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.

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.

The answer is No.

Let us assume that the claim you are asking about is true and let us try to arrive at contradiction.

Cardinality 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.

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.