Existence of a Borel measurable function below any positive function Let $f\colon \mathbb R\to (0,\infty)$ be a function taking positive values.
Does there exist a Borel measurable function $g\colon \mathbb R\to (0,\infty)$ taking positive values as well such that $g(x)\leq f(x)$ for all $x\in\mathbb R$?
 A: The answer is No.
Let us assume that the claim you are asking about is true and let us try to arrive at a contradiction.
Cardinality of the set of all Borel measurable functions is $\mathfrak c=2^{\aleph^0}$; see here.
Let us well-order all positive Borel functions as $\{g_\alpha; \alpha<\mathfrak c\}$. Let us consider some well-ordering $\mathbb R=\{x_\alpha; \alpha<\mathfrak c\}$ of real numbers. Let us define a new function $f\colon\mathbb R\to(0,\infty)$ as
$$f(x_\alpha)=\frac{g_\alpha(x_\alpha)}2.$$
Clearly, there is no $\alpha$ such that $g_\alpha\le f$. This contradicts the given claim. 
A: 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.
