Prove or construct counterexample for statement about measure. Let (S,S,u) be a measure space and f,g $\in$ $L^0$ satisfy u(x $\in S$ : $f(x) < g(x)) > 0$.  Prove or construct a counterexample for the following statement. There exists constants a, b $\in$ R s.t. u({x $\in$ S : f(x) $\leq$ a < b $\leq$ g(x)}) > 0.
My first instinct was to try to find a counterexample. I tried to do this by finding a situation where f(x) is always increasing s.t. for any choice of a, all f(x) < a would yield only a countable set of points, hence obtaining measure 0, but I am having a difficult time doing this. I also tried proving the statement using lim sups and infs, but I don't know if they even necessarily exist.

$\{x:f(x)<g(x)\}=\bigcup_{p \in Q} \bigcup_{q \in \Bbb{Q}}\{x:f(x) \leq p<q \leq g(x)\}$
There exist $p_0,q_0 \in \Bbb{Q}$ such that $u(\{x:f(x) \leq p<q \leq g(x)\})>0$
because if all these sets had measure zero then by subadditivity you would have that $u(\{x:f(x)<g(x)\})=0$
