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.

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.