Lebesgue measure has the Darboux property Let $A$ be a measurable Lebesgue set, with $\lambda(A)>0$($\lambda$ is the Lebesgue measure). Then, for every $b \in (0,\lambda(A))$, there exists a set $B$ measurable Lebesgue, $B\subset A$, with $\lambda(B)=b$. Can you give me a suggestion? I have no idea how to "build" the set $B$. It seems to me very likely to the Darboux property of functions..can it be used here?
 A: I assume $A$ is defined to be a subset of $\mathbb{R}$. 
Define $f(x) = \lambda(A \cap [-x,x])$. Then $f(x)$ is continuous, $f(0) = 0$, and $\lim_{x \to +\infty} f(x) = \lambda(A)$, so by the intermediate value theorem, there is some value of $x$ for which $f(x) = b$.
A: Hint: The Function $x \mapsto \lambda(A \cap (-\infty, x])$ is Lipschitz-continuous.
A: Suppose $A$ is bounded. Thus, there is an interval $[a,c]$ such that $A\subseteq [a,c]$. Moreover, $\lambda(A)<\infty$. Let $f:[a,c]\to\mathbb{R}$ given by $f(t)=\lambda([a,x]\cap A)$. Thus $f(x)<\infty$ for all $x\in[a,b]$.
On the other hand, $f(a)=\lambda([a,a]\cap A)\le\lambda([a,a])=0$ and $f(c)=\lambda([a,c]\cap B)=\lambda(B)$. 
Moreover, let $x,y\in[a,c]$. Set $u=\min\{x,y\}$, $v=\max\{x,y\}$. Therefore
$\begin{eqnarray}
|f(x)-f(y)|&=&|f(u)-f(v)|\\
&=&|\lambda([a,u]\cap A)-\lambda([a,v]\cap A)|\mbox{ and using $\lambda(A)<\infty$}\\
&=&\lambda([u,v]\cap A)\\
&\le&\lambda([u,v])\\
&=&v-u\\
&=&|x-y|.
\end{eqnarray}$
So $f$ is uniformly continuous in $[a,c]$. Then, there is $t\in(a,c)$ such that $f(t)=b$. Set $B=[a,t]\cap A$. The last means that $\lambda([a,t]\cap B)=b$.
This shows the case $A$ bounded. Can you do for $A$ unbounded?
EDIT If $A$ is unbounded, cover it by $[-n,n]$ and apply the last.
