Prove that if $f$ is defined as having a positive discontinuity at $c$ and $0$ otherwise on [a,b], it is Darboux integrable and its integral is 0.
$\forall \epsilon>0,$ choose $\delta=\frac{\epsilon}{2f(c)}$ if $\frac{\epsilon}{2f(c)}<(a-b)$ and choose $\delta=a-b$ if otherwise.
For Case 1: Choose a Partition P $\{a=x_1,x_2,..., x_n=b\}$ where size of any interval is $\delta$.
Then, $U(f;P)\leq\sum \sup f(x)\delta + 2f(c)\delta$ since c may lie on the boundary of an interval. Which is $=\epsilon.$
For Case 2: Use the Partition with only 1 subinterval, [a,b].
Then, $U(f;P)=f(c)(a-b)\leq\frac{\epsilon}{2}\leq\epsilon$ Reason: $\frac{\epsilon}{2f(c)}\geq(a-b)$ implies $\epsilon /2 \geq f(c)(a-b)$
Q.E.D
Does the above proof works?