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

(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 this) and thus Riemann integrable (see this) 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

(2)

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):

(3)