Bounded above kac moody weight modules are generated by primitive vectors

Let $N=U(\mathfrak{g}){\mathfrak{n}}^{+}v$. Every submodule killing the image of ${\mathfrak{n}}^{+}v$ contains $N$. Hence a nonzero highest image of $v$ exists exactly when $v\mathrm{\notin }N$ (use the quotient by $N$ itself for sufficiency). PBW writes $N=U({\mathfrak{n}}^{-})U(\mathfrak{h})U({\mathfrak{n}}^{+}){\mathfrak{n}}^{+}v$. Positive words have definite weights on $v$, so their Cartan factors act as scalars; and $U({\mathfrak{n}}^{+}){\mathfrak{n}}^{+}=U_0({\mathfrak{n}}^{+})$. Therefore $N=U({\mathfrak{n}}^{-})U_0({\mathfrak{n}}^{+})v$. This proves both implications of the criterion.

For a support weight $\mu$ put $d(\mu )=\mathrm{\mid }\{\eta \in \mathrm{supp}V:\eta \ge \mu \}\mathrm{\mid }$, a positive finite integer by F1. If $\eta >\mu$ is in the support, its upper set is a proper subset of this set, since it excludes $\mu$, so $d(\eta )<d(\mu )$. Induct on this integer. A primitive $v\in V_{\mu }$ already lies in the desired span. Otherwise step 1.1 expresses it as a finite sum of negative words applied to vectors $zv$ with $z$ a nonempty positive homogeneous word. Each nonzero $zv$ has weight strictly above $\mu$, so is in the required span by induction. Applying further negative words keeps it there. At $d(\mu )=1$, the positive words all kill $v$, so step 1.1 says $v$ is primitive. Zero vectors and finite sums of weight vectors finish the assertion.