A counterexample to "the integral of the limit is the limit of the integral"

[pasmith]

I was going to append this to @chwala's thread here, but thought it deserved a new thread.

For $n \geq 1$, define $$f_n : [0,1] \to \mathbb{R} : x \mapsto \begin{cases} 1 & x = 0, \\ n(1 - nx) & x \in (0, \tfrac 1n], \\ 0 & x \in (\tfrac 1n, 1]. \end{cases}$$ (Note that $f_n$ is discontinuous at 0 for $n \geq 2$.)

This sequence has the pointwise limit $$ f: [0,1] \to \mathbb{R} : x \mapsto \begin{cases} 1 & x = 0, \\ 0 & x \in (0,1] \end{cases}$$ since if $x > 0$ then there exists $N \in \mathbb{N}$ such that $x > \frac 1N$ and thus $f_n(x) = 0$ for all $n > N$.

The convergence is non-uniform since $\sup |f_n - f| = n$.

This example is constructed so that$$\lim_{n \to \infty} \int_0^1 f_n(x)\,dx = \frac12 \neq 0 = \int_0^1 \lim_{n \to \infty} f_n(x)\,dx.$$

EDIT: We can instead set $f_n(0) = 0$, so that the pointwise limit is simply the zero function.

 

[pasmith]

A generalization is to fix $g: [0,1] \to \mathbb{R}$ and let $$f_n : [0,1] \to \mathbb{R} : x \mapsto \begin{cases} ng(nx) & x < \frac 1n \\ 0 & x \geq \frac 1n \end{cases}.$$ Then $$\int_0^1 f_n(x)\,dx = \int_0^1 g(x)\,dx.$$ The pointwise limit is again zero for $x > 0$, but at zero we have $f_n(0) = ng(0)$ which diverges unless $g(0) = 0$. The pointwise limit is then everywhere zero. Convergence is still non-uniform, since $\sup |f_n - f| = n \sup |g|$.

A suitable $g$ is $$g : x \mapsto \begin{cases} 2x & x < \frac 12, \\ 2(1 - x) & x \geq \frac12. \end{cases}$$