Uniform limit of integrable functions is integrable

Let $f_n:[a,b]\to\mathbb{R}$ be Riemann integrable for each $n$ and suppose $f_n\to f$ uniformly on $[a,b]$.

Proposition:

• The limit function $f$ is Riemann integrable on $[a,b]$.
• The integrals converge to the integral of the limit: $\lim_{n\to\infty}\int_a^b f_n(x)\,dx=\int_a^b f(x)\,dx.$

This is the standard mechanism for proving integrability of limits and justifies passing limits through integrals under uniform convergence.

Proof sketch: Uniform convergence gives $\|f-f_n\|_\infty\to 0$. Then $\left|\int_a^b f-\int_a^b f_n\right| \le \int_a^b |f-f_n| \le (b-a)\|f-f_n\|_\infty\to 0,$ which implies the integral identity and forces $f$ to be integrable because the integrals of $f_n$ are finite and the upper /lower sums of $f$ are controlled by those of $f_n$ plus a uniform error.