proof : product of two riemann integrals are integrable

first show you only need to consider squares of functions as $f\cdot g = \frac{1}{4} \left[(f+g)^2 - (f-g)^2\right]$. Show then that you only need to consider only positive valued functions becuase $f(x)\cdot g(x)=\left|f(x)\right|^2$. then, if $0 \leq f(x) \leq M$ on $\left[a,b\right]$ show that $f^2(x) - f^2(y) \leq 2M \left(\,f(x)-f(y)\right)$.

does anyone know how I would answer this ??

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This is homework, right? (Can you capitalize your sentences, by the way? It makes for much more pleasant reading!) –  Mariano Suárez-Alvarez Sep 29 '10 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. –  Matt E Sep 30 '10 at 7:58

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.

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Is there a lower level proof? I mean using Calculus. –  Américo Tavares Sep 29 '10 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. –  Aryabhata Sep 29 '10 at 14:17
@Moron: Thanks! –  Américo Tavares Sep 29 '10 at 14:39
@Americo: You are welcome :-) –  Aryabhata Sep 30 '10 at 5:32