Dominance is necessary for an integrable highest weight module

Fix $i$ and put $e=e_i$, $f=f_i$, $h=h_i$, $a=\lambda (h_i)$. The relation $hf=f(h-2)$ gives $h{f}^{m}v=(a-2m)f^{m}v$ by induction. Since $ev=0$, induction also gives $e{f}^{m}v=m(a-m+1)f^{m-1}v$ for $m\ge 1$: for $m=1$ this is $efv=hv=av$, and from the formula at $m$ we get $e{f}^{m+1}v=fe{f}^{m}v+h{f}^{m}v=(m(a-m+1)+a-2m)f^{m}v=(m+1)(a-m)f^{m}v$.

By F2 there is a least positive integer $N$ such that $f^{N}v=0$. Minimality gives $f^{N-1}v\mathrm{\ne }0$, including $N=1$ because $v\mathrm{\ne }0$. Apply $e$ to the zero vector $f^{N}v$ and use 1.1: $0=N(a-N+1)f^{N-1}v$. In characteristic zero, $N\mathrm{\ne }0$ and the vector is nonzero, so $a=N-1\in {\mathbb{Z}}_{\ge 0}$. Minimality gives every nonzero power through $a$ and the first zero at $a+1$.

Apply 2.1 to each simple index; F1 gives $\lambda \in P^{+}$. For label zero, $N=1$ and the string is precisely the single nonzero vector $v$, with $fv=0$. The zero module is excluded by the nonzero highest vector hypothesis; if there are no simple indices, all asserted label conditions are vacuous. Only least integers and a fixed finite index set were used, so no choice assumption enters. F3 supplies the stated specialization to $L_A(\lambda )$.