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

${\displaystyle {\int }_a^b}f(x)𝑑x=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle {\int }_a^b}{f}_n(x)𝑑x.$(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

${\displaystyle {\int }_a^b}\underset{n\to \mathrm{\infty }}{lim}{f}_n(x)dx=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle {\int }_a^b}{f}_n(x)𝑑x,$(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):

${\displaystyle {\int }_a^b}{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{f}_n(x)dx={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle {\int }_a^b}{f}_n(x)𝑑x.$(3)