# integral of limit function

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

 $\displaystyle\int_{a}^{b}\!f(x)\,dx\;=\;\lim_{n\to\infty}\int_{a}^{b}\!f_{n}(x% )\,dx.$ (1)

Proof.  Let  $\varepsilon>0$.  The uniform continuity implies the existence of a positive integer $n_{\varepsilon}$ such that

 $|f_{n}(x)\!-\!f(x)|\;<\;\frac{\varepsilon}{b\!-\!a}\quad\forall x\in[a,\,b]% \qquad\mbox{when}\;\;n\,>\,n_{\varepsilon}.$

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

 $\left|\int_{a}^{b}\!f_{n}(x)\,dx\!-\!\int_{a}^{b}\!f(x)\,dx\right|\,=\,\left|% \int_{a}^{b}\!(f_{n}(x)\!-\!f(x))\,dx\right|\,\leqq\,\int_{a}^{b}\!|f_{n}(x)\!% -\!f(x)|\,dx\,<\,\frac{\varepsilon}{b\!-\!a}(b\!-\!a)\,=\,\varepsilon$

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

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

 $\displaystyle\int_{a}^{b}\!\lim_{n\to\infty}f_{n}(x)\,dx\;=\;\lim_{n\to\infty}% \int_{a}^{b}\!f_{n}(x)\,dx,$ (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 $\sum_{n=1}^{\infty}f_{n}(x)$ with continuous terms and converging uniformly on  $[a,\,b]$,  one gets from the theorem the result analogous to (2):

 $\displaystyle\int_{a}^{b}\!\sum_{n=1}^{\infty}f_{n}(x)\,dx\;=\;\sum_{n=1}^{% \infty}\int_{a}^{b}\!f_{n}(x)\,dx.$ (3)
Title integral of limit function IntegralOfLimitFunction 2013-03-22 19:01:41 2013-03-22 19:01:41 pahio (2872) pahio (2872) 10 pahio (2872) Theorem msc 26A15 msc 40A30 TermwiseDifferentiation