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)|<M for every x\in[a,b] and |g(x)|<N 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})|




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


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.

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