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 $x\in\mathbb R$?



Existence of a Borel measurable function below any positive function
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](https://math.stackexchange.com/questions/369859/cardinality-of-the-borel-measurable-functions).


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. 


