First, I want to figure out why, for the simple case where $ g:[a,b]\rightarrow\mathbb{R} $ is bounded and continuous except at some point $ x_0\in[a,b], $ $ g$ is Riemann integrable on [a,b].
I know the Riemann integrability condition that there must exist some partition $ P $ for which $ U(P,f) - L(P,f)\leq\epsilon, \forall \epsilon>0 $.
For my attempt at a proof, I said:
Let $ P=\{x_0,x_1,\cdots,x_n:x_0=a
EDIT: Using ncmathsadist's advice,
Let $\epsilon > 0$. Let $M = \sup_{x\in[a,b]} f$.
Let $ D=\{x_0,x_1,\cdots,x_{2n}:x_0
Let $ C $ be a subpartition containing all the other points. By visiting the proof that a continuous function is Riemann integrable, I can construct a $ C $ so that:
$ U(C\bigcup D,f)-L(C\bigcup D,f)< \frac{\epsilon}{2M}\times M+\frac{\epsilon}{2}=\epsilon $
This is because $ g $ is bounded, and any contribution by $ g $ to the sum from the discontinuous point must be less than the maximum, $ M $.