For this construction we need a countable disjoint family of sets $A_n \subset [0,1]$ of (Lebesgue) outer measure $1$. (See note below.)
Define function $f$ so that $f(x) = 1/n$ for $x \in A_n$. We claim there is no Lebesgue measurable $g$ with $0 < g \le f$. Indeed, $g(x) \le 1/n$ on set $A_n$, which has outer measure $1$, and $g$ is Lebesgue measurable, so $g(x) \le 1/n$ a.e. on $[0,1]$. This is true for all $n$, so $g(x) \le 0$ a.e. on $[0,1]$.
note
How to construct sets $A_n$? Follow the usual transfinite recursion construction for the Bernstein set. (For example, in my answer here.) But instead of choosing two points at each stage, choose countably many.
