Is there a simple functions between $2$ measurable functions? Assume that $f,g$ are real-valued Lebesgue-measurable functions on a measure space $(\Omega, \mathcal F, \mu)$ such that $f \le g$ almost surely. I would like to ask if there exists a simple function $h$ such that $f \le h \le g$ almost surely.
Update: Steven Stadnicki suggests that a counter-example is $f(x)=g(x)=x$. Now I would like to assume $f<g$ a.s. It's intuitive that $f(x)= x$ and $f(x)=x+1$ is still a counter-example. Is there an "easy" proof for it?

Here's a simpler counter-example to the revised question, one that works over a compact interval: take $f(x)=x$, $g(x)=2x$ over $0\leq x\leq 1$. Then $g(x)\gt f(x)$ a.e. (the point $x=0$ being the sole exception). Break the unit interval into subintervals $I_i=[2^{-(i+1)},2^{-i})$: any $h()$ satisfying $g\leq h\leq f$ a.e. must do so on at least one point on each of these intervals (since they have positive measure), so must take on some value $v_i$ satisfying $2^{-(i+1)}\leq v_i\leq 2^{-(i-1)}$. But now the sequence $v_1, v_4, \ldots, v_{3n+1}$ is a sequence of values $h$ takes on, with all its members lying in disjoint intervals and therefore all distinct.
