proof of continuous functions are Riemann integrable

Recall the definition of Riemann integral. To prove that $f$ is integrable we have to prove that $\lim_{\delta\to 0^{+}}S^{*}(\delta)-S_{*}(\delta)=0$. Since $S^{*}(\delta)$ is decreasing and $S_{*}(\delta)$ is increasing it is enough to show that given $\epsilon>0$ there exists $\delta>0$ such that $S^{*}(\delta)-S_{*}(\delta)<\epsilon$.

So let $\epsilon>0$ be fixed.

By Heine-Cantor Theorem $f$ is uniformly continuous i.e.

 $\exists\delta>0\ |x-y|<\delta\Rightarrow|f(x)-f(y)|<\frac{\epsilon}{b-a}.$

Let now $P$ be any partition of $[a,b]$ in $C(\delta)$ i.e. a partition $\{x_{0}=a,x_{1},\ldots,x_{N}=b\}$ such that $x_{i+1}-x_{i}<\delta$. In any small interval $[x_{i},x_{i+1}]$ the function $f$ (being continuous) has a maximum $M_{i}$ and minimum $m_{i}$. Since $f$ is uniformly continuous and $x_{i+1}-x_{i}<\delta$ we have $M_{i}-m_{i}<\epsilon/(b-a)$. So the difference between upper and lower Riemann sums is

 $\sum_{i}M_{i}(x_{i+1}-x_{i})-\sum_{i}m_{i}(x_{i+1}-x_{i})\leq\frac{\epsilon}{b% -a}\sum_{i}(x_{i+1}-x_{i})=\epsilon.$

This being true for every partition $P$ in $C(\delta)$ we conclude that $S^{*}(\delta)-S_{*}(\delta)<\epsilon$.

Title proof of continuous functions are Riemann integrable ProofOfContinuousFunctionsAreRiemannIntegrable 2013-03-22 13:45:34 2013-03-22 13:45:34 paolini (1187) paolini (1187) 7 paolini (1187) Proof msc 26A42