= Solution
{c}
Let $\mathcal G=T^{-1}(\mathcal F)=\{T^{-1}(A):A\in\mathcal F\}$. <Inverse images> commute with <complements> and <countable unions>, so this is a <sub-sigma-algebra> of $\mathcal F$. The <random variable> $X\circ T$ is $\mathcal G$-<measurable>.
The assumption $\mathbb P\circ T^{-1}=\mathbb P$ says that $T$ is a <measure-preserving transformation>. Consequently $X\circ T$ has the same <probability distribution> as $X$ and is <integrable>. It also implies $\mathbb E[W\circ T]=\mathbb E[W]$ for every <integrable random variable> $W$, first for <indicator functions>, then for <simple functions>, and then by positive and negative approximation.
To identify a <conditional expectation>, take any $A\in\mathcal F$. The assumed identity with the bounded <random variable> $Z=\mathbf1_A$ gives
$$
\begin{aligned}
\mathbb E[X\mathbf1_{T^{-1}(A)}]
&=\mathbb E[X(\mathbf1_A\circ T)]
=\mathbb E[X\mathbf1_A]\\
&=\mathbb E[(X\mathbf1_A)\circ T]
=\mathbb E[(X\circ T)\mathbf1_{T^{-1}(A)}].
\end{aligned}
$$
Every member of $\mathcal G$ has the displayed form $T^{-1}(A)$, so this verifies all defining <conditional expectation> identities. Therefore
$$
\mathbb E[X\mid\mathcal G]=X\circ T\quad\text{almost surely}.
$$
Apply part (iii), since this <conditional expectation> has the same <probability distribution> as $X$. We conclude
$$
\boxed{X=X\circ T\quad\mathbb P\text{-almost surely}.}
$$
This is <invariance from integral duality for a measure-preserving transformation>; neither invertibility of $T$ nor a <second moment> of $X$ is required.
Back to article page
