## Spring 2003 #1

Problem Statement: (a) State a definition of a Riemann Integral. (b) Use this definition to prove if $f(x), g(x)$ are both bounded, Riemann Integrable, and defined on $[a,b]$, then $fg(x)$ is Riemann Integrable on $[a,b]$.

(a) A bounded function $f$ is Riemann Integrable on $[a,b]$ if for every $\varepsilon>0$ there exists a $\delta>0$ such that for $\pi=\{x_0,x_1,\dots,x_n\}$ a partition of $[a,b]$ with $||\pi||<\delta$ it follows that $|\sum\limits_{k=1}^n (\sup f(x)_{x\in[x_{k-1},x_k]}-\inf f(x)_{x\in[x_{k-1},x_k]})(x_k-x_{k-1})|<\varepsilon$.

Proof: Since $f,g$ are both bounded there exist $M,N\in\mathbb{R}$ such that $|f(x)| for every $x\in[a,b]$ and $|g(x)| for every $x\in[a,b]$.

Let $\varepsilon>0$. Since both $f, g$ are Riemann Integrable on $[a,b]$ we know that there exist $\delta_1,\delta_2$ such that for any partition $\pi_1$ of $[a,b]$ with $||\pi||<\delta_1$ it follows that $|\sum\limits_{k=1}^s (\sup f(x)_{x\in[x_{k-1},x_k]}-\inf f(x)_{x\in[x_{k-1},x_k]})(x_k-x_{k-1})|<\dfrac{\varepsilon}{2N}$. Similarly for any partition $\pi_2$ with $||\pi_2||<\delta_2$ it follows that $|\sum\limits_{k=1}^t (\sup g(x)_{x\in[x_{k-1},x_k]}-\inf g(x)_{x\in[x_{k-1},x_k]})(x_k-x_{k-1})|<\dfrac{\varepsilon}{2M}$. Let $\delta=min\{\delta_1,\delta_2\}$ and $\pi=\{x_0,x_1,\dots,x_n\}$ be a partition of $[a,b]$ such that $||\pi||<\delta$. Then it follows that

$|\sum\limits_{k=1}^n(\sup fg(x)_{x\in[x_{k-1},x_k]}-\inf fg(x)_{x\in[x_{k-1},x_k]})(x_k-x_{k-1})|$

For space reasons I am going to drop the subscript notation on the sup and inf.

$=|\sum\limits_{k=1}^n(\sup f(x)-\inf f(x))\sup g(x)+\inf f(x)(\sup g(x)-\inf g(x))(x_k-x_{k-1})|$

$\leq|\sum\limits_{k=1}^n N(\sup f(x)-\inf f(x))(x_k-x_{k-1})|+|\sum\limits_{k=1}^n M(\sup g(x)-\inf g(x))(x_k-x_{k-1})|$

$

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

$=\varepsilon$

Thus, $fg(x)$ is Riemann Integrable on $[a,b]$.

$\Box$

Reflection: When I first approached this problem my first thought was to use the sup-inf definition of R.I. since we were told that the functions were both bounded. The part I got stuck on was how to relate $\sup fg-\inf fg$ to $\sup f-\inf f$ and $\sup g-\inf g$. I figured there was some trick to it, something like adding and subtracting the same thing to break the pieces apart. Ultimately that’s what we did but we got there in a different way. We wrote down what we wanted to end up with and worked backwards to figure out what we needed to add and subtract. Now I feel like I could come up with this trick on the actual exam if I needed to.