---
tag: 0BWGW68B66
kind: statement
role: theorem
name: Uniform limits commute with integration
msc: [40A30, 26A42]
uses: [08FPABDQ4C, 0QM4TXT3DR, 0MBWQKM8QR, 017DH8WXTE]
sources:
  - url: https://en.wikipedia.org/wiki/Uniform_convergence
    locator: To integrability
---

## 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|08FPABDQ4C]] 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|017DH8WXTE]].
By [[monotonicity|0MBWQKM8QR]], $\big| \int_a^b f_n - \int_a^b f \big| \le \varepsilon_n (b - a) \to 0$.
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.