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Suppose that $g:[a,b]\to\mathbb{R}$ is continuous except at $r_1,r_2\in(a,b)$. Prove that $g$ is Riemann integrable on $[a,b]$.

Intuitively, I know this is true because the upper sum and lower sum will only differ by the contribution from $f(r_1)$ and $f(r_2)$, and you refine the partition until it disappears. But I'm not quite sure how to turn that into a formal proof.

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  • $\begingroup$ MAybe you can do it in a similar way that I do in this answer: math.stackexchange.com/questions/2980279/rieman-integration/… $\endgroup$ – Tito Eliatron Nov 2 '18 at 20:07
  • $\begingroup$ You also need to assume that $g$ is bounded on $[a, b]$. The result can be easily generalized: if the function $g:[a, b]\to\mathbb {R} $ is bounded on $[a, b] $ and if the set $D$ of its discontinuities on $[a, b] $ has a finite number of limit points then $g$ is Riemann integrable on $[a, b]$. $\endgroup$ – Paramanand Singh Nov 3 '18 at 2:46
  • $\begingroup$ The crux of the argument is to show that a single discontinuity does not matter and without loss of generality one can take this single discontinuity at end point of interval of integration. Start by proving that if $g$ is bounded on $[a, b] $ and continuous on $(a, b] $ then $f$ is Riemann integrable on $[a, b] $. $\endgroup$ – Paramanand Singh Nov 3 '18 at 2:51

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