Proving $fg$ and $f+g$ is Riemann integrable through the easy and hard way.

Question

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.

Answer 1

You are stuck with the last steps of

$$\begin{align} U(fg,P) - L(fg,P) &\leq U(f,P)U(g,P) - L(f,P)U(g,P) + L(f,P)U(g,P) - L(f,P)L(g,P) \\ &=U(g,P)[U(f,P) - L(f,P)] + L(f,P)[U(g,P) - L(g,P)]\\ &\leq \epsilon/2[U(g,P) + L(f,P)]\\ &\leq \epsilon/2[ U(g,P) + \sup \{L(f,P) \}]. \end{align}$$

A key idea is that since both $f$ and $g$ are integrable, the Riemann partition sums $U(g$ and $L(f)$ exist and are finite (I suppress all partition notation - but they're always there). That is, there are some $I_f$ and $I_g$ such that $L(f) \leq I_f \leq U(f)$ and $L(g) \leq I_g \leq U(g)$ for all sufficiently fine partitions, and hence we know the values of $L(f), U(f), L(g), U(g)$ within $\epsilon$.

In particular, $\lvert U(g) \rvert \leq I_g + \epsilon$ and $\lvert L(f) \rvert \leq I_f$. So we finish with the triangle inequality:

$$\begin{align} \frac{\epsilon}{2} \lvert U(g) + L(f) \rvert &\leq \frac{\epsilon}{2} \lvert U(g) \rvert + \frac{\epsilon}{2} \lvert L(f) \rvert \\ &\leq \frac{\epsilon}{2} \lvert I_g + \epsilon \rvert + \frac{\epsilon}{2} \lvert I_f \rvert \\ &\leq K\epsilon \end{align}$$ for a particular but absolute (independent of the choice of partition) constant $K$, and thus is arbitrarily small, concluding the proof.

Answer 2

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 $\|P\|\rightarrow 0$.

Answer 3

Given a function $f:\ [a,b]\to{\mathbb R}$ and a subinterval $Q\subset[a,b]$ write $\mu(Q)$ for the length of $Q$, and put $$\|\Delta f\|_Q:=\sup_{x,y\in Q}|f(y)-f(x)|\qquad\bigl(=\sup_{x\in Q} f(x)-\inf_{x\in Q} f(x)\bigr)\ .$$ This $f$ is Riemann integrable over $[a,b]$ if for any $\epsilon>0$ we can find a partition $P$ of $[a,b]$ into finitely many subintervals $Q_k$ such that $$\bigl(U(f,P)-L(f,P)=\bigr)\qquad D_P(f):=\sum_k \|\Delta f\|_{Q_k}\>\mu(Q_k)<\epsilon\ .$$ When $f$ passes this test it is automatically bounded. Furthermore we note that a refinement of $P$ makes $D_P(f)$ smaller.

Now assume that two integrable functions $f$ and $g$ are given. In the following I shall deal with $fg$ only, since $f+g$ is simpler. Then there is an $M>0$ such that both $f$ and $g$ are globally bounded by $M$. From $$f(y)g(y)-f(x)g(x)=f(y)\bigl(g(y)-g(x)\bigr)+g(x)\bigl(f(y)-f(x)\bigr)$$ it follows that $$|f(y)g(y)-f(x)g(x)|\leq M\bigl(|f(y)-f(x)|+|g(y)-g(x)|\bigr)\qquad\forall \ x,\>y\in[a,b]\ ,$$ and this implies that for any subinterval $Q\subset[a,b]$ we have $$\|\Delta (fg)\|_Q\leq M\bigl(\|\Delta f\|_Q+\|\Delta g\|_Q\bigr)\ .\tag{1}$$ Now let an $\epsilon>0$ be given. By assumption there are partitions $P'$ and $P''$ of $[a,b]$ such that $$D_{P'}(f)<{\epsilon\over 2M},\quad D_{P''}(g)<{\epsilon\over 2M}\ .$$ Let $P$ be a common refinement of $P'$ and $P''$. Then from $(1)$ it follows that $$D_P(fg)\leq M\bigl(D_P(f)+D_P(g)\bigr)\leq M\bigl(D_{P'}(f)+D_{P''}(g)\bigr)<\epsilon\ .$$