The following question appeared on the Northwestern University Spring 2013 preliminary exam in analysis which is freely available online.
Show that, if $f:\mathbb{R}\rightarrow\mathbb{R}$ is measurable, then the set $\{x\in\mathbb{R}\mid m(f^{-1}(x))>0\}$ has measure zero.
Let this set be denoted by $S$. I claim that $S$ is countable and hence has Lebesgue measure zero. $S=\bigcup_{n\in\mathbb{Z}}\{x\in\mathbb{R}\mid 10^n\leq m(f^{-1}(x))\leq10^{n+1}\}$. Suppose $S$ is uncountable. Since a countable union of countable sets is countable, at least one of the sets in the union is uncountable. In particular, for some $\delta>0$, there are uncountably many $x\in S$ with $m(f^{-1}(x))>\delta$. Further, each of the sets $f^{-1}(x)$ are disjoint.
My intuition is that since $\mathbb{R}$ can be covered by countably many disjoint line segments with length $\delta$ there is not enough room to accommodate uncountably many of them, no matter how they're arranged. I'm having trouble formalizing this idea.