# 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 $\u03f5>0$ there exists $\delta >0$ such that $$.

So let $\u03f5>0$ be fixed.

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

$$ |

Let now $P$ be any partition^{} of $[a,b]$ in $C(\delta )$ i.e. a partition $\{{x}_{0}=a,{x}_{1},\mathrm{\dots},{x}_{N}=b\}$ such that $$. In any small interval $[{x}_{i},{x}_{i+1}]$ the function $f$ (being continuous^{}) has a maximum ${M}_{i}$ and minimum ${m}_{i}$. Since $f$ is uniformly continuous and $$ we have $$. So the difference^{} between upper and lower Riemann sums is

$$\sum _{i}{M}_{i}({x}_{i+1}-{x}_{i})-\sum _{i}{m}_{i}({x}_{i+1}-{x}_{i})\le \frac{\u03f5}{b-a}\sum _{i}({x}_{i+1}-{x}_{i})=\u03f5.$$ |

This being true for every partition $P$ in $C(\delta )$ we conclude that $$.

Title | proof of continuous functions are Riemann integrable |
---|---|

Canonical name | ProofOfContinuousFunctionsAreRiemannIntegrable |

Date of creation | 2013-03-22 13:45:34 |

Last modified on | 2013-03-22 13:45:34 |

Owner | paolini (1187) |

Last modified by | paolini (1187) |

Numerical id | 7 |

Author | paolini (1187) |

Entry type | Proof |

Classification | msc 26A42 |