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](http://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.