Failure of exchange of integral and limit of function $f_n(x)=ne^{-nx}$

Question

Suppose you have a sequence of functions indexed by $n\in\mathbb{N}$, for $x\in[0,1]$, defined by $f_n(x)=ne^{-nx}$. Is it true that $$\lim_{n\to\infty}\int_0^1f_n(x)dx= \int_0^1\lim_{n\to\infty}f_n(x)dx.$$

The answer is no. I post the question as a teaching example for people unsure of what could go wrong, especially because $f_n(x)$ is an expression that is fairly common in econometrics/economics and I've seen people just recklessly exchange limit and integral.

Answer 1

The answer is that it cannot be exchanged because $f_n$ is neither monotonic (decreasing or increasing) nor there exists a dominant function such that $|f_n(x)|\leq g(x)$ for all $n$ and $x$ ($\{f_n\}_{n\in\mathbb{N}}$ is not uniformly integrable, either). To see it:

1. Monotonicity: One could be tempted to say that, for each $x\in[0,1)$ ($x=1$ is of measure zero and thus irrelevant), there is $n_x$ such that if $n\geq n_x$, $0\leq f_n(x)<f_{n+1}(x)$, and then try to apply this version of Monotone Convergence Theorem on the tail of the sequence. However, $n_x$ grows unboundedly in $x$, namely, as $x\to 1$, $n_x\to\infty$, which means that there is no $n^*$ such that if $n\geq n^*$, $0\leq f_n(x)<f_{n+1}(x)$ for every $x\in[0,1)$. So, no hope of using MCT.
2. Dominance: no hope of finding $g$ such that $|f_n(x)|\leq g(x)$ for all $n$ and $x$, because $f_n(0)\to\infty$ as $n\to \infty$. So no hope of using DCT either.

What happens is that we have $\int_0^1f_n(x)dx=1-e^{-n}$ and $\lim_{n\to\infty}f_n(x)=0$, so

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