= Solution
As printed, \b[$\theta_1(x)=1+x$ 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,
$$
A=\{0<x<1:\ \{\log_2(-\log x)\}\in[0,1/2)\}
$$
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 \b[$\theta_2$ 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 \b[$\theta_4$ is ergodic] despite having period two.
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