Let $\Omega$ be a measurable subset of $\Bbb R^n$, and let $W$ be an open subset of $\Bbb R^m$. If $f:\Omega\to W$ is measurable, and $g:W\to\Bbb R^p$ is continuous, then $g\circ f:\Omega\to\Bbb R^p$ is measurable.
MY ATTEMPT
Let $V\subseteq\Bbb R^p$ be an open set. Then we have that $g^{-1}(V)$ is open. Since $f$ is measurable, $f^{-1}(g^{-1}(V))$ is also measurable.
In other words, for every open set $V\subseteq\Bbb R^p$, the set $(g\circ f)^{-1}(V)$ is measurable, and the result holds.
I am a little bit unsure about the wording of the proof. Could someone please check my arguments?