Let $f$ be a bounded function on $[a,b]$ whose set of discontinuities has measure zero. Show that $f$ is measurable. Then show that the same holds without the assumption of boundedness.
This problem comes from the section in Royden and Fitzpatrick's Real Analysis where Lebesgue's theorem about Riemann integrable functions is presented, which, to my mind, suggests that I should be using that theorem or other theorems appearing in the same section. Indeed, this solution (see page 15, problem 16) appears to use Lebesgue's theorem and certain "comments preceding [it]." But this seems like overkill. Why can't I just use the following two basic theorems appearing in one of the preceding chapters:
Proposition 3: A real-valued function that is continuous on its measurable domain is measurable
Proposition 5: Let $f$ be an extended real-valued function on $E$. For a measurable subset $D$ of $E$, $f$ is measurable on $E$ if and only if the restriction of $f$ to $D$ and $E-D$ are measurable.
Letting $D_f \subseteq [a,b]$ denote all the points at which $f$ is discontinuous, the hypothesis states that $m(D_f)=0$, so that $f$ is clearly measurable over it. Since $f$ is continuous over $[a,b]-D_f$, proposition 3 says it must be measurable over it. By proposition 5, $f$ must be measuralbe over $[a,b]$. Notice that neither boundedness nor the fact that $f$'s domain is a compact interval was used.
Does this seem right, or have I made some silly mistake somewhere?