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Ok I think I've got it, following the partition argument indicated by Michael Greinecker in the commentscomments above. Thank you Michael! Please let me know if you spot any errors.

Lemma. Let $G$ be a locally compact group with Borel $\sigma$-algebra $\mathcal{B}(G)$. For each measurable function $k : G \to G$ and the push-forward measure $\kappa := k_* \eta = \eta \circ k^{-1}$ on $G$, there exists a disintegration $\eta_k(\mathrm{d} g|g')$ such that for any $B \in \mathcal{B}(G)$, \begin{equation} \eta(B) = \int_G \eta_k(B|g') \kappa(\mathrm{d} g') = \int_G \eta_k(B|k(g)) \eta(\mathrm{d} g). \end{equation}

Proof. We construct the global disintegration by a local partition argument. By local compactness of $G$, there exists a countable partition $\mathcal{C}$ of disjoint pre-compact sets of finite Haar measure, whose union equals $G$.

For each $C \in \mathcal{C}$, define the probability measure $\eta^C := \frac{1}{\eta(C)} 1_C \eta$. Each $\eta^C$ is Radon, so by the disintegration theorem (cf. Theorem 3.1 of Theorem 3.1 of Leao et al. 2004Leao et al. 2004), there exists a regular conditional probability through $k$, i.e., a measurable measure-valued function $g' \mapsto \eta^C_k(\mathrm{d} g|g')$ such that the disintegration equation holds for each $B \in \mathcal{B}(G)$: \begin{equation} \frac{\eta(C \cap B)}{\eta(C)} = \eta^C(B) = \int_G \eta^C_k(B|g') \kappa(\mathrm{d} g') = \int_G \eta_k^C(B|k(g)) \eta(\mathrm{d} g). \end{equation}

We now define $\eta_k$ by combining across partition sets. Suppose $B \in \mathcal{B}(G)$ has finite Haar measure $\eta(B) < \infty$. Then: \begin{equation} \eta(B) = \sum_C \eta(C \cap B) = \int_G \sum_{C \in \mathcal{C}} \eta(C) \eta^C_k(B|g') \kappa(\mathrm{d} g') =: \int_G \eta_k(B|g') \kappa(\mathrm{d} g'), \end{equation} where we define $\eta_k(B|g') := \sum_{C \in \mathcal{C}} \eta(C) \eta^C_k(B|g')$.

Ok I think I've got it, following the partition argument indicated by Michael Greinecker in the comments above. Thank you Michael! Please let me know if you spot any errors.

Lemma. Let $G$ be a locally compact group with Borel $\sigma$-algebra $\mathcal{B}(G)$. For each measurable function $k : G \to G$ and the push-forward measure $\kappa := k_* \eta = \eta \circ k^{-1}$ on $G$, there exists a disintegration $\eta_k(\mathrm{d} g|g')$ such that for any $B \in \mathcal{B}(G)$, \begin{equation} \eta(B) = \int_G \eta_k(B|g') \kappa(\mathrm{d} g') = \int_G \eta_k(B|k(g)) \eta(\mathrm{d} g). \end{equation}

Proof. We construct the global disintegration by a local partition argument. By local compactness of $G$, there exists a countable partition $\mathcal{C}$ of disjoint pre-compact sets of finite Haar measure, whose union equals $G$.

For each $C \in \mathcal{C}$, define the probability measure $\eta^C := \frac{1}{\eta(C)} 1_C \eta$. Each $\eta^C$ is Radon, so by the disintegration theorem (cf. Theorem 3.1 of Leao et al. 2004), there exists a regular conditional probability through $k$, i.e., a measurable measure-valued function $g' \mapsto \eta^C_k(\mathrm{d} g|g')$ such that the disintegration equation holds for each $B \in \mathcal{B}(G)$: \begin{equation} \frac{\eta(C \cap B)}{\eta(C)} = \eta^C(B) = \int_G \eta^C_k(B|g') \kappa(\mathrm{d} g') = \int_G \eta_k^C(B|k(g)) \eta(\mathrm{d} g). \end{equation}

We now define $\eta_k$ by combining across partition sets. Suppose $B \in \mathcal{B}(G)$ has finite Haar measure $\eta(B) < \infty$. Then: \begin{equation} \eta(B) = \sum_C \eta(C \cap B) = \int_G \sum_{C \in \mathcal{C}} \eta(C) \eta^C_k(B|g') \kappa(\mathrm{d} g') =: \int_G \eta_k(B|g') \kappa(\mathrm{d} g'), \end{equation} where we define $\eta_k(B|g') := \sum_{C \in \mathcal{C}} \eta(C) \eta^C_k(B|g')$.

Ok I think I've got it, following the partition argument indicated by Michael Greinecker in the comments above. Thank you Michael! Please let me know if you spot any errors.

Lemma. Let $G$ be a locally compact group with Borel $\sigma$-algebra $\mathcal{B}(G)$. For each measurable function $k : G \to G$ and the push-forward measure $\kappa := k_* \eta = \eta \circ k^{-1}$ on $G$, there exists a disintegration $\eta_k(\mathrm{d} g|g')$ such that for any $B \in \mathcal{B}(G)$, \begin{equation} \eta(B) = \int_G \eta_k(B|g') \kappa(\mathrm{d} g') = \int_G \eta_k(B|k(g)) \eta(\mathrm{d} g). \end{equation}

Proof. We construct the global disintegration by a local partition argument. By local compactness of $G$, there exists a countable partition $\mathcal{C}$ of disjoint pre-compact sets of finite Haar measure, whose union equals $G$.

For each $C \in \mathcal{C}$, define the probability measure $\eta^C := \frac{1}{\eta(C)} 1_C \eta$. Each $\eta^C$ is Radon, so by the disintegration theorem (cf. Theorem 3.1 of Leao et al. 2004), there exists a regular conditional probability through $k$, i.e., a measurable measure-valued function $g' \mapsto \eta^C_k(\mathrm{d} g|g')$ such that the disintegration equation holds for each $B \in \mathcal{B}(G)$: \begin{equation} \frac{\eta(C \cap B)}{\eta(C)} = \eta^C(B) = \int_G \eta^C_k(B|g') \kappa(\mathrm{d} g') = \int_G \eta_k^C(B|k(g)) \eta(\mathrm{d} g). \end{equation}

We now define $\eta_k$ by combining across partition sets. Suppose $B \in \mathcal{B}(G)$ has finite Haar measure $\eta(B) < \infty$. Then: \begin{equation} \eta(B) = \sum_C \eta(C \cap B) = \int_G \sum_{C \in \mathcal{C}} \eta(C) \eta^C_k(B|g') \kappa(\mathrm{d} g') =: \int_G \eta_k(B|g') \kappa(\mathrm{d} g'), \end{equation} where we define $\eta_k(B|g') := \sum_{C \in \mathcal{C}} \eta(C) \eta^C_k(B|g')$.

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Tom LaGatta
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Ok I think I've got it, following the partition argument indicated by Michael Greinecker in the comments above. We may bootstrap our way up through conditioning based on functions ofThank you Michael! Please let me know if you spot any errors.

Lemma. Let $G$ be a locally compact group with Borel $\sigma$-algebra $\mathcal{B}(G)$. For each measurable function $k : G \to G$ and the formpush-forward measure $\alpha(g') = \alpha(\gamma(g^{-1} \theta)^{-1} g)$$\kappa := k_* \eta = \eta \circ k^{-1}$ on $G$, there exists a disintegration $\eta_k(\mathrm{d} g|g')$ such that for any $\int_G \alpha(g) \eta(\mathrm{d} g) = 1$$B \in \mathcal{B}(G)$, and disintegrate \begin{equation} \eta(B) = \int_G \eta_k(B|g') \kappa(\mathrm{d} g') = \int_G \eta_k(B|k(g)) \eta(\mathrm{d} g). \end{equation}

Proof. We construct the probability measuresglobal disintegration by a local partition argument. By local compactness of $\alpha \eta$ through$G$, there exists a countable partition $g' = k_\gamma^\theta$$\mathcal{C}$ of disjoint pre-compact sets of finite Haar measure, whose union equals $G$. 

For each such $\alpha$ and$C \in \mathcal{C}$, define the probability measure $\theta \in \Theta$$\eta^C := \frac{1}{\eta(C)} 1_C \eta$. Each $\eta^C$ is Radon, so by the disintegration theorem of(cf. LeaoTheorem 3.1 of Leao et al. 2004 ensures that $\alpha \eta$ admits), there exists a measurable disintegrationregular conditional probability through $(\alpha \eta)_{k_\gamma^{\theta}}(\mathrm{d}g|g')$$k$, i.e., a measurable measure-valued function \begin{equation} \int_G X(g) \alpha(\gamma(g^{-1} \theta)^{-1} g) \eta(\mathrm{d} g) = \int_G \int_G X(g) (\alpha \eta)_{k_\gamma^\theta}(\mathrm{d} g | g') \kappa_\gamma^\theta(\mathrm{d} g'). \end{equation} In particular,$g' \mapsto \eta^C_k(\mathrm{d} g|g')$ such that the disintegration equation holds for each Borel set $B \in \mathcal{B}(G)$, we may consider $X(g) := 1_B(g) \frac{1}{\alpha(\gamma(g^{-1} \theta)^{-1} g)}$. Consequently, \begin{equation} \eta(B) = \int_G \frac{(\alpha \eta)_{k_\gamma^\theta}(B|g')}{\alpha(g')} \kappa_\gamma^\theta(\mathrm{d} g'). \end{equation}: Therefore any such $\eta_{k_\gamma^\theta}(\mathrm{d} g | g') := \frac{(\alpha \eta)_{k_\gamma^\theta}(\mathrm{d} g|g')}{\alpha(g')}$ can serve as the improper posterior, and this does not depend on specific choice of\begin{equation} \frac{\eta(C \cap B)}{\eta(C)} = \eta^C(B) = \int_G \eta^C_k(B|g') \kappa(\mathrm{d} g') = \int_G \eta_k^C(B|k(g)) \eta(\mathrm{d} g). \end{equation}

We now define $\alpha$, up to$\eta_k$ by combining across partition sets of. Suppose $\eta$-measure zero$B \in \mathcal{B}(G)$ has finite Haar measure $\eta(B) < \infty$.

Thank you Michael Greinecker for your help! Please let me know if you see any errors Then: \begin{equation} \eta(B) = \sum_C \eta(C \cap B) = \int_G \sum_{C \in \mathcal{C}} \eta(C) \eta^C_k(B|g') \kappa(\mathrm{d} g') =: \int_G \eta_k(B|g') \kappa(\mathrm{d} g'), \end{equation} where we define $\eta_k(B|g') := \sum_{C \in \mathcal{C}} \eta(C) \eta^C_k(B|g')$.

Ok I think I've got it. We may bootstrap our way up through conditioning based on functions of the form $\alpha(g') = \alpha(\gamma(g^{-1} \theta)^{-1} g)$ for $\int_G \alpha(g) \eta(\mathrm{d} g) = 1$, and disintegrate the probability measures $\alpha \eta$ through $g' = k_\gamma^\theta$. For each such $\alpha$ and $\theta \in \Theta$, the disintegration theorem of Leao et al. ensures that $\alpha \eta$ admits a measurable disintegration $(\alpha \eta)_{k_\gamma^{\theta}}(\mathrm{d}g|g')$, i.e., \begin{equation} \int_G X(g) \alpha(\gamma(g^{-1} \theta)^{-1} g) \eta(\mathrm{d} g) = \int_G \int_G X(g) (\alpha \eta)_{k_\gamma^\theta}(\mathrm{d} g | g') \kappa_\gamma^\theta(\mathrm{d} g'). \end{equation} In particular, for each Borel set $B \in \mathcal{B}(G)$, we may consider $X(g) := 1_B(g) \frac{1}{\alpha(\gamma(g^{-1} \theta)^{-1} g)}$. Consequently, \begin{equation} \eta(B) = \int_G \frac{(\alpha \eta)_{k_\gamma^\theta}(B|g')}{\alpha(g')} \kappa_\gamma^\theta(\mathrm{d} g'). \end{equation} Therefore any such $\eta_{k_\gamma^\theta}(\mathrm{d} g | g') := \frac{(\alpha \eta)_{k_\gamma^\theta}(\mathrm{d} g|g')}{\alpha(g')}$ can serve as the improper posterior, and this does not depend on specific choice of $\alpha$, up to sets of $\eta$-measure zero.

Thank you Michael Greinecker for your help! Please let me know if you see any errors.

Ok I think I've got it, following the partition argument indicated by Michael Greinecker in the comments above. Thank you Michael! Please let me know if you spot any errors.

Lemma. Let $G$ be a locally compact group with Borel $\sigma$-algebra $\mathcal{B}(G)$. For each measurable function $k : G \to G$ and the push-forward measure $\kappa := k_* \eta = \eta \circ k^{-1}$ on $G$, there exists a disintegration $\eta_k(\mathrm{d} g|g')$ such that for any $B \in \mathcal{B}(G)$, \begin{equation} \eta(B) = \int_G \eta_k(B|g') \kappa(\mathrm{d} g') = \int_G \eta_k(B|k(g)) \eta(\mathrm{d} g). \end{equation}

Proof. We construct the global disintegration by a local partition argument. By local compactness of $G$, there exists a countable partition $\mathcal{C}$ of disjoint pre-compact sets of finite Haar measure, whose union equals $G$. 

For each $C \in \mathcal{C}$, define the probability measure $\eta^C := \frac{1}{\eta(C)} 1_C \eta$. Each $\eta^C$ is Radon, so by the disintegration theorem (cf. Theorem 3.1 of Leao et al. 2004), there exists a regular conditional probability through $k$, i.e., a measurable measure-valued function $g' \mapsto \eta^C_k(\mathrm{d} g|g')$ such that the disintegration equation holds for each $B \in \mathcal{B}(G)$: \begin{equation} \frac{\eta(C \cap B)}{\eta(C)} = \eta^C(B) = \int_G \eta^C_k(B|g') \kappa(\mathrm{d} g') = \int_G \eta_k^C(B|k(g)) \eta(\mathrm{d} g). \end{equation}

We now define $\eta_k$ by combining across partition sets. Suppose $B \in \mathcal{B}(G)$ has finite Haar measure $\eta(B) < \infty$. Then: \begin{equation} \eta(B) = \sum_C \eta(C \cap B) = \int_G \sum_{C \in \mathcal{C}} \eta(C) \eta^C_k(B|g') \kappa(\mathrm{d} g') =: \int_G \eta_k(B|g') \kappa(\mathrm{d} g'), \end{equation} where we define $\eta_k(B|g') := \sum_{C \in \mathcal{C}} \eta(C) \eta^C_k(B|g')$.

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Tom LaGatta
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Ok I think I've got it. We may bootstrap our way up through conditioning based on functions of the form $\alpha(g') = \alpha(\gamma(g^{-1} \theta)^{-1} g)$ for $\int_G \alpha(g) \eta(\mathrm{d} g) = 1$, and disintegrate the probability measures $\alpha \eta$ through $g' = k_\gamma^\theta$. For each such $\alpha$ and $\theta \in \Theta$, the disintegration theorem of Leao et al. ensures that $\alpha \eta$ admits a measurable disintegration $(\alpha \eta)_{k_\gamma^{\theta}}(\mathrm{d}g|g')$, i.e., \begin{equation} \int_G X(g) \alpha(\gamma(g^{-1} \theta)^{-1} g) \eta(\mathrm{d} g) = \int_G \int_G X(g) (\alpha \eta)_{k_\gamma^\theta}(\mathrm{d} g | g') \kappa_\gamma^\theta(\mathrm{d} g'). \end{equation} In particular, for each Borel set $B \in \mathcal{B}(G)$, we may consider $X(g) := 1_B(g) \frac{1}{\alpha(\gamma(g^{-1} \theta)^{-1} g)}$. Consequently, \begin{equation} \eta(B) = \int_G \frac{(\alpha \eta)_{k_\gamma^\theta}(B|g')}{\alpha(g')} \kappa_\gamma^\theta(\mathrm{d} g'). \end{equation} Therefore any such $\eta_{k_\gamma^\theta}(\mathrm{d} g | g') := \frac{(\alpha \eta)_{k_\gamma^\theta}(\mathrm{d} g|g')}{\alpha(g')}$ can serve as the improper posterior, and this does not depend on specific choice of $\alpha$, up to sets of $\eta$-measure zero.

Thank you Michael Greinecker for your help! Please let me know if you see any errors.