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Suppose $\tilde{\mathbb{P}}$ is a probability measure with $d\tilde{\mathbb{P}}=Zd\mathbb{P}$, so $Z\ge 0$ and $\mathbb{E}Z=1$. For a sub-sigma-algebra $\mathcal{G}$, write $Z_{\mathcal{G}}=\mathbb{E}[Z\mid \mathcal{G}]$. If $X$ is integrable under $\tilde{\mathbb{P}}$, then on $\{Z_{\mathcal{G}}>0\}$,This is Bayes formula for conditional expectation. It changes the underlying probability measure as well as conditioning on the available information. The set where $Z_{\mathcal{G}}=0$ has zero $\tilde{\mathbb{P}}$-probability, so a version there can be chosen arbitrarily.

To verify the formula, let $Y$ denote the ratio. For $A\in \mathcal{G}$, measurability of $Y$ and the tower property give ${\int }_{A}Yd\tilde{\mathbb{P}}=\mathbb{E}[1_{A}ZY]=\mathbb{E}[1_A{Z}_{\mathcal{G}}Y]=\mathbb{E}[1_{A}ZX]={\int }_{A}Xd\tilde{\mathbb{P}}$. These are exactly the defining conditional-expectation identities. For a filtration, choosing $\mathcal{G}={\mathcal{F}}_t$ gives the usual time-indexed form. The denominator is the likelihood weight after reducing the full information to $\mathcal{G}$, not the original pointwise weight $Z$.

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 $Z=1_A$ givesEvery member of $\mathcal{G}$ has the displayed form $T^{-1}(A)$, so this verifies all defining conditional expectation identities. ThereforeApply part (iii), since this conditional expectation has the same probability distribution as $X$. We concludeThis is invariance from integral duality for a measure-preserving transformation; neither invertibility of $T$ nor a second moment of $X$ is required.