Theorem 17.3.1469E7D1FEF207

Proof. Both conditions are invariant under adding a constant to $u$, so we may assume $u\leq \phi $. Put $u_ k=\max \{ u,\phi -k\} $ and $A_ k=\{ u>\phi -k\} $. The potentials $u_ k$ belong to $\mathcal E^\infty (X,\theta ,\phi )$ and have mass $m$. Since $u_ k\leq \phi $, all terms in (17.3.3) are nonpositive, and keeping only the term $j=n$ gives

Suppose that $E_\theta ^\phi (u)>-\infty $. Since $E_\theta ^\phi (u)\leq E_\theta ^\phi (u_ k)\leq E_\theta ^\phi (\phi )=0$, the right side is bounded by $C=-(n+1)E_\theta ^\phi (u)$. On $X\setminus A_ k$ the weight $\phi -u_ k$ equals $k$ outside a pluripolar set; on the plurifine open set $A_ k$ we have $u_ k=u$, so $\mu _{u_ k}=\mu _ u$ there by locality. Hence

The sets $A_ k$ increase to the set where $u$ is finite, which carries $\mu _ u$, so $\int _ X\langle \theta _ u^ n\rangle \geq m$; since $u\leq \phi $, mass monotonicity (theorem 13.8.7) gives the reverse inequality, and therefore $u\in \mathcal E(X,\theta ,\phi )$. Locality also gives $\int _{A_ k}(\phi -u)\mu _ u\leq C$, and monotone convergence yields $\int _ X(\phi -u)\mu _ u\leq C$. This proves that $u\in \mathcal E^1(X,\theta ,\phi )$ and the first inequality in (17.3.15).

Conversely, suppose $u\in \mathcal E^1(X,\theta ,\phi )$, and let $I=\int _ X(\phi -u)\mu _ u<\infty $. The measures $\mu _{u_ k}$ and $\mu _ u$ have the same total mass $m$ and agree on $A_ k$, so $\mu _{u_ k}(X\setminus A_ k)=\mu _ u(X\setminus A_ k)$. Thus

The lower tangent bound in (17.3.10), applied to $u_ k$ and $\phi $, gives $E_\theta ^\phi (u_ k)\geq -I$. Letting $k\to \infty $ and using proposition 17.3.12, we obtain $E_\theta ^\phi (u)\geq -I>-\infty $, which is the second inequality in (17.3.15). The first inequality was proved above. $\square$