Proving that there exists a saturated set with given highest weight

From the definition of the set $\Pi$ it is immediate that $\sigma(\Pi)=\Pi$ for all $\sigma\in W$. By Lemma 13.2A we have $\sigma\mu\prec\mu$ for all dominant weights $\mu$ and all $\sigma\in W$. This allows us to redescribe the set $\Pi$ as $$ \Pi=\{\mu\in\Lambda\mid \sigma\mu\prec\lambda\ \text{for all $\sigma\in W$}\}.\qquad(*) $$ Let then $\mu\in\Pi$ and $\alpha\in\Phi$ be arbitrary. Consider the $\alpha$-chain of weights connecting $\mu$ and $\mu-m\alpha$, $m=\langle\mu,\alpha\rangle$. Let us fix an integer $i$ between $0$ and $m$, and study the weight $\mu'=\mu-i\alpha$.

The claim is that $\mu'\in \Pi$. We are going to apply description $(*)$, so let $\sigma\in W$ be arbitrary. The upshot is that $\sigma(\mu')$ is on the $\sigma(\alpha)$-chain from $\sigma\mu$ to $\sigma(\mu-m\alpha)=\sigma(\mu)-m\sigma(\alpha)$. So according to the sign of $m$ and the positivity of $\sigma(\alpha)$ we have either $$ \sigma\mu\prec\sigma\mu'\prec\sigma\mu-m\sigma(\alpha) $$ or $$ \sigma\mu-m\sigma(\alpha)\prec\sigma\mu'\prec\sigma\mu. $$

But here $\sigma\mu$ and $\sigma(\mu-m\alpha)=\sigma s_\alpha\mu$ are both in the $W$-orbit of $\mu$, so according to $(*)$ they are both $\prec\lambda$. So irrespective of which alternative holds we get that $\sigma(\mu')\prec\lambda$.

So $\sigma(\mu')\prec\lambda$ for all $\sigma\in W$, and thus by $(*)$ we have that $\mu'\in\Pi$. Q.E.D.