Let $f: X \to [0, \infty)$ and $p:X \to \mathbb{R}$ be measurable functions. Prove that $g(x) = [f(x)]^{p(x)}$ is a measurable function.
My plan was invoking the following theorem:
For $(X,M)$ a measurable space, $(Y, \tau_Y), (Z, \tau_Z)$ topological spaces, if $f: X \to Y$ is measurable and $g: Y \to Z$ is continuous, then $h = g \circ f$ is measurable.
Now, I defined the function
$$\phi: [0,\infty)\times \mathbb{R} \to [0, \infty)$$ $$\phi(f(x),p(x)) := [f(x)]^{p(x)}$$
Now I just have to prove continuity of $\phi$, but I am having trouble doing this.
Your function $\phi$ is not continuous!. In fact, $(\frac 1 n, -1) \to (0,-1)$ but $\phi (\frac 1 n, -1) \to \infty$. Instead of this approach write $g(x)$ as $e^{p(x)\log (f(x))}$ and note that $\log (f(x))$ is an extended real valued measurable function. Can you complete the argument now?