$\DeclareMathOperator\PSH{PSH}$This question is about Proposition 9.25 page 252 from the book "Degenerate Complex Monge-Ampère Equations" by Vincent Guedj and Ahmed Zeriahi (see picture below).
Let $X$ be a compact complex Kähler manifold and $\omega$ a closed $(1, 1)$ form which is semi positive and smooth. One can define the set $\PSH(X, \omega)$ as the set of $\phi \in L^{1}$ such that the current $\omega + dd^{c} \phi$ is positive. If $\phi$ is locally bounded one defines, for every positive close current $T$ of bidegree $(n-1, n-1)$, the positive current $(\omega + dd^{c} \phi) \wedge T := \omega \wedge T + dd^{c}(\phi T)$. Hence, the positive distribution $MA(\phi) := (\omega + dd^{c} \phi)^{n}$ which is a positive measure is well defined.
Fact : if $B$ is a small open in $X$, for all $\phi \in \PSH(X, \omega)$, their exists a unique $\psi \in \PSH(X, \omega)$ such that $\psi \ge \phi$ in $X$ and such that $MA(\psi) = 0$ in $B$ and $\phi = \psi $ in $X - B$.
Now, if $h$ is $C^{2}$ smooth on $x$ with real values, then one can defined the envelope $P(H)(x) := \sup ( \phi(x) | \phi \in \PSH(X, \omega), ~ \phi \leq h )$.
I send you as an image the proof I don't understand : when they say Then $v_{j} \ge u_{j}$ and we still have $v_{j} \leq h$ by continuity if B is small enough.
How can they choose $B$ small enough uniformly in $j$?
By unicity of the previous fact, it's easy done for a fixed $j$. But I don't see how to do it uniformly.