Finitely many discontinuities implies Riemann integrable

Finitely many discontinuities implies Riemann integrable: Let $f:[a,b]\to\mathbb{R}$ be bounded . If $f$ has only finitely many discontinuities on $[a,b]$, then $f$ is Riemann integrable on $[a,b]$.

This theorem illustrates that Riemann integration tolerates “isolated bad points.” It is a stepping stone toward the full characterization via the size of the discontinuity set.

Proof sketch: Let $D=\{x_1,\dots,x_m\}$ be the discontinuity set. Cover each $x_j$ by a small interval $I_j$ so that the total length $\sum |I_j|$ is very small. On the complement $K=[a,b]\setminus\bigcup I_j$, the function $f$ is continuous and hence uniformly continuous , so choose a partition fine enough on $K$ to make oscillations small. The contribution to $U-L$ from $\bigcup I_j$ is controlled by boundedness of $f$ and the small total length, so overall $U-L$ can be made $<\varepsilon$.