Lemma 5.19.12 (Siu's decomposition)F956E74CD8C5

Proof. By the proved divisorial decomposition, \[T=R+\sum_D\nu(T,D)[D],\qquad R\geq0,\] where the series converges weakly and every summand is positive. Remove the summand with $D=E$; the remaining weakly convergent sum together with $R$ is positive and closed. It is exactly $T-\nu(T,E)[E]$. If $\nu(T,E)=0$, the assertion reduces to positivity of $T$.