I am new on this site. I am wondering if my solution is correct.
I know we can do this using the fact that composition of continuous functions with a measurable one is measurable, but I want to see if my reasoning is correct for this specific example. Let X be a set, with sigma algebra $A$. Suppose $f: X\to \mathbb{R}$, and suppose $f^3$ is measurable. Then show that $f$ is measurable.
My attempt: Let $a \in \mathbb{R}$, we will show $\{x: f(x)<a\}$ is measurable. Now $\{x: f(x)<a\}=\{x: f^3(x)<a^3\}$. But $\{x: f^3(x)<a^3\}$ is measurable as $f^3$ is measurable. Hence since $a$ is arbitrary and the intervals $(-\infty,b)$ generate the borel sigma algebra, $f$ is measurable.
Thanks in advance.
And if its not correct, please tell me why.
Edit: The sigma algebra on R is assumed to be the Borel sigma algebra.