I have to show the following:
Suppose $A$ is a measurable subset of $\mathbb{R}$ and $B$ an open subset of $\mathbb{R}$. If $f: A \to B$ is measurable, and $g:B \to \mathbb{R}$ continuous, then $g \circ f : A \to \mathbb{R}$ is measurable.
We have defined a measurable function $f$ as satisfying $\{x \in A: f(x)>a \}$ is measurable for any $a \in \mathbb{R}$. I also know of a few sets that are measurable.
$\textbf{My Work Thus Far:}$ I realize that I need to show $(g \circ f)^{-1} ((a, \infty)) = f^{-1}(g^{-1}(a,\infty))$ is measurable for any $a \in \mathbb{R}$. Since $g$ is continuous, we see that $g^{-1}((a,\infty))$ is an open set, which we can call X. We can write X as a countable union of open intervals. So, I need to show that $f^{-1}(X)$ is measurable, where X is an open set in $\mathbb{R}$ which can be written as the countable union of open intervals in $\mathbb{R}$ and $f$ measurable.
$\textbf{Edit:}$ Since $f$ is measurable, we see that $ Y = f^{-1}((a, \infty))$ is measurable for any $a \in \mathbb{R}$. We can find $a \in \mathbb{R}$ so that $X \subset Y$. Since Y is measurable, X is measurable as a subset of $X$.