PomeExample 20.2.17.2272C87A04CE Assume that $\varphi \in \mathrm{PSH}(X,\theta )$ for some closed smooth real $(1,1)$-form $\theta $ on $X$, $\nu (\varphi ,Y)=0$ and
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\begin{equation} \lim _{\epsilon \searrow 0} \int _{\widetilde{Y}} \left( \theta |_{\widetilde{Y}} +\epsilon \omega |_{\widetilde{Y}}+\mathrm{dd}^{\mathrm{c}}\operatorname{Tr}_ Y^{\theta +\epsilon \omega }(\varphi )\right)^{\dim Y} >0 \end{equation}
for any arbitrary choice of a Kähler form $\omega $ on $X$. Then it follows from proposition 16.4.4 that $\operatorname{Tr}_ Y^{\theta }(\varphi )$ is defined, and its mass is exactly the above limit.
As a consequence, we have the following formula:
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\begin{equation} \int _{\widetilde{Y}} \left(\theta |_{\widetilde{Y}}+\mathrm{dd}^{\mathrm{c}}\operatorname{Tr}_ Y^{\theta }(\varphi )\right)^{\dim Y}=\lim _{\epsilon \searrow 0} \int _{\widetilde{Y}} \left( \theta |_{\widetilde{Y}} +\epsilon \omega |_{\widetilde{Y}}+\mathrm{dd}^{\mathrm{c}}\operatorname{Tr}_ Y^{\theta +\epsilon \omega }(\varphi )\right)^{\dim Y}, \end{equation}
where the left-hand side is understood as $0$ if $\operatorname{Tr}_ Y^{\theta }(\varphi )$ is not defined.
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