On the set $\Omega = [0,1)^2$ define \[ \theta(x,y) = \begin{cases}(2x,y/2), & \text{if }x < 1/2\\ (2-2x,1-y/2), & x \ge 1/2\end{cases} \] Then the Lebesgue measure is invariant under $\theta$. The transformation $\theta$ is known as Baker's Transformation.

It suffices to prove that $\lambda([\theta \in A]) = \lambda(A)$ for all sets of the form $A = [0,a)\times [0,b)$ ($0\le a\le 1$, $0\le b\le 1$), where $\lambda$ is the two-dimensional Lebesgue measure. Note that $\theta$ maps $[0,\frac12)\times [0,1)$ onto $[0,1)\times [0,\frac12)$ and $[\frac12, 1)\times [0,1)$ onto $(0,1]\times (\frac12, 1]$. Since $A$ can be written as the union of two sets of the form $B = [0,a)\times [0,b)$ and $C = [0,a)\times [\frac12, \frac12 + c)$ for some $0\le a\le 1$ and $0\le b\le \frac12$, $0\le c\le \frac12$, it suffices to consider these sets.

If $B = [0,a)\times [0,b)$, then $[\theta\in B] = [0,\frac{a}{2})\times [0,2b)$, so $\lambda([\theta\in B]) = \frac{a}{2}\cdot 2b = ab = \lambda(B)$.

If $C = [0,a)\times [\frac12, \frac12 + c)$, then $[\theta\in C] = (1-\frac{a}{2}, 1)\times (1-2c,1)$ (note that $\theta(1-\frac{a}{2},1-2c) = (a,\frac12 + c)$ and $\theta(1,1) = (0,\frac12)$ if we allow $(1,1)$ as an argument of $\theta$) and thus $\lambda([\theta\in C]) = \frac{a}{2}\cdot 2c = ac = \lambda(C)$.