PomeRemark 19.5.36.49B696578E15 More generally, the same argument shows the following result: Let $k=0,\ldots ,n$ and $T\in \mathcal{Z}_+(X,\alpha )$ such that $\nu (T,Y_ k)=0$. Assume that $\operatorname{Tr}^{\alpha |_{Y_ k}}_{Y_ k}(T)$ is defined. Then
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\begin{equation} \left\{ y\in \mathbb {R}^{n-k}: (0,\ldots ,0,y)\in \Delta _{Y_{\bullet }}(T)\right\} =\Delta _{Y_ k\supseteq \cdots \supseteq Y_ n}\left(\operatorname{Tr}^{\alpha |_{Y_ k}}_{Y_ k}(T)\right). \end{equation}
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