Counterexample to $f_ng_n \not\to fg$ in measure I'm looking for a pair of sequences $f_n \to f$ in measure, $g_n \to g$ in measure, where $f_ng_n \not\!\to fg$ in measure.
I've tried a number of things with characteristic functions that move around with $n$, and nothing seems to pan out. I've also tried looking at non-Lebesgue measures, like the counting measure. At least, I realize that whatever measure one uses must not be finite, or else $f_ng_n \to fg$ is always true.
Any hints?

The classic counterexample I'm aware of is with the Lebesgue measure on $\mathbb{R}$, $f_n=g_n=x+\frac{1}{n}$, and $f=g=x$. Then obviously we have
$f_n\to f$ and $g_n\to g$ in measure, but 
$$f_ng_n=x^2+\tfrac{2}{n}x+\tfrac{1}{n^2}\not\to x^2=fg$$
in measure because for any $\epsilon>0$ and $n\in\mathbb{N}$.
 $$\{x\in \mathbb{R}\mid \tfrac{2}{n}x+\tfrac{1}{n^2}\geq\epsilon\}=[\tfrac{n^2\epsilon-1}{2n},\infty)$$
has infinite measure, so for any $\epsilon>0$,
$$\lim_{n\to\infty}\mu(\{x\in\mathbb{R}\mid |f_ng_n(x)-fg(x)|\geq\epsilon\})=\lim_{n\to\infty}\infty=\infty$$
I'm sorry I don't know how to give a good hint for this one; I didn't see it on my own either, someone pointed it out to me. In retrospect the intuition is that we set things up so that the gap between the $f_n$'s and $f$, and $g_n$'s and $g$, when multiplied, becomes "amplified" (since $\frac{2}{n}x+\frac{1}{n^2}$ depends on $x$, whereas $\frac{1}{n}$ doesn't).
