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?

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.
