# Disintegration of a push forward measure

Let $$\mu$$ and $$\nu$$ two probabilites on $$\mathbb{R}^{d}$$.

Let $$T : \mathbb{R}^{d} \rightarrow \mathbb{R}^{d}$$ a mesurable map such that $$T_{\#} \mu = \nu$$. I can disintegrate $$\gamma := (id,T)_{\#} \mu$$ according to $$(h, h_{\#} \gamma$$), I get a familly of measure $$\gamma_{y}$$ concentred on $$h^{-1}(\{y\})$$ as usually. But this time $$\gamma$$ have a special form. So do you think it can give me information about $$\gamma_{y}$$ ?

And with more generalities what do we get when we disintegrate a push forward measure $$\gamma := f_{\#}\mu$$ according $$(h, h_{\#} \gamma)$$ ?

Any help would be apprecieted, thanks and regards.

EDIT : As I have no answer, I'll try to be more specific, my question is, do we have $$\gamma_{y} := (id,S)_{\#}\omega$$ with $$\omega$$ a measure. I think $$\omega$$ could be a disintegration of $$\mu$$.

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• It's a bit unclear what you are asking. What is $h$? What does 'disintegration according to $(h, h_{\#}\gamma)$' mean? – Steve Oct 15 at 14:36
• What about $T_{\#}\mu = \nu$? Wouldn't $T_{\#}\mu$ live on $\mathbb{R}$, not $\mathbb{R}^d$? – Nik Weaver Oct 15 at 14:42
• Oh I see, I edited, $T : \mathbb{R}^{d} \rightarrow \mathbb{R}^{d}$ not $T : \mathbb{R}^{d} \rightarrow \mathbb{R}$ obiously. – CechMS Oct 15 at 16:09
• Thanks. You still need to address Steve's questions. – Nik Weaver Oct 15 at 16:15
• The map $(id,T)$ is an isomorphism of measure spaces, so that its conditional measures are just $\delta$-measures. – R W Oct 15 at 16:56

If $$\gamma = \gamma_{y} \oplus \alpha_{\#} \gamma$$ the disintegration of $$\gamma$$ according to $$(\alpha, \alpha_{\#})$$. Then $$\beta_{\#}\gamma = \beta_{\#} \gamma_{y} \oplus \alpha_{\#} \gamma$$
Now let $$\beta_{\#}\gamma = \mu_{y} \oplus \delta_{\#} \beta_{\#}\gamma$$ the disintegration of $$\beta_{\#} \gamma$$ according to $$(\delta, \delta_{\#} \beta_{\#} \gamma)$$ then
$$\mu_{y} = \beta_{\#} \gamma_{y}$$