# integral of limit function

Theorem. If a sequence ${f}_{1},{f}_{2},\mathrm{\dots}$ of real functions, continuous^{} on the interval $[a,b]$, converges uniformly on this interval to the limit function^{} $f$, then

${\int}_{a}^{b}}f(x)\mathit{d}x=\underset{n\to \mathrm{\infty}}{lim}{\displaystyle {\int}_{a}^{b}}{f}_{n}(x)\mathit{d}x.$ | (1) |

*Proof.* Let $\epsilon >0$. The uniform continuity implies the existence of a positive integer
${n}_{\epsilon}$ such that

$$ |

The function $f$ is continuous (see http://planetmath.org/node/7191this) and thus Riemann integrable^{} (http://planetmath.org/RiemannIntegral) (see http://planetmath.org/node/4461this) on the interval. Utilising the estimation theorem of integral, we obtain

$$ |

as soon as $n>{n}_{\epsilon}$. Consequently, (1) is true.

Remark 1. The equation (1) may be written in the form

${\int}_{a}^{b}}\underset{n\to \mathrm{\infty}}{lim}{f}_{n}(x)dx=\underset{n\to \mathrm{\infty}}{lim}{\displaystyle {\int}_{a}^{b}}{f}_{n}(x)\mathit{d}x,$ | (2) |

i.e. under the assumptions of the theorem, the integration and the limit process can be interchanged.

Remark 2. Considering the partial sums of a series ${\sum}_{n=1}^{\mathrm{\infty}}{f}_{n}(x)$ with continuous terms and converging uniformly on $[a,b]$, one gets from the theorem the result analogous to (2):

${\int}_{a}^{b}}{\displaystyle \sum _{n=1}^{\mathrm{\infty}}}{f}_{n}(x)dx={\displaystyle \sum _{n=1}^{\mathrm{\infty}}}{\displaystyle {\int}_{a}^{b}}{f}_{n}(x)\mathit{d}x.$ | (3) |

Title | integral of limit function |
---|---|

Canonical name | IntegralOfLimitFunction |

Date of creation | 2013-03-22 19:01:41 |

Last modified on | 2013-03-22 19:01:41 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 10 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 26A15 |

Classification | msc 40A30 |

Related topic | TermwiseDifferentiation |