## Suz and Mike had a question

Problem Statement: If $f(x)$ is bounded with finitely many discontinuities on $[a,b]$ then $f$ is Riemann Integrable on $[a,b]$.

Proof: Let $N$ be the number of discontinuities of $f$ on $[a,b]$ and let $M\in\mathbb{R}$ be such that $\left|f(x)\right| for every $x\in [a,b]$. We know such an $M$ exists since $f$ is bounded. Since $f$ is bounded we will be using the sup-inf definition of Riemann Integrable.

Denote a discontinuity by $d_j$ where $d_1.  Let $d=min\{\dfrac{\left|d_j-d_k\right|}{3}: j,k\in[1,N]$ and $j\neq k\}$. Let $I_1=[a,d_1-d], I_2=[d_1+d,d_2-d],\dots I_{N+1}=[d_N,b]$. By construction there are no discontinuities in any $I_j$ and so $f$ is Riemann Integrable one each $I_j$.

Let $\varepsilon>0$. Then for each $I_j$ there exists a $\delta_j>0$ such that for any partition $\pi$ of $I_j$ with $\|\pi\|<\delta_j$ it follows that $\left|\sum\limits_{k=1}^{n}(sup-inf)\delta x_k\right|<\dfrac{\varepsilon}{2(N+1)}$.

Let $\delta=min\{\delta_1,\dots\delta_{N+1},d,\dfrac{\varepsilon}{4NM}\}$ and let $\pi$ be a partition of $[a,b]$ with $\|\pi\|<\delta$. Consider the sum below, we may separate the sum into parts, intervals which contain a discontinuity and those which do not.

$\left|\sum\limits_{k=1}^{n}(sup-inf)\delta x_k\right|=\left|\sum\limits_{k=1}^{n_1}(sup-inf)\delta x_k+\sum\limits_{k=n_1}^{n_1+1}(sup-inf)\delta x_k+\dots+\sum\limits_{k={n_N+1}}^{n}(sup-inf)\delta x_k\right|$

$<\dfrac{\varepsilon}{2(N+1)}+2M\delta+\dots+2M\delta+\dfrac{\varepsilon}{2(N+1)}$

Note that there are $N+1$ $I_j$ intervals and so there are $N+1$ terms less than $\dfrac{\varepsilon}{2(N+1)}$. Also note that we have bounded each of the intervals that contain a discontinuity by $2M\delta$, this is since the largest possible difference between the sup and inf is $2M$. Since there are $N$ discontinuities there are $N$ such terms in our inequality. Now we will use the fact that $\delta<\dfrac{\varepsilon}{4NM}$ in order to simplify our inequality further.

$<\dfrac{\varepsilon}{2(N+1)}(N+1)+N\left(\dfrac{2M\varepsilon}{4NM}\right)$

$=\dfrac{\varepsilon}{2}+\dfrac{\varepsilon}{2}=\varepsilon$

Thus, we have shown that $f$ is Riemann Integrable on $[a,b]$.

$\Box$

Thank goodness $f(x)$ is founded, or this proof wouldn’t have anything to stand on! 😉