A uniform limit of continuous functions is continuous, and ∫fₙ → ∫f; without uniform convergence, the limit of the integrals can differ from the integral of the limit.
Why prefer uniform convergence? Because it lets you swap limits with the operations of calculus. The proofs are short, and the counterexamples, where the convergence is only pointwise, show that each hypothesis is needed. Integrals bring in the fundamental theorem.
🔑 Key idea
If $f_n \to f$ uniformly on $[a, b]$ and each $f_n$ is continuous, then $f$ is continuous, and $\int_a^b f_n \to \int_a^b f$. For derivatives the hypothesis must be on the derivatives: if $f_n' \to g$ uniformly (and $f_n(x_0)$ converges at one point), then $f_n \to f$ with $f' = g$.
💡 Why it works
Continuity (the "$\frac{\varepsilon}{3}$ argument"). Given $\varepsilon > 0$, choose $n$ with $\|f_n - f\| < \frac{\varepsilon}{3}$, then $\delta$ from the continuity of $f_n$ at $c$. For $|x - c| < \delta$: $$|f(x) - f(c)| \le |f(x) - f_n(x)| + |f_n(x) - f_n(c)| + |f_n(c) - f(c)| < \varepsilon.$$ Integrals. $$\left|\int_a^b f_n - \int_a^b f\right| \le \int_a^b |f_n - f| \le (b - a)\|f_n - f\| \to 0. \quad \blacksquare$$ Both proofs use that one $n$ controls every $x$ at once, which is exactly what pointwise convergence cannot provide.
Derivatives follow from integrals when the $f_n'$ are continuous: write $f_n(x) = f_n(x_0) + \int_{x_0}^{x} f_n'$. As $n \to \infty$, the right side tends to $f(x_0) + \int_{x_0}^{x} g$, by the integral result applied to $f_n' \to g$. So $f(x) = f(x_0) + \int_{x_0}^{x} g$, and the fundamental theorem gives $f' = g$.
$f_n(x) = nx(1 - x^{2})^{n}$: every $f_n$ has area close to $\frac{1}{2}$, but the bump slides toward $0$ and gets taller, and the pointwise limit is $0$.
Worked examples
✏️ Worked example: The limit of the integrals is not the integral of the limit
For $f_n(x) = nx(1 - x^{2})^{n}$ on $[0, 1]$, find $\lim \int_0^1 f_n$ and $\int_0^1 \lim f_n$.
Answer: Both are $\frac{1}{2}$, as the theorem guarantees.
✏️ Worked example: Derivatives need more
$f_n(x) = \frac{\sin(nx)}{n} \to 0$ uniformly on $\mathbb{R}$. Do the derivatives converge to $0$?
$f_n'(x) = \cos(nx)$, and $f_n'(0) = 1$ for every $n$.
Answer: No. Uniform convergence of $f_n$ does not control $f_n'$.
⚠️ Common mistake
Swapping without checking. Moving a limit inside an integral, or differentiating a series term by term, is a swap of two limits, and it can fail. Before writing $\lim \int f_n = \int \lim f_n$, check uniform convergence (or another theorem's hypotheses). A quick warning sign is a bump that gets taller and thinner: the sup norm does not go to $0$.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.