Uniform limits commute with integration

Statement

Let $a < b$, and let $f_n \colon [a, b] \to \mathbb{R}$ be Riemann integrable for each $n$. If $f_n \to f$ uniformly on $[a, b]$, then $f$ is Riemann integrable on $[a, b]$, and $$\lim_{n \to \infty} \int_a^b f_n(x) \, dx = \int_a^b f(x) \, dx.$$

Remarks

Let $\varepsilon_n = \sup |f_n - f|$. Then $f_n - \varepsilon_n \le f \le f_n + \varepsilon_n$. So the upper and lower sums of $f$ differ by at most $2 \varepsilon_n (b - a)$ more than those of $f_n$. This gives integrability, by Riemann's criterion. By monotonicity, $\big| \int_a^b f_n - \int_a^b f \big| \le \varepsilon_n (b - a) \to 0$. Pointwise convergence is not enough. On $[0, 1]$, the functions $f_n(x) = n^2 x (1 - x^2)^n$ tend to $0$ pointwise, but $\int_0^1 f_n = \frac{n^2}{2n + 2} \to \infty$.

Explanations

L1 human

If $f_n$ is uniformly close to $f$, the areas under their graphs are close too. So we may take the limit inside the integral.