5.29.14 Convergence of potentials and measures97C040D849DD

Proof. Choose open sets $O_ j$ of capacity at most $2^{-j}$ such that the restriction of $f$ to $X\setminus O_ j$ is continuous, and set $G_ j=\bigcup _{k\geq j}O_ k$. Then the sets $G_ j$ decrease, their capacities tend to zero, and $f$ is continuous on the closed complement of each $G_ j$. Choose $M\geq \sup _ Xf$ and define

At a boundary point of $G_ j$, the values of $f_ j$ on the closed complement tend to the value of $f$ there, while the values inside $G_ j$ are at least that value. Thus $f_ j$ is lower semicontinuous. The functions $f_ j$ decrease to $f$ outside $\bigcap _ jG_ j$, a set of capacity zero. Since $f_ j-f$ is supported in $G_ j$, they also converge to $f$ in capacity. $\square$

Proof. Translate the $v_ i$ so that $0\leq v_ i\leq 2A$, and put $q=(v_1+\cdots +v_ k)/B$, where $B\geq \max \{ 2k,2kA,1\} $. Then $0\leq q\leq 1$ and

where all coefficients are positive. Expanding $\omega _ q^ n$ shows that the given mixed product is bounded by a constant multiple of $\omega _ q^ n$, and the definition of capacity proves the first claim. The case $k=0$ follows by testing with $v=0$. For a bounded function $z$, split its integral into the parts over $\{ |z|>t\} $ and over its complement; first use convergence in capacity for fixed $t>0$, and then let $t\downarrow 0$. $\square$

Proof. On a relatively compact coordinate ball, choose a fixed quadratic function $q$ such that the functions $v_ j+q$ are plurisubharmonic; they still have common bounds. Expand

The absolute value of the $\ell $-th summand is bounded by a fixed constant times

where $\kappa _0$ is a fixed positive Euclidean form. This positive mixed measure is bounded, uniformly in $j$, by a constant times the local capacity: add a fixed strictly plurisubharmonic quadratic function to $v_ j+q$, expand the top power of its Hessian, and rescale its uniformly bounded range into $[0,1]$ to apply the definition of capacity. This also covers $\ell =0$. Summing over a finite cover gives the asserted control of the total variations. The ordinary decreasing continuity of Bedford–Taylor products gives weak convergence term by term. For a bounded quasi-continuous coefficient, remove an open set of small capacity and extend the restriction of the coefficient to the compact complement to a continuous function on $X$. The variation bound controls the resulting error uniformly for the sequence and its limit, and letting the capacity of the removed set tend to zero proves the assertion. $\square$

Proof. If the first assertion failed, there would be $\epsilon >0$ and Borel sets $E_ j$ of capacity at most $2^{-j}$ with $\mu (E_ j)\geq \epsilon $. By countable subadditivity of capacity,

Hence $\limsup _ jE_ j$ is pluripolar and has $\mu $-measure zero. On the other hand, since $\mu $ is finite, continuity from above applied to the sets $\bigcup _{k\geq j}E_ k$ shows that $\limsup _ jE_ j$ has measure at least $\epsilon $, a contradiction. If $u_ j\to u$ in capacity, applying the first assertion to the sets $\{ |u_ j-u|>\delta \} $ gives the consequence. $\square$

Proof. In the decreasing case $u$ is a common lower bound for the sequence and its limit; in the increasing case $u_1$ is one. Choose a Hermitian form $\Omega \geq \theta $ and truncate all functions at $-T$. The truncations are bounded, monotone and $\Omega $-plurisubharmonic, so the convergence in capacity of locally bounded monotone sequences, recalled in section 5.27, applies to them on the balls of a finite cover. By quasi-continuity of the common lower bound, there is an open set of arbitrarily small capacity containing its pole set such that the restriction of the lower bound to the compact complement is continuous. There it is finite and hence bounded below, so the capacity of the sublevel set of the lower bound at height $-T$ tends to zero as $T\to \infty $. Since the functions agree with their truncations outside this sublevel set, letting $T\to \infty $ proves the lemma. $\square$

Proof. For $n=1$, Stokes' theorem gives $\int _ X\theta _ v=\int _ X\theta $, so assume $n\geq 2$. Choose a fixed Hermitian form $\Omega \geq \theta $. All currents on the left-hand side are positive, and $\Omega _\psi \geq \theta _\psi $. Hence Stokes' theorem for bounded potentials gives

The first term on the right is bounded independently of $v$. To estimate the second, differentiate:

where the second term is absent for $n=2$. Since $\mathrm{dd}^{\mathrm{c}}\Omega $ and $\mathrm d\Omega \wedge \mathrm d^{\mathrm c}\Omega $ are smooth forms, the absolute values of these terms are bounded by fixed multiples of the positive measures $\Omega ^2\wedge \Omega _\psi ^{n-2}$ and $\Omega ^3\wedge \Omega _\psi ^{n-3}$. The weighted local Chern–Levine–Nirenberg inequality of section 5.27 bounds the integrals of $|v|$ against these measures. Indeed, on each larger coordinate ball, add fixed smooth corrections to $v$ and $\psi $, shift the corrected $v$ so that it becomes negative, expand in the corrected positive Hessian of $\psi $, and apply that inequality on a smaller ball. The resulting bound is

By theorem 5.25.2, normalized $\theta $-plurisubharmonic functions have uniformly bounded $L^1$ norms, and summing the local bounds proves the assertion. $\square$