1:36 PM
in Mathematics, 29 mins ago, by user193319
Problem: If $g$ is strictly monotone on $\Bbb{R}$, and $f$ is a measurable function on the measurable set $E \subseteq \Bbb{R}$, then $g \circ f$ is measurable.
@user193319 Can't you simply use the fact that for any $a\in\mathbb R$ the set $g^{-1}(a,\infty)=I$ is an interval.
So for $h=g\circ f$ you get that $h^{-1}(a,\infty) = f^{-1}(g^{-1}(a,\infty))= f^{-1}(I)$ is a measurable set.
I see that you get basically the same response in the main chatroom. :-)
in Mathematics, 3 mins ago, by Ryan Unger
@user193319 Actually you don't need to worry about the discontinuities. Recall that for measurability it suffices to check if $f^{-1}((a,\infty])$ is measurable for all $a\ge -\infty$. But for a monotone (increasing) function, $f^{-1}((a,\infty])$ is always an interval.
in Mathematics, 2 mins ago, by Ryan Unger
So the preimage by $g$ gives an interval, then use the measurability of $f$...