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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

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If this is a homework problem, then -- as stated in the FAQ mathoverflow.net/faq#whatnot -- your question would be better suited to one of the sites mentioned there. –  Yemon Choi Sep 29 '10 at 7:58
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This question has been reasked on MSE, so could be closed here without loss. –  Mariano Suárez-Alvarez Sep 29 '10 at 8:07
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2 Answers 2

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
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So this was a homework question? –  Robin Chapman Sep 29 '10 at 18:23
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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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