If $T$ is measure preserving, then $f\mapsto f\circ T$ is an isometry on $L^\infty$

Question

Let $(E,\mathcal E,\mu)$ be a probability space and $T:E\to E$ be $(\mathcal E,\mathcal E)$-measurable with $$T_\ast\mu=\mu\tag1.$$ How can we show that $$\mathcal L^p(\mu)\ni f\mapsto f\circ T\tag2$$ is an isometry on $\mathcal L^p(\mu)$ for all $p\in[1,\infty]$?

This is trivial, using $(1)$, when $p<\infty$. But I've no idea how we can show it when $p=\infty$. Maybe in the following: Let $f\in\mathcal L^\infty(\mu)$. Assume first that $f(E)$ is finite, i.e. $$f=\sum_{i=1}^ka_i1_{A_i}\tag3$$ for some $k\in\mathbb N$, $a_1,\ldots,a_k\in\mathbb R$ and disjoint $A_1,\ldots,A_k\in\mathcal E$. Then $$f\circ T=\sum_{i=1}^ka_i1_{A_i}\tag4.$$ If the sets $A_i$ would be $T$-invariant (i.e. $1_{A_i}\circ T=1_{A_i}$) we could conclude $f\circ T=f$. But this would mean that $f$ is measurable wrt the $\sigma$-algebra of $T$-invariant sets ...

However, if we could prove the claim in the case where $f(E)$ is finite, we could easily conclude by denseness of the elementary functions in $L^\infty(\mu)$.

Answer 1

First I will show that for any $f \in L^\infty(\mu)$, we have that $\|f\|_\infty \leq \|f \circ T\|_\infty$.

For this, take a null set $A$ such that $\|f \circ T\|_\infty = \sup_{\Omega \setminus A} |f \circ T|$. Then $$\mu[T(\Omega \setminus A)] = \mu[T^{-1}(T(\Omega \setminus A))] \geq \mu(\Omega \setminus A) = 1$$ so that $\mu[T(\Omega \setminus A)] = 1$. As a result, we have that $$\|f\|_\infty \leq \sup_{T(\Omega \setminus A)} |f| = \sup_{\Omega \setminus A} |f \circ T| = \|f \circ T\|_\infty.$$

For the other inequality, take a null set $B$ such that $\|f\|_\infty = \sup_{\Omega \setminus B} |f|$. We would like to say that $$\sup_{\Omega \setminus B} |f| = \sup_{T^{-1}(\Omega \setminus B)} |f \circ T| $$ so that we could immediately conclude using the fact that $\|f \circ T\|_\infty \leq \sup_{T^{-1}(\Omega \setminus B)} |f \circ T|$. Unfortunately, since $T$ need not be surjective, $TT^{-1}(\Omega \setminus A) = (\Omega \setminus A) \cap T(\Omega)$ may be strictly smaller than $\Omega \setminus A$. The saving grace is that the difference is a measure $0$ set.

Indeed, by a similar argument as given earlier, $\mu[T(\Omega)] = 1$, so we must have that $$\sup_{\Omega \setminus B} |f| = \sup_{(\Omega \setminus B) \cap T(\Omega)} |f|$$ since otherwise $B \cup T(\Omega)^c$ is a null set such that $\|f\|_\infty > \sup_{\Omega \setminus(B \cup T(\Omega)^c)} |f|$ which contradicts the definition of $\|f\|_\infty$. We have that $$\mu(T^{-1}((\Omega \setminus B) \cap T(\Omega))) = \mu((\Omega \setminus B) \cap T(\Omega)) = 1$$ This implies that $$\|f \circ T\|_\infty \leq \sup_{T^{-1}((\Omega \setminus B) \cap T(\Omega))} |f \circ T| = \sup_{(\Omega \setminus B) \cap T(\Omega)} |f| = \sup_{\Omega \setminus B} |f| = \|f\|_\infty$$ which completes the proof.