Swapping a Limit and an Integral: A Counterexample

A common mistake in analysis is to assume that pointwise convergence allows you to interchange a limit and an integral. In general, it does not.

The basic comparison is

$$\lim_{n\to\infty}\int f_n\,d\mu \qquad\text{versus}\qquad \int\left(\lim_{n\to\infty} f_n\right)\,d\mu.$$

These expressions are equal under useful hypotheses such as the dominated convergence theorem or the monotone convergence theorem. Without such a condition, they can be different.

A counterexample on [0,1]

Define

$$f_n(x)= \begin{cases} n, & 0<x<\frac{1}{n},\\ 0, & \text{otherwise}, \end{cases} \qquad 0\le x\le 1.$$

Each $f_n$ is a narrow spike with height $n$ and width $1/n$. Its area remains constant.

Step 1: Find the pointwise limit

Fix any $x\in[0,1]$.

If $x=0$, then $f_n(0)=0$ for every $n$. If $x>0$, choose $N>1/x$. For every $n\ge N$, we have $1/n<x$, so $x$ is outside the interval $(0,1/n)$ and therefore $f_n(x)=0$.

Thus, for every fixed $x$,

$$\lim_{n\to\infty}f_n(x)=0.$$

Consequently,

$$\int_0^1\lim_{n\to\infty}f_n(x)\,dx =\int_0^1 0\,dx =0.$$

Step 2: Integrate before taking the limit

For each $n$,

$$\int_0^1 f_n(x)\,dx =\int_0^{1/n}n\,dx =n\left(\frac{1}{n}\right)=1.$$

Therefore,

$$\lim_{n\to\infty}\int_0^1f_n(x)\,dx =\lim_{n\to\infty}1 =1.$$

The two results are different:

$$\boxed{ \lim_{n\to\infty}\int_0^1 f_n(x)\,dx=1 \ne 0=\int_0^1\lim_{n\to\infty}f_n(x)\,dx }.$$

This is a complete example showing that pointwise convergence alone is insufficient.

Why the example works

The mass of $f_n$ does not disappear; it moves into a smaller and smaller neighborhood of $0$. At any fixed point, the spike eventually vanishes, which produces the pointwise limit zero. However, the total area under the spike is always one.

The sequence is not dominated by a single integrable function. Indeed, if an integrable function $g$ satisfied $f_n(x)\le g(x)$ almost everywhere for every $n$, then $g$ would need to accommodate arbitrarily high values near $0$. The concentration of the mass is precisely what prevents the usual dominated-convergence argument.

The convergence is pointwise everywhere, and even almost everywhere, but it is not uniform: the maximum of $f_n$ is $n$, so the functions do not become uniformly small.

When the interchange is valid

Dominated convergence theorem

Suppose the $f_n$ are measurable, $f_n\to f$ almost everywhere and there is an integrable function $g$ such that

$$|f_n(x)|\le g(x)$$

for every $n$ and almost every $x$. Then

$$\lim_{n\to\infty}\int f_n\,d\mu =\int f\,d\mu.$$

For example, let $f_n(x)=x^n$ on $[0,1]$. For every $x<1$, $x^n\to0$, and the exceptional point $x=1$ does not affect the integral. Also, $0\le x^n\le1$, and the constant function $1$ is integrable on $[0,1]$. Hence dominated convergence gives

$$\lim_{n\to\infty}\int_0^1x^n\,dx =\int_0^1\lim_{n\to\infty}x^n\,dx =0.$$

Directly, the left-hand side is

$$\lim_{n\to\infty}\frac{1}{n+1}=0.$$

Monotone convergence theorem

If $f_n$ are measurable, $0\le f_n\le f_{n+1}$ and $f_n\to f$ pointwise, then

$$\lim_{n\to\infty}\int f_n\,d\mu =\int f\,d\mu,$$

even if the integral is infinite. The nonnegative, increasing assumption is essential; the spike sequence above is nonnegative but not increasing in $n$ at each point.

Uniform convergence on a finite interval

For Riemann integrals, if $f_n\to f$ uniformly on $[a,b]$ and each function is integrable, then

$$\lim_{n\to\infty}\int_a^b f_n(x)\,dx =\int_a^b f(x)\,dx.$$

A direct estimate explains this:

$$\left|\int_a^b f_n(x)\,dx-\int_a^b f(x)\,dx\right| \le (b-a)\sup_{x\in[a,b]}|f_n(x)-f(x)|,$$

and the right-hand side tends to zero.

Common mistakes

• Confusing pointwise and uniform convergence: Pointwise convergence controls one fixed point at a time. Uniform convergence controls all points simultaneously.
• Checking only that each integral exists: The existence of every $\int f_n$ does not guarantee that the limiting operation is interchangeable.
• Using dominated convergence without a dominator: The bound must come from one fixed integrable function, not a function $g_n$ that changes with $n$.
• Ignoring the measure space: A constant bound is integrable on a finite-measure interval, but generally not on an infinite-measure space.
• Forgetting cancellation: For signed functions, large positive and negative values can cancel. The dominated convergence theorem controls the absolute values and avoids this problem.

The safe procedure is to identify the convergence mode and verify a theorem’s hypotheses before swapping the limit and the integral.