Transport de mesure

Let's disintegrate $\mu$ and $\nu$, two probabilities on $\mathbb{R}^{d}$ , according to $$

\pi_{k} (x_{1},...,x_{d}) = (x_{k},...,x_{d})

$$ We get a family of measures and each measure $\mu_{k,d}^{+} =: \mu^{+}$ (resp $\nu^{+}_{k,d} =: \nu^{+}$) is concentrated on $\mathbb{R}^{k-1} \times \{x_{k} \} \times ... \times \{ x_{d} \}$ (resp $\mathbb{R}^{k-1} \times \{y_{k} \} \times ... \times \{ y_{d} \}$).

For each $(x_{+},y_{+}) := (x_{k},...,x_{d},y_{k},...,y_{d})$ we consider $\gamma^{+,+}$ the transport measure between $\mu^{+}$ and $\nu^{+}$. Which means $$

\gamma^{+,+} = (\text{id},T)_{\#} \mu^{+}

$$ with $T_{\#} \mu^{+} = \nu^{+}$ a mesurable function.

Finally let's consider $\gamma = \gamma^{+,+} \otimes \eta \in \mathcal{P}(\mathbb{R}^{2d})$ with $\eta \in \mathcal{P}(\mathbb{R}^{d-k+1} \times \mathbb{R}^{d-k+1})$ and $$

\gamma(A) = \int{\gamma^{++}(A) d\eta}

$$ In this definition we have something to prove,that's $\gamma$ is indeed a measure. The only thing to prove is that $$

(x_{+},y_{+}) \rightarrow \gamma^{++}(A)

$$ is measurable.