Uniform Convergence and Integration

In analysis, you often encounter sequences of functions where each function $f_n$ can be integrated. A fundamental question arises: if $f_n$ converges to a limit function $f$, does the integral of $f_n$ converge to the integral of $f$? In other words, when can you legitimately swap the limit and integral operations? The answer hinges critically on the type of convergence. While pointwise convergence is intuitively appealing, it is treacherously weak for this purpose. The powerful and sufficient condition is uniform convergence.

The Fundamental Problem: Swapping Limits and Integrals

Consider a sequence of functions $f_n:[a,b]\to \mathbb{R}$ defined on a closed, bounded interval. If this sequence has a pointwise limit $f(x)={\lim }_{n\to \infty }{f}_n(x)$, it is tempting to conclude that $$\underset{n\to \infty }{\lim }{\int }_a^b{f}_n(x)dx={\int }_a^b\underset{n\to \infty }{\lim }{f}_n(x)dx={\int }_a^{b}f(x)dx.$$ This interchange is a common maneuver in calculus, but it is not justified by pointwise convergence alone. The core issue is that pointwise convergence controls the function's value at each individual point $x$ as $n$ grows, but it says nothing about the rate of convergence across different points. A small area of "bad behavior" in the function can persist and travel around the interval, potentially preventing the integral from stabilizing to the expected limit. Uniform convergence, by demanding that all points converge at a similar rate, eliminates this possibility.

Pinpointing the Difference: Pointwise vs. Uniform Convergence

It is essential to distinguish these two modes of convergence clearly.

A sequence of functions $\{f_n\}$ converges pointwise to $f$ on a set $E$ if, for each fixed $x\in E$ and for every $ϵ>0$, there exists an integer $N$ such that $|f_n(x)-f(x)|<ϵ$ for all $n\ge N$. Crucially, $N$ can depend on both $ϵ$ and the point $x$.

In contrast, $\{f_n\}$ converges uniformly to $f$ on $E$ if, for every $ϵ>0$, there exists an integer $N$ such that $|f_n(x)-f(x)|<ϵ$ for all $n\ge N$ and for all $x\in E$ simultaneously. Here, $N$ depends only on $ϵ$, not on $x$. You can visualize uniform convergence as the entire graph of $f_n$ eventually lying within an $ϵ$-band around the graph of $f$.

A classic example illustrates the distinction. Define $f_n(x)=x^n$ on $[0,1]$. This sequence converges pointwise to the function $f(x)$ which is $0$ for $x\in [0,1)$ and $1$ at $x=1$. However, the convergence is not uniform. For any $n$, there are points near $1$ where $f_n(x)$ is close to $1$, far from the limit value $0$. No single $N$ works for all $x$ if $ϵ<1$.

The Main Theorem: Uniform Convergence Preserves Integrability and Allows Interchange

The central positive result provides clear, sufficient conditions for interchanging the limit and the integral.

Theorem: Let $\{f_n\}$ be a sequence of Riemann-integrable functions on a closed, bounded interval $[a,b]$. If $f_n$ converges uniformly to $f$ on $[a,b]$, then the limit function $f$ is Riemann-integrable on $[a,b]$, and $$\underset{n\to \infty }{\lim }{\int }_a^b{f}_n(x)dx={\int }_a^{b}f(x)dx.$$ If we further assume each $f_n$ is continuous, the proof is more straightforward and highlights the core ideas.

Proof Sketch (for continuous $f_n$):

1. Continuity of the Limit: A uniform limit of continuous functions is continuous. Since each $f_n$ is continuous on $[a,b]$ and $f_n\to f$ uniformly, $f$ is continuous. Any continuous function on a closed interval is Riemann-integrable.
2. Interchanging the Limit: We want to show ${\lim }_{n\to \infty }{\int }_a^b{f}_n={\int }_a^{b}f$. Consider the difference:

$$\left|{\int }_a^b{f}_n(x)dx-{\int }_a^{b}f(x)dx\right|=\left|{\int }_a^b(f_n(x)-f(x))dx\right|.$$ By the properties of the integral, $$\left|{\int }_a^b(f_n(x)-f(x))dx\right|\le {\int }_a^b|f_n(x)-f(x)|dx.$$

1. Applying Uniform Convergence: Because the convergence is uniform, for any $ϵ>0$, we can find an $N$ such that for all $n\ge N$ and for all $x\in [a,b]$, $|f_n(x)-f(x)|<ϵ/(b-a)$.
2. Final Estimate: Plugging this uniform bound into the integral gives:

$${\int }_a^b|f_n(x)-f(x)|dx<{\int }_a^b\frac{ϵ}{b-a}dx=ϵ.$$ Therefore, for $n\ge N$, $\left|{\int }_a^b{f}_n-{\int }_a^{b}f\right|<ϵ$, which is precisely the definition that ${\int }_a^b{f}_n$ converges to ${\int }_a^{b}f$.

The power of uniform convergence is evident in step 3: it allows us to bound the integrand $|f_n-f|$ by a single, small number everywhere, making the integral of the difference arbitrarily small.

Instructive Counterexamples: The Failure of Pointwise Convergence

Pointwise convergence alone cannot guarantee the interchange. These counterexamples are not pathologies; they are essential learning tools.

Counterexample 1: The Shrinking Triangle (Limit of Integrals ≠ Integral of Limit) Define $f_n$ on $[0,1]$ as a moving, shrinking triangle: $$f_n(x)=\{\begin{array}{ll}{n}^{2}x& 0\le x\le 1/n,\\ 2n-n^{2}x& 1/n<x\le 2/n,\\ 0& x>2/n.\end{array}$$ The graph is a triangle with base $[0,2/n]$ and height $n$. For any fixed $x>0$, once $n>2/x$, we have $f_n(x)=0$. At $x=0$, $f_n(0)=0$ always. So $f_n\to 0$ pointwise. However, the area under each triangle (the integral) is constant: ${\int }_0^1{f}_n(x)dx=1$ (area of a triangle: $\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\cdot \frac{2}{n}\cdot n=1$). Thus, $$\underset{n\to \infty }{\lim }{\int }_0^1{f}_n(x)dx=1\text{but}{\int }_0^1\left(\underset{n\to \infty }{\lim }{f}_n(x)\right)dx={\int }_0^{1}0dx=0.$$ The pointwise limit of the integrals is not the integral of the pointwise limit. The convergence is not uniform because the "bump" of area becomes arbitrarily tall to preserve area while narrowing, violating the requirement that the entire graph stay within a thin band around the limit function.

Counterexample 2: Loss of Integrability Consider a sequence on $[0,1]$ where $f_n(x)$ is defined to be $1$ at $n$ distinct rational points and $0$ elsewhere, arranged so that as $n$ increases, you "mark" more and more rationals. Each $f_n$ is integrable (it has only finitely many discontinuities). The sequence converges pointwise to the Dirichlet function $f(x)$, which is $1$ on rationals and $0$ on irrationals. The pointwise limit $f$ is not Riemann-integrable. Here, pointwise convergence fails to preserve the fundamental property of integrability itself.

Common Pitfalls

1. Assuming Pointwise Convergence is Sufficient: This is the most critical error. Always suspect that interchanging a limit and an integral under only pointwise convergence is invalid. Your first instinct should be to test for uniform convergence.
2. Misjudging Uniform Convergence with Graphical Intuition: The "shrinking triangle" example shows that a sequence can appear to visually converge to zero (pointwise) while failing uniformly because a small set of points stubbornly maintains large function values. A rigorous $ϵ$-$N$ argument or the use of the supremum norm $||f_n-f|{|}_{\infty }={\sup }_{x\in [a,b]}|f_n(x)-f(x)|$ is necessary.
3. Overlooking the Conditions on the Domain: The standard theorem applies to a closed and bounded interval $[a,b]$. For unbounded intervals (e.g., $[0,\infty )$) or improper integrals, uniform convergence alone is insufficient; you need additional conditions, often involving domination by an integrable function (as in the Lebesgue Dominated Convergence Theorem).
4. Applying the Theorem to Derivatives: A common related mistake is to assume uniform convergence of functions implies convergence of their derivatives. This is false. The interchange ${\lim }_{n\to \infty }{f}_n^{\prime }=({\lim }_{n\to \infty }{f}_n{)}^{\prime }$ requires uniform convergence of the derivatives $f_n^{\prime }$, not of the functions themselves.

Summary

• The core problem is determining when the limit of integrals equals the integral of the limit: $\lim \int f_n=\int \lim f_n$.
• Pointwise convergence is insufficient. It does not control the behavior of the sequence across all points simultaneously, allowing "bad behavior" to persist in a way that can disrupt the integral.
• Uniform convergence is a key sufficient condition. For a sequence of Riemann-integrable (or continuous) functions on a closed interval $[a,b]$, uniform convergence to $f$ guarantees that $f$ is integrable and that the limit and integral can be interchanged.
• The proof hinges on bounding the integral of the difference $|f_n-f|$. Uniform convergence provides a constant, small bound for this difference over the entire interval.
• Counterexamples are essential. They demonstrate how pointwise convergence can fail to preserve integrability or the value of the integral, underscoring the necessity of the stronger uniform condition.
• Always check the domain. The standard theorem is specific to bounded intervals. Extending these ideas to more general domains requires more advanced tools and theorems.