Theorem 10.13.27B5F62BF9296

Proof. Step 1: reduction to analytic singularities. By the analytic regularization of 5.36.3, a Kähler current $T\in \alpha $ can be approximated by currents in $\alpha $ with analytic singularities and smooth remainder away from their poles. After a sufficiently small loss of positivity these are still Kähler currents, and their positive Lelong sets are contained in $E_+(T)$. Hence the intersection defining $E_{\mathrm{nK}}(\alpha )$ may be taken over Kähler currents with analytic singularities and smooth remainder.

Step 2: intersections. Let $\theta +\mathrm{dd}^{\mathrm{c}}u_1$ and $\theta +\mathrm{dd}^{\mathrm{c}}u_2$ be two such currents. By lemma 5.24.1, $\max (u_1,u_2)$ is the potential of a current dominating $\delta \omega $ for a common $\delta >0$, and its Lelong number at each point is the minimum of those of $u_1$ and $u_2$. Regularizing this current as in Step 1 gives a Kähler current with analytic singularities whose positive Lelong set is contained in the intersection of the two previous sets.

Step 3: realization. The positive Lelong sets of Step 1 are analytic subsets of $X$. By the descending chain condition for analytic subsets of a compact complex manifold, some finite intersection of them is minimal, and it then equals the full intersection $E_{\mathrm{nK}}(\alpha )$. Repeated use of Step 2 gives a single Kähler current with analytic singularities whose polar set, which is also its positive Lelong set, is contained in this finite intersection. Since $E_{\mathrm{nK}}(\alpha )$ is contained in the positive Lelong set of every Kähler current in $\alpha $, equality holds.

Step 4: the two consequences. For a positive current $T\in \alpha $ and $\varepsilon >0$, we have $T\in \alpha [-\varepsilon \omega ]$, so minimality gives $\nu (T_{\min ,\varepsilon },x)\leq \nu (T,x)$. Taking the supremum over $\varepsilon $ and then the intersection over Kähler currents $T$ gives $E_{\mathrm{nn}}(\alpha )\subset E_{\mathrm{nK}}(\alpha )$. If the non-Kähler locus is empty, the current of Step 3 has no poles and is therefore a Kähler form in $\alpha $. Conversely, if $\alpha $ contains a Kähler form, using it in the defining intersection gives $E_{\mathrm{nK}}(\alpha )=\varnothing $. $\square$