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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.

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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?

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  • $\begingroup$ What is it with sequences of the form $\frac{1}{n}$ (or very similar) that they keep popping up all over the place in measure theory? $\endgroup$ – The Bosco Feb 1 at 8:48
  • $\begingroup$ They pop up everywhere in Mathematics, not just measure Theory! $\endgroup$ – Kavi Rama Murthy Feb 1 at 8:52
  • $\begingroup$ But I've seen them way more often in this branch! Also, what you meant was that I should substitute my $\phi$ function with $g$ or the $g$ in the theorem I wrote? $\endgroup$ – The Bosco Feb 1 at 9:11

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