# f & g Riemann integrable, show fg is integrable

• Mar 15th 2010, 10:37 PM
kingwinner
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]}
$\displaystyle M_i(f,P)$ = sup{f(x): $\displaystyle x_{i-1}$ ≤ x ≤ $\displaystyle x_i$}
$\displaystyle m_i(f,P)$ = inf{f(x): $\displaystyle x_{i-1}$ ≤ x ≤ $\displaystyle x_i$} where P is a partition of [a,b]

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

Any help is appreciated!

[also under discussion in math links forum]
• Mar 15th 2010, 10:42 PM
Drexel28
Quote:

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]}
$\displaystyle M_i(f,P)$ = sup{f(x): $\displaystyle x_{i-1}$ ≤ x ≤ $\displaystyle x_i$}
$\displaystyle m_i(f,P)$ = inf{f(x): $\displaystyle x_{i-1}$ ≤ x ≤ $\displaystyle x_i$} where P is a partition of [a,b]

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

Any help is appreciated!

[also under discussion in math links forum]

You've discussed that $\displaystyle f,g\in\mathcal{R}\implies f+g\in\mathcal{R}$ and $\displaystyle f\circ g\in\mathcal{R}$, right? If so $\displaystyle fg=\frac{1}{4}\left[(f+g)^2-(f-g)^2\right]$
• Mar 15th 2010, 10:51 PM
kingwinner
Quote:

Originally Posted by Drexel28
You've discussed that $\displaystyle f,g\in\mathcal{R}\implies f+g\in\mathcal{R}$ and $\displaystyle f\circ g\in\mathcal{R}$, right? If so $\displaystyle fg=\frac{1}{4}\left[(f+g)^2-(f-g)^2\right]$

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?
• Mar 15th 2010, 10:54 PM
Drexel28
Quote:

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.
• Feb 12th 2011, 09:19 PM
skxsephiroth
Quote:

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]}
$\displaystyle M_i(f,P)$ = sup{f(x): $\displaystyle x_{i-1}$ ≤ x ≤ $\displaystyle x_i$}
$\displaystyle m_i(f,P)$ = inf{f(x): $\displaystyle x_{i-1}$ ≤ x ≤ $\displaystyle x_i$} where P is a partition of [a,b]

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