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first show you only need to consider squares of functions as f.g = 1/4 [(f+g)sqr - (f-g)sqr]. show then that you only need to consider only positive valued functions becuase f(x).g(x)=|f(x)|sqr. then , if 0 <=f(x) <= M on [a,b] show that f sqr(x) - f sqr(y) <= 2M (f(x)-f(y)).

does anyone know how i would answer this ??

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closed as too localized by Yemon Choi, Mariano Suárez-Alvarez, Robin Chapman, S. Carnahan Sep 29 '10 at 8:33

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

If this is a homework problem, then -- as stated in the FAQ -- your question would be better suited to one of the sites mentioned there. – Yemon Choi Sep 29 '10 at 7:58
This question has been reasked on MSE, so could be closed here without loss. – Mariano Suárez-Alvarez Sep 29 '10 at 8:07

It follows from Lebesgue's characterization of Riemann integrable functions as bounded functions continuous outside a set of Lebesgue measure zero.

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thanks for your advice, but is there a simpler approach because i am only a second year student and we have not covered Lebesgue's characterization of Riemann integrable functions at all. – sam Sep 29 '10 at 7:53
So this was a homework question? – Robin Chapman Sep 29 '10 at 18:23

If $f$ and $g$ are Riemann integrable over the interval $[a,b]$ then there is an $M$ such that $|f|$ and $|g|$ are both $\le M$ on $[a,b]$. The Riemann integrability of $f g$ then immediately follows from the inequality $$|f(x)g(x)-f(x')g(x')|\le |f(x)-f(x')||g(x)|+|f(x')||g(x)-g(x')|$$ $$\le M(|f(x)-f(x')| +|g(x)-g(x')|) $$ for all $x, x'\in [a,b]$.

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