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?
 A: 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).
A: You can also do it with counting measure on $\mathbb{N}$ and functions from $\mathbb{N}$ to $\mathbb{R}$. Let 
$$\begin{align*}
x_n&:\mathbb{N}\to\mathbb{R}:k\mapsto 2^{-n}\;,\\
x&:\mathbb{N}\to\mathbb{R}:k\mapsto 0\;,\text{ and}\\
y,y_n&:\mathbb{N}\to\mathbb{R}:k\mapsto 2^k\;.
\end{align*}$$
For $m,n\in\mathbb{N}$ we have 
$$\begin{align*}
\mu\big(\{k\in\mathbb{N}:|x_n(k)-x(k)|\ge 2^{-m}\}\big)&=\mu\big(\{k\in\mathbb{N}:2^{-n}\ge 2^{-m}\}\big)\\
&=\begin{cases}
\infty,&n\le m\\
0,&n>m\;,
\end{cases}
\end{align*}$$
so $\langle x_n:n\in\mathbb{N}\rangle\to x$ in measure, and obviously $\langle y_n:n\in\mathbb{N}\rangle\to y$ in measure. But 
$$x_ny_n:\mathbb{N}\to\mathbb{R}:k\mapsto 2^{-n}2^k=2^{k-n},$$
so for $m,n\in\mathbb{N}$ we have
$$\begin{align*}
\mu\big(\{k\in\mathbb{N}:|(x_ny_n)(k)-(xy)(k)|\ge 2^{-m}\}\big)&=\mu\big(\{k\in\mathbb{N}:2^{k-n}\ge 2^{-m}\}\big)\\
&=\mu\big(\{k\in\mathbb{N}:k\ge n-m\}\big)\\
&=\infty\;,
\end{align*}$$
and $\langle x_ny_n:n\in\mathbb{N}\rangle$ does not converge to $xy$ in measure.
