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Let $(\Omega, S)$ be a measurable space. If $f:\Omega \rightarrow \mathbb{R}$ is a strictly positive measurable function and $g:\Omega \rightarrow \mathbb{R}$ is measurable show that $f^g$ is measurable.

I think I should use the theorem on the measurability of the composed function to solve this exercise but I don't realize how.

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Hint: consider the function $f:(0,\infty)\times\mathbb R\to\mathbb R$, with $f(x,y)=x^y$.

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I will give it a try ,.... since f is strictly positive , we can see that the function $ln(f)$ is measurable. Now $f^g$ can be written as $e^{g*ln(f)}$. The product $g*ln(f) $ is again measurable. Let's set it equal to $h$. We then have $e^h$ , which is a composition with the exponential function which is measurable.

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