uniform convergence preserves integrals

Here is the precise statement that lets you slide a limit symbol inside a Riemann integral: if functions converge uniformly on a closed bounded interval, then the limit of their integrals equals the integral of their limit. The area under f_n marches steadily to the area under f, because uniform convergence stops any area from leaking out at a stray point.

Theorem: let f_n be Riemann integrable on [a, b] and suppose f_n -> f uniformly on [a, b]. Then f is Riemann integrable on [a, b] and lim_{n->infinity} integral over [a,b] of f_n = integral over [a,b] of f = integral over [a,b] of (lim_n f_n). The estimate behind it is short: |integral of f_n - integral of f| <= integral over [a,b] of |f_n - f| <= (b - a) sup over [a,b] of |f_n - f| -> 0.

The two caveats are worth stating plainly. First, uniformity is sufficient but not necessary — many sequences with the integral passing to the limit are not uniformly convergent, and the Lebesgue convergence theorems (dominated, monotone) handle far more cases. Second, the interval must be finite: on an unbounded interval the factor (b - a) is infinite and uniform convergence alone does NOT force the integrals to converge, as mass can escape to infinity.