# How do I prove that a function with a finite number of discontinuities is Riemann integrable over some interval?

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 $P=\{x_0,x_1,\cdots,x_n:x_0=a<x_1<\cdots<x_n=b\}$ be a partition of $[a,b]$. Let $x_0\in[x_{k-1},x_k].$ Then normally for an everywhere continuous function, since $[a,b]$ is a compact set, the function is uniformly continuous and therefore $\exists\delta(\epsilon):|x-y|<\delta\implies|f(x)-f(y)|<\epsilon,\forall\epsilon>0$. Using this, we can make any of the subintervals in the partition arbitrarily small, and eventually make the difference between the upper and lower sums arbitrarily small by using the property that a continuous function will attain its maximum and minimum. But how do I do something similar for $g$? Furthermore, how can I extend that to the case in which $g$ is discontinuous at $x_0, x_1, \cdots, x_n$?

Let $\epsilon > 0$. Let $M = \sup_{x\in[a,b]} f$.

Let $D=\{x_0,x_1,\cdots,x_{2n}:x_0<x_1<\cdots<x_{2n}\}$ be a subpartition containing all the points $x_0,x_1,\cdots,x_n$ where $g$ is discontinuous, such that $\sum_{i=1}^{2n}(x_i-x_{i-1}) < \frac{\epsilon}{2M}$

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, $M$.

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 This is looking good. To make things a bit clearer, let $K$ denote the union of the subintervals containing the discontinuities and $L$ denote the union of the subintervals that do not contain the discontinuities. You know $f$ is uniformly continuous on $L$. – ncmathsadist Jan 28 '12 at 13:07

Let $\epsilon > 0$. Put $M = \sup_{x\in[a,b]} f$. Now choose a partition so that the total length of the intervals containing the discontnuities of $f$ is smaller than $\epsilon/2M$. The contribution from the intervals with discontinuities of $f$ is smaller than $\epsilon/2$. Since $f$ is continuous on the rest, it is fairly easy to engineer an argument that supplies a partition whose upper and lower sums differ by less than $\epsilon/2$.