I have proved the following statement and I would like to know if my proof is correct and if it could be improved.
"Suppose $(X, \mathcal{S})$ is a measurable space and $f, g: X → \mathbb{R}$ are $\mathcal{S}$-measurable functions. Prove that if $f(x) > 0$ for all $x\in X$, then $f^g$ (which is the function whose value at $x\in X$ equals $f(x)^{g(x)}$) is an $\mathcal{S}$-measurable function."
My proof:
The function $f^g$ can be equivalently seen as the function $h:X\to\mathbb{R},\ h(x):=e^{g(x)\ln (f(x))}=\exp\circ (g\cdot (\ln\circ f))$.
We first note that if $f_{\mathcal{B}}:\mathbb{R} \to\mathbb{R}, f_{\mathcal{S}}:X\to\mathbb{R}$ are respectively a Borel-measurable and an $\mathcal{S}$-measurable function then $f_{\mathcal{B}}\circ f_{\mathcal{S}}:X\to\mathbb{R}$ is an $\mathcal{S}$-measurable function in fact if we take $B\in\mathcal{B}$ then $(f_{\mathcal{B}}\circ f_{\mathcal{S}})^{-1} (B)=(f_{\mathcal{S}}^{-1}\circ f_{\mathcal{B}}^{-1})(B)=f_{\mathcal{S}}^{-1}(\underbrace{f_{\mathcal{B}}^{-1}(B)}_{\in\mathcal{B}})\in\mathcal{S}$.
Now, $\ln$ is a continuous real-valued function defined on $(0,\infty)\in\mathcal{B}$ hence a Borel-measurable function and $f$ is $\mathcal{S}$-measurable so $\ln\circ f$ is $\mathcal{S}$-measurable: since $g$ is also $\mathcal{S}$-measurable (and knowing that the product of $\mathcal{S}$-measurable functions is $\mathcal{S}$-measurable) we have that $g\cdot (\ln\circ f)$ is $\mathcal{S}$-measurable too.
Similarly, $\exp$ is a continuous real-valued function defined on $\mathbb{R}\in\mathcal{B}$ hence a Borel-measurable function so $h=\exp\circ (g\circ (\ln\circ f))$ is $\mathcal{S}$-measurable too and this concludes the proof. $\square$