I'm supposed to prove this one using only the definition of a measurable function (I can use other results about measurable sets, though)
My Attempt:
Let $\{x_1,...,x_n\}$ be the discontinuities of $f$, and let $E=[a,b]-\{x_1,...,x_n\}$.
Let $g$ denote the restriction of $f$ to $E$, so that $g$ is continuous on its domain.
By definition of measurability, we need to show that $g^{-1}((c,\infty))$ is measurable $\forall c\in \mathbb{R}$.
However, $g$ is continuous, and $(c,\infty)$ is open. Thus $g^{-1}((c,\infty))$ is open and thus is measurable.
If there is no $x_i$ such that $f(x_i)\in (c,\infty)$, then $f^{-1}((c,\infty)) = g^{-1}((c,\infty))$ is measurable.
Otherwise there are some $\{x_i,...,x_j\}$ such that $f(\{x_i,...,x_j\})\subseteq (c,\infty)$, and thus $f^{-1}((c,\infty)) = g^{-1}((c,\infty)) \cup \{x_i,...,x_j\}$. Since $\{x_i,...,x_j\}$ has measure zero, and $g^{-1}((c,\infty))$ is measurable, it follows that $f^{-1}((c,\infty))$ is measurable.
This seems fine to me, but one of my hypotheses is that $f$ is bounded, which didn't show up in this proof.
Any thoughts are greatly appreciated.
Thanks.