Which counterexamples to keep in mind when studying (Lebesgue) integral convergence theorems

Question

When studying the classical theorems regarding the convergence of a sequence of integrals, to the integral of the limit function, e.g.

• Beppo Levi

• Dominated convergence

• Vitalli's theorem

which counterexamples should one keep in mind?

Answer 1

Here are my favorite counterexamples:

• On $\mathbb{R}$, $f_n = 1_{[n,n+1]}$. This sequence converges to 0 pointwise but the integrals converge to 1.

• On $\mathbb{R}$, $f_n = \frac{1}{n} 1_{[0,n]}$. This sequence converges to $0$ uniformly but the integrals converge to $1.$

• On $[0,1]$, $f_n = n 1_{(0, 1/n)}$. This sequence converges to $0$ pointwise but the integrals converge to $1.$

Any of these examples can be modified to give a sequence where the integrals converge to infinity instead. You can turn them into a Beppo-Levi counterexample by considering $g_n = f_n - f_{n-1}$, so that $\sum g_n = \lim f_n$ as a telescoping sum.

Also useful to keep in mind:

• On $\mathbb{R}$, $f_n = 1_{[n, n+1/n]}$. This sequence converges to 0 pointwise and in $L^1$ but has no dominating function.

• On $\mathbb{N}$, the sequence $f_n = 1_{\{n\}}$. This sequence is $L^1$ bounded, convergent pointwise, and uniformly absolutely continuous (try $\delta = 1/2$), but not uniformly integrable (depending on your exact definitions).

• The typewriter sequence which converges to 0 in $L^1$ but diverges pointwise everywhere.

• The sequence $f_n(x) = e^{2 \pi i n x}$ on $[0,1]$, which diverges pointwise and in $L^p$ but converges weakly in $L^2$ to $0$.

Answer 2

$$ f_n(x) = \begin{cases} n & \text{if } 0<x<1/n, \\ 0 & \text{if } x\le 0 \text{ or } x\ge 1/n. \end{cases} $$ Then you have $$ \lim_{n\to\infty} \int_{\mathbb R} f_n(x)\,dx = 1 \ne 0 = \int_{\mathbb R} \lim_{n\to\infty} f_n(x) \,dx. $$ The reason this doesn't contradict the dominated convergence theorem is of course that $$ \int_{\mathbb R} \sup_n f_n(x) \,dx = +\infty, $$ so there is no dominating function whose integral is finite.