# Prove that composition of functions is measurable

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?
• 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? – The Bosco Feb 1 at 8:48
• 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? – The Bosco Feb 1 at 9:11