Antidominant regular Verma modules are simple

Suppose a proper nonzero submodule $N\subset M(\lambda )$ exists. By [L1], it contains a nonzero singular vector $v_{\mu }$ of weight $\mu =\lambda -\beta$ for some $\beta \in Q^{+}$. The highest-weight space has dimension one, so $\beta \mathrm{\ne }0$: otherwise $v_{\mu }$ would generate $M(\lambda )$, contrary to $N$ being proper. The cyclic submodule $U(\mathfrak{g})v_{\mu }$ is a highest-weight module of weight $\mu$.

By [L2], the Casimir acts on $M(\lambda )$ with scalar $(\lambda ,\lambda +2\rho )$ and on its cyclic submodule $U(\mathfrak{g})v_{\mu }$ with scalar $(\mu ,\mu +2\rho )$. These scalars agree on the nonzero vector $v_{\mu }$. Substituting $\mu =\lambda -\beta$ and expanding gives $2(\lambda +\rho ,\beta )=(\beta ,\beta )$.

Write $\beta ={\sum }_i{n}_i{\alpha }_i$ in the simple-root basis, where $n_i\in {\mathbb{Z}}_{\ge 0}$ and some $n_i>0$. By [L3], $2(\lambda +\rho ,{\alpha }_i)=({\alpha }_i,{\alpha }_i)⟨\lambda +\rho ,{\alpha }_i^{\vee }⟩<0$ for every $i$. Therefore $(\lambda +\rho ,\beta )<0$, whereas positive definiteness gives $(\beta ,\beta )>0$, contradicting step 2.1. Thus $M(\lambda )$ has no proper nonzero submodule and is simple.