The product of two Riemann integrable functions is integrable

Question

The goal is to show that the product of two Riemann integrable functions is integrable.

First step is to use the identity $f\cdot g = \frac{1}{4} \left[(f+g)^2 - (f-g)^2\right]$ so that we only need to consider squares of functions.

The second step is to reduce to positive valued functions because $f(x)^2=\left|f(x)\right|^2$.

The third step is to use that if $0 \leq f(x) \leq M$ on $\left[a,b\right]$, $$f^2(x) - f^2(y) \leq 2M \left(\,f(x)-f(y)\right)$$

How should I go about implementing the above steps?

Answer 1

This is not a problem I would assign as homework (at least, not without substantial guidance). Rather, it is one of the fundamental results of the subject -- the subject being advanced calculus / elementary real analysis -- and as such I would expect any instructor / textbook to supply a proof. For instance, Rudin's Principles of Mathematical Analysis covers this. Or see for instance the chapter on integration here.

As Robin says, the result also follows from Lebesgue's criterion of Riemann integrability: now that's something -- I mean the deduction from Lebesgue's Criterion, not the proof of Lebesgue's Criterion! -- I would leave as an exercise, since finding this short argument on one's own helps to drive home the power of the Lebesgue criterion.

Answer 2

This follows from Lebesgue's characterization of Riemann integrable functions as bounded functions continuous outside a set of Lebesgue measure zero. This characterization is usually the swiftest way of deciding on the Riemannn integrability of a function.

Answer 3

For sake of completeness I will prove the statement in a more elementary way. Let's start by showing that $f^2$ is Riemann-integrable:

We know that $f:[a,b]\to \mathbb{R}$ is Riemann-integrable so it follows that $|f|$ is Riemann-integrable as well. It holds that $|f|^2=f^2$, so if $\int\limits_a^b f(x)^2dx $ exists then we have the equality $\int\limits_a^b |f(x)|^2dx =\int\limits_a^b f(x)^2dx$.

As $|f|$ is Riemann-integrable (and hence bounded) it follows that for an arbitrary $\epsilon>0$ there exists a partitition $P:=\{t_0,t_1\cdots, t_n\}$ of $[a,b]$ such that:

$$ \sum\limits_{i=1}^{n}(M_i-m_i)(t_i-t_{i-1})<\frac{\epsilon}{2\sup(|f|)},\\ \text{where } M_i:=\sup\{|f(x)|\mid x\in[t_{i-1},t_i]\}\\ \text{and } m_i:=\inf\{|f(x)|\mid x\in[t_{i-1},t_i]\}. $$

Since $|f([a,b])|\geq 0$, we know that $\sup(|f|^2)=\sup(|f|)^2$ and $\inf(|f|^2)=\inf(|f|)^2$, respectively.

We use these results to give an upper bound of the difference of the Darboux-sums of $f^2$:

$$ \sum\limits_{i=1}^{n}(M'_i-m'_i)(t_i-t_{i-1})=\sum\limits_{i=1}^{n}(M_i^2-m_i^2)(t_i-t_{i-1})=\sum\limits_{i=1}^{n}(M_i-m_i)(M_i+m_i)(t_i-t_{i-1})\\ \leq 2\sup(|f|)\sum\limits_{i=1}^{n}(M_i-m_i)(t_i-t_{i-1})<2\sup(|f|)\frac{\epsilon}{2\sup(|f|)}=\epsilon\\ \text{where } M'_i:=\sup\{f(x)^2\mid x\in[t_{i-1},t_i]\}\\ \text{and } m'_i:=\inf\{f(x)^2\mid x\in[t_{i-1},t_i]\}. $$

This shows that $f^2$ is Riemann-integrable. Applying the ordinary rules of Riemann-integrable functions (regarding addition of functions and multiplying factors) and the hint that $fg=\frac{1}{4}((f+g)^2-(f-g)^2)$ one immediately sees that $fg$ is Riemann-integrable.

Answer 4

For a more elementary proof, you might want to show that if $f$ is Riemann-integrable on $[a,b]$ with $m\leq f(x)\leq M$ and $\phi:[m,M]\rightarrow\Bbb R$ is a continuous function, then $\phi\circ f$ is Riemann-integrable on $[a,b]$.

In particular if $\phi=(x\mapsto x^2)$ and $f$ is Riemann-integrable on $[a,b]$, we will get that $f^2$ is Riemann-integrable on $[a,b]$, and that gets you where you want to go.

Since this is a pretty old question and the existing answers are hints, I'll be pretty terse in the outline of this proof and let the interested fill in the details.

• You want to use the fact that a continuous function on a closed interval is...
• Recall that a function is Riemann-integrable on an interval $[a,b]$ if and only if for all $\epsilon>0$ there is a partition $P$ of $[a,b]$ such that $U(f,P)-L(f,P)<\epsilon$.
• You'll want to split up the underlying partition of the interval $[a,b]$ you're using into two sets. One set will use the fact that $\phi$ is ..., and the other set will use the fact that $\phi$ is bounded and $f$ is Riemann-integrable.