If f is continous on [a,b],then f belongs to R[a,b],R-RIEMANN INTEGRAL

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- Mar 24th 2008, 02:41 AMyarlagaddatrinathriemann integral
If f is continous on [a,b],then f belongs to R[a,b],R-RIEMANN INTEGRAL

- Mar 24th 2008, 08:34 AMJhevon
- Mar 24th 2008, 09:34 AMThePerfectHacker
A continous function on [a,b] is always Riemann integrable. Since $\displaystyle f$ is continous it is bounded by extreme value theorem. Furthermore, continous functions on compact (in particular closed) sets are uniformly continous. To prove Riemann integrability we need to show $\displaystyle S_P(f) - s_P(f) < \epsilon$ where $\displaystyle P$ is a partition of $\displaystyle [a,b]$. Given any $\displaystyle \epsilon > 0$ there is a $\displaystyle \delta > 0$ such that $\displaystyle |x-y|<\delta \mbox{ and }x,y\in [a,b]\implies |f(x)-f(y)|<\epsilon$. Pick a partition $\displaystyle P$ such that $\displaystyle (x_i - x_{i-1}) < \delta$. Then $\displaystyle f$ assumes its maximum and minimum on each interval and so $\displaystyle S_P(f) - s_P(f) < \epsilon(b-a)$, and this quantity can be made arbitrary small.