# The product of two Riemann integrable functions is integrable

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?

• This is homework, right? (Can you capitalize your sentences, by the way? It makes for much more pleasant reading!) Commented Sep 29, 2010 at 8:05
• You seem to have outlined a proof sketch in your question. Namely, you have outline how to reduce to the case of squares, and then how to show that the difference of the square of the values of $f$ at nearby points is bounded by a scalar times the difference of the values of $f$. The next step will be to look at the Riemann sums for $f^2$, and control them in terms of the Riemann sums for $f$, using the bound you have proved. Commented Sep 30, 2010 at 7:58

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.

• Please note that the above referenced document is no longer available at the given location. Given that it has been cited a number of times on MO and math.SE, please consider restoring it at the original location, if possible. Thank you. Commented Aug 12, 2014 at 13:25

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.

• Is there a lower level proof? I mean using Calculus. Commented Sep 29, 2010 at 10:56
• @Americo: Yes there is a lower level proof. I suppose Robin didn't want to spoil a possible homework question. In any case, this is a useful characterization to know. Commented Sep 29, 2010 at 14:17

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.

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.