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

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

$$ \mu_{y} = \beta_{\#} \gamma_{y} $$

I hope someone will answer...

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