Let $X$ be a set with a finite Lebesgue measure (e.g. a subset of the unit interval), and $u,v$ two measures such that:
$$v(X) > u(X)$$
Does there exist a subset $Y\subset X$, with Lebesgue measure exactly $1/2$ that of $X$, such that:
$$v(Y) > u(Y)$$ ?
The intuition the pigeonhole principle: if $v>u$ on the subset $X$, then, if we take $Y$ as that half of $X$ on which $v$ is largest and $u$ is smallest, we will surely have $v>u$. Is this correct?
Let $X$ be a set with finite Lebesgue measure and $u$ and $v$ two measures on Borel algebra of $X$ with $v(X) > u(X)$.
Now, for $Y \subset X$ measurable, let $Z = X \backslash Y$. Then if $u(Y) \ge v(Y)$ and $u(Z) \ge v(Z)$ we have
\begin{align*}
u(X) = u(Y)+u(Z) \ge v(Y)+v(Z)= v(X) > u(X).
\end{align*}
Thus $u(Y) <v(Y)$ or $u(Z) <v(Z)$.
As we can construct a set with Lebesgue measure $\mu \lambda(X)$ for any $\mu \in [0,1]$, see here, if follows that there exists a $Y\subset X$ such that $v(Y) > u(Y)$ and the Lebesgue measure of $Y$ is half that of $X$.
