\to \infty}F(x)=s$ by continuity from above/below)

Answer 2

Overkill answer: If $\lambda$ is Lebesgue measure, $\nu:A\mapsto \lambda(A\cap A)$ is an atomless measure, hence an atomless vector measure, and hence its range $L$ is closed and convex, by Lyapunov's Theorem. Since $\nu(\phi)=0$ and $\nu(\mathbb R)=s$, we know $L=[0,s]$, and hence for each $m\in [0,s]$ there exists an $A\subset S$ with $\nu(A)=m$.