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][1].)  But instead of choosing two points at each stage, choose countably many.


  [1]: https://math.stackexchange.com/a/303721/442