Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 4 / 25J / b

As printed, ${\theta }_1(x)=1+x$ is not a self-map of $[0,1]$, so it is not an ergodic transformation of the specified space. No reduction modulo one is stated; if it were imposed it would instead be the identity, also nonergodic.

The map ${\theta }_2(x)=x^2$ is not Lebesgue measure-preserving: the inverse image of $[0,1/4]$ is $[0,1/2]$, of a different measure. It also has nontrivial invariant sets if ergodicity is defined using invariant sets alone. For example,is invariant under squaring because the logarithm inside the fractional part increases by one. Both it and its complement contain intervals of positive length, so their measures are positive. Thus ${\theta }_2$ is not ergodic under either convention.

Reflection ${\theta }_3(x)=1-x$ preserves Lebesgue measure, but $[0,1/4]\cup [3/4,1]$ is invariant of measure $1/2$, so it is not ergodic. Finally ${\theta }_4$ exchanges the two equally weighted atoms. The only invariant subsets are the empty and full sets, hence ${\theta }_4$ is ergodic despite having period two.