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.