On the set $\O=[0,1)^2$ define $$ \theta(x,y)=\left\{\begin{array}{cl} (2x,y/2)&\mbox{if $x < 1/2$}\\ (2-2x,1-y/2)&\mbox{$x\geq1/2$} \end{array}\right. $$ Then the Lebesgue measure is invariant under $\theta$. The transformation $\theta$ is known as Baker's Transformation
\begin{eqnarray*} \l(\{(x,y):\theta(x,y)\in[0,s)\times[0,t)\}) &=&\l(x\leq s/2,y\leq 2t\wedge1)+\l(x\geq1-s/2,y\geq2-2t)\\ &=&\tfrac12s(2t\wedge1+(1-(2-2t)\wedge1) \end{eqnarray*} Both for $t\leq1/2$ and for $t > 1/2$ this gives: $st$