Real Analysis II - Riemann Integral - Education - Nairaland
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| Real Analysis II - Riemann Integral by Danieljnrs(op): 6:02pm On Mar 02, 2021 |
Please someone should help me with this? Thank you
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| Re: Real Analysis II - Riemann Integral by plessis: 4:22pm On Mar 03, 2021 |
Danieljnrs:Proof. Let U and L denote the upper and lower riemann sums of the functions f and |f|. U(f)≤U(|f|). L(f)≤L(|f|). Hence, U(|f|)-L(|f|)<U(f)-L(f)<€ This implies that U(|f|)-L(|f|)<€ Therefore |f| is Riemann integrable. Since Every Riemann integrable function is continuous and every continuous function is bounded. Then |f| is continuous. Therefore |f| is bounded. 1 Like |
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