# Thread: f & g Riemann integrable, show fg is integrable

1. ## f & g Riemann integrable, show fg is integrable

If f and g are both Riemann integrable on [a,b], prove that fg is also Riemann integrable.

This is what I've got so far...
Let ||f||∞= sup{|f(x)|: x E [a,b]}
$M_i(f,P)$ = sup{f(x): $x_{i-1}$ ≤ x ≤ $x_i$}
$m_i(f,P)$ = inf{f(x): $x_{i-1}$ ≤ x ≤ $x_i$} where P is a partition of [a,b]

Let x,t E [ $x_{i-1}, x_i$]
Then |f(x)g(x)-f(t)g(t)| |f(x)| |g(x)-g(t)| + |f(x)-f(t)| |g(t)| ≤ ||f||∞ [ $M_i(g,P)-m_i(g,P)$] + [ $M_i(f,P)-m_i(f,P)$] ||g||∞

Any help is appreciated!

[also under discussion in math links forum]

2. Originally Posted by kingwinner
If f and g are both Riemann integrable on [a,b], prove that fg is also Riemann integrable.

This is what I've got so far...
Let ||f||∞= sup{|f(x)|: x E [a,b]}
$M_i(f,P)$ = sup{f(x): $x_{i-1}$ ≤ x ≤ $x_i$}
$m_i(f,P)$ = inf{f(x): $x_{i-1}$ ≤ x ≤ $x_i$} where P is a partition of [a,b]

Let x,t E [ $x_{i-1}, x_i$]
Then |f(x)g(x)-f(t)g(t)| |f(x)| |g(x)-g(t)| + |f(x)-f(t)| |g(t)| ≤ ||f||∞ [ $M_i(g,P)-m_i(g,P)$] + [ $M_i(f,P)-m_i(f,P)$] ||g||∞

Any help is appreciated!

[also under discussion in math links forum]
You've discussed that $f,g\in\mathcal{R}\implies f+g\in\mathcal{R}$ and $f\circ g\in\mathcal{R}$, right? If so $fg=\frac{1}{4}\left[(f+g)^2-(f-g)^2\right]$

3. Originally Posted by Drexel28
You've discussed that $f,g\in\mathcal{R}\implies f+g\in\mathcal{R}$ and $f\circ g\in\mathcal{R}$, right? If so $fg=\frac{1}{4}\left[(f+g)^2-(f-g)^2\right]$
I've seen the result that if f and g are Riemann integrable, then f+g is Riemann intergrable, but I haven't seen the composition one. So I think we may as well prove this problem for the product fg from first principles.

Is there any way to prove that fg is Riemann integrable from basic definitions of upper and lower sums, and perhaps the integrability criterion?

4. Originally Posted by kingwinner
I've seen the result that if f and g are Riemann integrable, then f+g is Riemann intergrable, but I havn't seen the composition one. So I think we may as well prove this problem for the product fg from first principles.

Is there any way to prove that fg is Riemann integrable from basic definitions of upper and lower sums, and perhaps the integrability criterion?
I am not writing it out.

5. Originally Posted by kingwinner
If f and g are both Riemann integrable on [a,b], prove that fg is also Riemann integrable.

This is what I've got so far...
Let ||f||∞= sup{|f(x)|: x E [a,b]}
$M_i(f,P)$ = sup{f(x): $x_{i-1}$ ≤ x ≤ $x_i$}
$m_i(f,P)$ = inf{f(x): $x_{i-1}$ ≤ x ≤ $x_i$} where P is a partition of [a,b]

Let x,t E [ $x_{i-1}, x_i$]
Then |f(x)g(x)-f(t)g(t)| |f(x)| |g(x)-g(t)| + |f(x)-f(t)| |g(t)| ≤ ||f||∞ [ $M_i(g,P)-m_i(g,P)$] + [ $M_i(f,P)-m_i(f,P)$] ||g||∞

Any help is appreciated!

[also under discussion in math links forum]
I was motivated by your steps to write the proof. Sorry I do not know how to use LATEX so I have attached the proof as a .doc document.

The proof uses the notation of oscillation of a function on an interval, which is a very simple concept. Basically, it is the difference between the supremum and infimum of f on a given interval.
This notation is very useful because of how the upper Darboux and lower Darboux sum is defined.

The proof assumes a simple lemma about the oscillation of f on an interval, ie.
the oscillation of f on an interval = sup|f(x)-f(y)| for all x,y in the interval.
A proof of this lemma is also attached for your reference.

Hope this is useful (: