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In wide generality, the disintegration theorem says that Radon probability measures admit disintegrations. I'm trying to understand the case when we weaken this to infinite measures, specifically infinite Haar measures on locally-compact groups. In statistics, these are considered improper priors, which can frequently result in posteriors which are probability measures. However, this is not guaranteed, and is generally a thorny endeavor, so I am trying to tread carefully. See Chang & Pollard 1997 for examples.

Let $\eta$ be an infinite Haar measure on a locally-compact group $G$. Consider a measurable parameter set $\Theta$ which $G$ acts measurably on, and consider a random variable $\gamma : \Theta \to G$. For each $\theta$, define $k_\gamma^\theta(g) := \gamma(g^{-1} \theta)^{-1} g$, and the push-forward measures $\kappa_\gamma^\theta := \eta \circ (k_\gamma^\theta)^{-1}$ on $G$, which are infinite when $\eta$ is infinite.

If we consider a weighted version of $\eta$, i.e., $\alpha \eta$ for integrable $\alpha$, then a disintegration of $\alpha \eta$ through $k_\gamma^\theta$ exists by the disintegration theorem, since $\alpha \eta$ is Radon. However, I don't believe this holds for the infinite measure $\eta$.

Does a disintegration exist for the infinite measure $\eta$? If not, then what additional assumptions must we make for there to exist a disintegration of $\eta$ through $k_\gamma^\theta$?

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    $\begingroup$ At least for $\sigma$-finite measures, you can take a partition into countably many measurable sets of finite measure, apply the disintegration theorem for each cell, and then combine the resulting disintegrations to one grand disintegration. $\endgroup$ Commented Feb 22, 2023 at 8:59
  • $\begingroup$ @MichaelGreinecker I like this idea very much but it sounds too good to be true. In Chang & Pollard Examples 11 & 12 are both $\sigma$-finite but admit non-$\sigma$-finite marginals. In your approach, do we need the assumption both that the original measure $\eta$ and its marginal $\kappa_\gamma^\theta$ are $\sigma$-finite, in which case we have this more general disintegration theorem? I'm surprised I haven't found this in the literature, given the wealth of literature on disintegrations of probability measures. P.S. So nice to see you here, hope you are well $\endgroup$ Commented Feb 23, 2023 at 6:06
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    $\begingroup$ I took a look at the two examples by Chang and Pollard. The problem there seems to be that disintegrations are not unique, not that they do not exist. Also, you may not be able to disintegrate with respect to the marginal. Take the constant function from $\mathbb{R}$, endowed with Lebesgue measure, to $0$. The $\{0\}$-marginal is an infinite point mass and useless for integrating, but you can disintegrate the measure on the product into $\delta_0$ and the kernel that is constant with Lebesgue measure. $\endgroup$ Commented Feb 23, 2023 at 6:47

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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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  • $\begingroup$ Don't forget to accept your own answer, if it is satisfactory! (Otherwise this post will keep getting bumped back to the front page from time to time.) $\endgroup$
    – LSpice
    Commented Mar 23 at 1:15

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