Problem: Suppose $f,g$ are Riemann integrable functions, show that $f+g$ and $fg$ are also Riemann integrable.
I know there is really easy to do this with measure theory, but I want to see if this method works as well. I will write an answer for $f+g$ using measures.
Write up 1: Denote $D_f, D_g$ to be the set of all discontinuities of $f$ and $g$. $f+g$ can only be integrable if and only if $D_f \cap D_g$ has measure zero, but without any loss of generality, we also have $D_f \cap D_g \subset D_f$, the set on the left has measure zero. So $f+g$ is Riemann integrable.
Okay this one is the longer one.
Write up 2: As $f$ and $g$ are integrable, there are partitions $P_f$ and $P_g$ such that $$U(f,P_f) – L(f,P_f) < \epsilon/2,$$
$$U(g,P_g) – L(g,P_g) < \epsilon/2.$$
Now we begin the estimation starting with $$L(f,P)L(g,P) \leq L(fg,P),$$ and $$U(f,P)U(g,P) \geq U(fg, P).$$
Therefore if we let $P \supset P_f \cup P_g$ then we get,\begin{align}
U(fg,P) – L(fg,P) &\leq U(f,P)U(g,P) – L(f,P)U(g,P) + L(f,P)U(g,P) – L(f,P)L(g,P) \\
&=U(g,P)[U(f,P) – L(f,P)] + L(f,P)[U(g,P) – L(g,P)]\\
&\leq \epsilon/2[U(g,P) + L(f,P)]\\
&\leq \epsilon/2[ U(g,P) + \sup \{L(f,P) \}]
\end{align}
I am stuck with the last step, I am not sure how to bound $U(g,P)$. May I get some pointers?
My futile idea is that I can do the following bound $U(g,P) < \epsilon/2 + L(g,P) < \epsilon/2 + \sup \{L(g,P) \}$
note: I know for $fg$, there is another short proof with $4fg = (f+g)^2 – (f – g)^2.$ I am not seeking that one either.
Best Answer
One of your inequalities is not true: $$ L(f,P)L(g,P) \le L(fg,P). $$ If $P$ is a partition of $[0,2]$, and $f$, $g$ are constant and equal to $1$ on $[0,2]$, then $$ 2*2=L(f,P)L(g,P)\not\le L(fg,P)=2. $$ Alternative Method: You have shown that $f+g$ is Riemann integrable because $f$ and $g$ are. One property you have failed to use is that Riemann integrable functions $f$, $g$ on $[a,b]$ are necessarily bounded in absolute value on $[a,b]$. Because of this, it is possible to reduce to the case where $f \ge 0$ and $g \ge 0$ by adding a constant $L$ to both $f$ and $g$: $$ fg = (f+L)(g+L)-L(f+g)-L^{2}. $$ So, without loss of generality, assume $0 \le f \le K$, $0 \le g \le K$ on the interval of integration for some constant $K$. If $h_{m}$ denotes the greatest lower bound of $h$ on an interval $I$ and $h_{M}$ denotes the least upper bound of $h$ on $I$, then $$ \begin{align} (fg)_{M}-(fg)_{m} & \le f_{M}g_{M}-f_{m}g_{m} \\ & =(f_{M}-f_{m})g_{M}+f_{m}(g_{M}-g_{m}) \\ & \le K(f_{M}-f_{m})+K(g_{M}-g_{m}). \end{align} $$ Therefore, $$ U(fg,P)-L(fg,P) \le K(U(f,P)-L(f,P))+K(U(g,P)-L(g,P)). $$ The right side tends to $0$ as $\|P\|\rightarrow 0$ because $f$ and $g$ are Riemann integrable. So, the left side also tends to $0$ as $\|P\|\rightarrow 0$.