# optimum transportation – Transport of touchstone Answer

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optimum transportation – Transport of touchstone

Let’s decompose $$mu$$ and $$nu$$, two possibilities on $$mathbb{R}^{d}$$ , in response to
$$pi_{okay} (x_{1},…,x_{d}) = (x_{okay},…,x_{d})$$
We get a household of measures and every touchstone $$mu_{okay,d}^{+} =: mu^{+}$$ (resp $$nu^{+}_{okay,d} =: nu^{+}$$) is focused on $$mathbb{R}^{k-1} occasions {x_{okay} } occasions … occasions { x_{d} }$$ (resp $$mathbb{R}^{k-1} occasions {y_{okay} } occasions … occasions { y_{d} }$$).

For every $$(x_{+},y_{+}) := (x_{okay},…,x_{d},y_{okay},…,y_{d})$$ we deem $$gamma^{+,+}$$ the Knothe exaltation touchstone between $$mu^{+}$$ and $$nu^{+}$$. We occupy it’s realized by a map $$T$$, i.e.
$$gamma^{+,+} = (textual content{id},T)_{#} mu^{+}$$
with $$T_{#} mu^{+} = nu^{+}$$

Finally let’s deem $$gamma = gamma^{+,+} otimes eta in mathcal{P}(mathbb{R}^{2nd})$$ with $$eta in mathcal{P}(mathbb{R}^{d-k+1} occasions mathbb{R}^{d-k+1})$$ and

$$gamma(A) = int{gamma^{++}(A) deta}$$

In this definition we’ve one thing to show,that is $$gamma$$ is certainly a touchstone. The solely factor to show is that

$$(x_{+},y_{+}) rightarrow gamma^{++}(A)$$
is measurable.

But I do not know how you can show it, do somebody know the way to try this ? I might breathe relieved to learn it.

Thanks and regards.

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