Let $(X,\mathcal{A},m,T)$ be a probability preserving transformation.
Prove that the operator $U:f\mapsto f\circ T$ satisfies $$ \|Uf\|_{p}=\|f\|_{p} $$ for every $1\le p<\infty$.
My idea: $$ \|Uf\|^p_p=\int\limits_X|f\circ T|^p\mathrm{d}m=\int\limits_X |f|^p\mathrm{d}T^*m=\int\limits_X |f|^p\mathrm{d}m $$
where in the last equality I used the fact that the measure $m$ is invariant.
However I can't justify the second passage in the equality above. Above, $T^*$ is the push-forward measure: $T^*(A):=m(T^{-1}A)$.
If $1_A$ is the characteristic function of some set, then $$1_A \circ T=1_{T^{-1}(A)}$$
Therefore, if $f =\sum a_i 1_{A_i}$ you have $$f \circ T = \sum a_i ( 1_{A_i} \circ T)= \sum a_i 1_{T^{-1}(A_i)} $$
Thus $$\int\limits_X|f\circ T|^p\mathrm{d}m=\int\limits_X|\sum a_i 1_{T^{-1}(A_i)}|^p\mathrm{d}m=\int\limits_X|\sum a_i 1_{A_i}|^p\mathrm{d} T^* m$$