Proof.
Step 1. We first assume that the $\mathbb D_ i^ j$'s and the $\mathbb D_ i$'s are all big.
Choose $\mathcal I$-good non-divisorial currents $T_ i^ j$ and $T_ i$ such that $\mathbb D(T_ i^ j)=\mathbb D_ i^ j$ and $\mathbb D(T_ i)=\mathbb D_ i$. Fix a Kähler form $\omega $ on $X$ and fix $\epsilon >0$. Since $\mathbb D_{i,X}^ j\to \mathbb D_{i,X}$, for all sufficiently large $j$ the class
\[ \mathbb D_{i,X}+\epsilon \{ \omega \} -\mathbb D_{i,X}^ j \]
is Kähler. Choose a Kähler form $\chi _ i^ j$ in this class and set $S_ i^ j=T_ i^ j+\chi _ i^ j$. Then $S_ i^ j$ lies in the fixed class $\mathbb D_{i,X}+\epsilon \{ \omega \} $ and has the same divisorial Lelong numbers as $T_ i^ j$. Hence, by theorem 18.1.42 and proposition 18.1.27 2,
\[ \varlimsup _{j\in J}\operatorname{vol}\Bigl (T_1^ j,\ldots ,T_ n^ j\Bigr )\leq \varlimsup _{j\in J}\operatorname{vol}\Bigl (S_1^ j,\ldots ,S_ n^ j\Bigr )\leq \operatorname{vol}\left(T_1+\epsilon \omega ,\ldots ,T_ n+\epsilon \omega \right). \]
Therefore,
\[ \varlimsup _{j\in J}\Bigl (\mathbb D_1^ j,\ldots ,\mathbb D_ n^ j\Bigr )\leq \Bigl (\mathbb D_1+\epsilon \mathbb D(\omega ),\ldots ,\mathbb D_ n+\epsilon \mathbb D(\omega )\Bigr ). \]
Letting $\epsilon \to 0+$ gives (20.1.48).
Step 2. Next we handle the general case.
Take a Kähler form $\omega $ on $X$. Then by 20.1.44, for any $\epsilon \in (0,1]$, we have
\[ \begin{aligned} \varlimsup _{j\in J}\left(\mathbb {D}_1^ j,\ldots ,\mathbb {D}_ n^ j \right) \leq & \varlimsup _{j\in J}\left(\mathbb {D}_1^ j+\epsilon \mathbb {D}(\omega ),\ldots ,\mathbb {D}_ n^ j+\epsilon \mathbb {D}(\omega ) \right)\\ \leq & \left( \mathbb {D}_1+\epsilon \mathbb {D}(\omega ),\ldots ,\mathbb {D}_ n+\epsilon \mathbb {D}(\omega ) \right)\\ \leq & \left( \mathbb {D}_1,\ldots ,\mathbb {D}_ n \right)+C\epsilon . \end{aligned} \]
Since $\epsilon $ is arbitrary, our assertion follows.
$\square$
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