Uniform Convergence and Integration

Term-by-term integration under uniform convergence underlies the rigorous derivation of elementary function identities from power series (e.g. $\ln$, $\arctan$, elliptic integrals), and is the justification behind computing definite integrals of special functions via their series expansions in applied mathematics and physics. In Fourier analysis, uniform convergence of a Fourier series (guaranteed under suitable smoothness of the target function) licenses integrating the series against test functions term by term, a step used constantly in solving the heat and wave equations by separation of variables. In probability and statistics, uniform convergence results (e.g. Glivenko–Cantelli-type theorems) control when expectations (integrals against a measure) of empirical approximations converge to true expectations, echoing the same "swap limit and integral" logic. In numerical analysis and scientific computing, uniform error bounds on function approximations (splines, Chebyshev approximations related to the Stone-Weierstrass Theorem) are exactly what certify that integrating an approximation gives a controllably accurate estimate of the true integral. The same interchange principle, generalized via dominated and monotone convergence, becomes a cornerstone of measure-theoretic integration and underlies rigorous justifications throughout mathematical physics, cryptographic hardness proofs relying on analytic estimates, and the convergence theory of numerical PDE solvers.

:::quiz[If $f_n \to f$ pointwise (not necessarily uniformly) on $[a,b]$ and all $f_n$ are Riemann integrable with $f$ Riemann integrable, then $\int_a^b f_n \to \int_a^b f$ always holds.]

• True, pointwise convergence always suffices
• True, but only if $[a,b]$ is replaced by all of $\mathbb{R}$
• False — the moving tent function gives a counterexample even though $f$ is integrable ✓
• False, because pointwise limits of integrable functions are never integrable

:::quiz[For the moving tent functions $f_n$ with base $1/n$ and height $n$ on $[0,1]$, which statement is correct?]

• $f_n \to 0$ uniformly and $\int_0^1 f_n \to 0$
• $f_n \to 0$ pointwise only, and $\int_0^1 f_n = 1/2$ for every $n$, so the limit of integrals is $1/2 \ne 0$ ✓
• $f_n$ does not converge pointwise anywhere
• $\int_0^1 f_n \to \infty$

:::quiz[Why does the proof require the interval $[a,b]$ to be bounded?]

• It doesn't; the theorem is stated and proved identically for unbounded intervals
• Because the key estimate $\int_a^b |f_n - f| \le \|f_n-f\|_\infty \cdot (b-a)$ needs $(b-a)$ finite to force the right side to $0$ ✓
• Because Riemann integrability is undefined on unbounded intervals
• Because uniform convergence is impossible on unbounded intervals

:::faq[If $f_n \to f$ uniformly on $[a,b]$, is the rate of convergence of $\int_a^b f_n$ to $\int_a^b f$ controlled explicitly?] Yes, and this quantitative bound is arguably more useful than the qualitative existence-of-a-limit statement. The proof shows directly that

so any explicit bound on the sup-norm error immediately yields an explicit bound on the integral error, scaled by the length of the interval. This is exactly how the numerical error estimate was obtained in the worked example: a bound $\|f_n - f\|_\infty \le 28/(n+1)$ translated into an explicit bound $56/(n+1)$ on the integral's error.

Ask the tutor: Why doesn't uniform convergence of $f_n$ alone guarantee that $f_n'$ converges to $f'$, the way it guarantees the integrals converge?

Picture the graph of $f_n$ as a ribbon around the graph of $f$, of vertical half-width $\varepsilon_n = \|f_n - f\|_\infty$. Uniform convergence says this ribbon squeezes down to zero width everywhere at once as $n \to \infty$: the same $\varepsilon_n$ works for every $x \in [a,b]$ simultaneously. Since the ribbon has width $2\varepsilon_n$ and the interval has length $b-a$, the area trapped between the graphs of $f_n$ and $f$ is at most $2\varepsilon_n(b-a)$, which forces

This one-line estimate is the proof; everything else is bookkeeping about integrability. Pointwise convergence gives no such ribbon: at each fixed $x$, $f_n(x)$ eventually gets close to $f(x)$, but the rate can be wildly different at different points, and in particular a "spike" of fixed area can travel around the interval, shrinking in width as $n$ grows, hitting height $\to \infty$ or staying bounded, while still converging to $0$ pointwise almost everywhere. The area under the spike need not vanish. The two pictures below contrast a genuine uniform squeeze with a pointwise-only "moving spike."

Example 1 — a rational sequence with an asymmetric limit. Let

We claim $f_n \to f(x) = x$ uniformly on $[1,3]$, and verify the interchange of limit and integral directly. Pointwise/uniform limit. Write

Careful regrouping: $nx^2 - nx = nx(x-1)$, and there's a leftover $-x^2$, so

For $x \in [1,3]$: $|nx(x-1)| \le n \cdot 3 \cdot 2 = 6n$, $|\sin(nx)| \le 1$, and $x^2 \le 9$. Also $n + x \ge n+1$. So

This bound does not go to $0$ — indeed $\frac{6n+10}{n+1} \to 6$ — so this crude estimate is too lossy; let's redo it more carefully by separating the exactly-linear part. Actually reconsider: $f_n(x) = \dfrac{nx^2 + \sin(nx)}{n+x} = \dfrac{nx^2}{n+x} + \dfrac{\sin(nx)}{n+x}$. Compare $\dfrac{nx^2}{n+x}$ to $x$:

Hmm, let's instead directly bound $\left|\dfrac{nx^2}{n+x} - x\right| = \left|\dfrac{nx^2 - nx - x^2}{n+x}\right| = \left|\dfrac{x(nx-n-x)}{n+x}\right|$. For fixed $x\in[1,3]$, as $n \to \infty$ this is $\dfrac{x \cdot n(x-1) - x^2}{n+x} \to x(x-1)$ if $x \ne 1$ — not zero. So in fact $f_n \not\to x$; let's find the correct limit. Compute the honest limit: for fixed $x$,

dividing numerator and denominator by $n$. So the correct pointwise limit is $f(x) = x^2$, not $x$. This is a good reminder that limits of this type must be computed by dividing through by the dominant power of $n$, not by naive algebraic cancellation. We now redo the estimate against $f(x) = x^2$.

For $x \in [1,3]$: $|\sin(nx) - x^3| \le 1 + 27 = 28$, and $n + x \ge n+1$. So

This proves $f_n \to f(x)=x^2$ uniformly on $[1,3]$ (the bound is independent of $x$ and $\to 0$). Each $f_n$ is continuous on $[1,3]$ (a quotient of continuous functions with nonvanishing denominator $n+x \ge n+1>0$), hence Riemann integrable. By the theorem, $f(x) = x^2$ is Riemann integrable on $[1,3]$ (as we already knew, being continuous) and

We can sanity-check the rate: the theorem's proof shows $\left|\int_1^3 f_n - \frac{26}{3}\right| \le (b-a)\|f_n - f\|_\infty \le 2 \cdot \dfrac{28}{n+1} = \dfrac{56}{n+1} \to 0$, giving an explicit, verifiable error bound — e.g. at $n = 100$, the integral of $f_{100}$ is guaranteed to lie within $56/101 \approx 0.554$ of $26/3 \approx 8.667$.

Example 2 — term-by-term integration of a non-power series. Consider on $[0,1]$ the series

We integrate term by term to evaluate $\int_0^1 g$, and separately identify $g$ in closed form to check the answer. Uniform convergence via the M-test. For $x \in [0,1]$,

Since $\sum_{k=1}^\infty \frac{1}{k\,2^k}$ converges (compare to the geometric series $\sum 2^{-k}$, or note it's $\ln 2$ exactly, in any case a convergent series of positive constants — ratio test gives ratio $\to 1/2 <1$), the Weierstrass M-Test gives that $\sum f_k$ converges uniformly on $[0,1]$. Term-by-term integration. Each $f_k$ is continuous, hence Riemann integrable on $[0,1]$, and

By the Corollary,

Closed form check. Recognize $g(x) = \sum_{k=1}^\infty \frac{(-1)^{k+1}(x/2)^k}{k} = \ln\!\left(1 + \frac{x}{2}\right)$, using the standard Taylor series $\ln(1+u) = \sum_{k\ge1} \frac{(-1)^{k+1}u^k}{k}$ for $|u|\le 1$ (valid at $u=x/2\in[0,1/2]\subset(-1,1]$, where the series converges). So

Substitute $u = 1 + x/2$, $du = dx/2$, limits $u: 1 \to 3/2$:

Numerically, $\ln(3/2) \approx 0.405465$, so $3 \ln(3/2) - 1 \approx 1.2164 - 1 = 0.2164$. This matches (and gives a closed form for) the series $\sum_{k=1}^\infty \dfrac{(-1)^{k+1}}{k(k+1)2^k}$ obtained by direct term-by-term integration — a nontrivial numerical series identity that the theorem converts into a two-line integral computation.