Uniform convergence on compact sets allows switching the limit and the integral.

Question

Why does uniform convergence on compact sets allows switching the limit and the integral?

Answer 1

In integral; if $f_n\to f$ uniformly on the compact $K$, then $|\int_Kf_n-\int_Kf|=|\int_K(f_n-f)|\leq \|f_n-f\|_{K}\int_K1\to 0$, hence $\int_Kf_n\to \int_Kf$.

Answer 2

To expand on Hamou's answer:

Suppose that a sequence $\{f_n\}$ of functions on a compact set is such that $f_n \to f$ uniformly. We can then state that for any $\epsilon>0$, there is an $N$ such that $\sup_{K}|f(x) - f_n(x)| < \epsilon/(\int_K 1\,dx)$ whenever $n>N$. It follows that for $n>N$, we have $$ \left|\int_K f(x)\,dx - \int_K f_n(x)\,dx \right| = \left|\int_K (f(x)- f_n(x))\,dx \right| \\ \leq \int_K |f(x)- f_n(x)|\,dx\\ \leq \int_K \frac{\epsilon}{\int_K 1\,dx}\,dx = \epsilon $$ Thus, we conclude that $$ \int_K f_n(x)\,dx \to \int_K f(x)\,dx $$ as $n \to \infty$.

Uniform convergence allows us to bound $f_n - f$, and compactness allows us to compute $\int_K 1\,dx$.