$\newcommand{\g}{\mathfrak{g}}$
Let $P(\g)$ be the weight lattice of $\g$ a semisimple Lie algebra over $\mathbb{C}$, and $P_{++}(\g)$ the set of dominant integral weights.
A subset $\Psi \subset P(\g)$ is $\Phi$-saturated if for all $\lambda \in \Psi$ and $\alpha \in \Phi$, $\lambda – k\alpha \in \Psi$ for all integers $k \in [0, \langle \lambda, \check{\alpha} \rangle]$.
An element $\lambda \in \Psi$ is called $\Phi$-extreme if for all $\alpha \in \Phi$, either $\lambda + \alpha \not\in \Psi$ or $\lambda – \alpha \not\in \Psi$.
I wish to show the following result: for $V$ a finite-dimensional irrep of $\g$ of highest weight $\lambda$, the orbit of $\lambda$ under the Weyl group is the set of $\Phi$-extreme elements of $\chi(V)$, where $\chi(V)$ is the weight lattice of $V$ (the set of all weights of $\g$ on $V$).
I know that $\chi(V)$ is the smallest $\Phi$-saturated subset of $P(\g)$ containing $\lambda$.
I am specifically confused in the following line of the proof of the previous result:
Since $W \cdot \Phi = \Phi$, the set of $\Phi$-extreme elements of $\chi(V)$ is invariant under $W$. Thus it suffices to show that if $\mu \in P_{++}(\g) \cap \chi(V)$ and $\mu$ is $\Phi$-extreme, then $\mu = \lambda$.
I understand how the bolded statement is proven, but I do not see how it implies the result, that $$ W \cdot \lambda = \{\Phi\text{-extreme elements of } \chi(V)\}.$$
I understand that the right-hand side can be described as all weights in $\chi(V)$ for which either adding or subtracting any $\alpha \in \Phi$ is no longer a weight, and I also know that any weight can be transformed into a unique dominant weight by some $w \in W$, however there is no guarantee this would also be an integral weight.
In fact, if $\mu$ is both dominant integral and also a weight of $V$, then $\mu = \lambda$ seems forced by the uniqueness of the highest weight: the assumption that it is also $\Phi$-extreme seems superfluous here.
What am I misunderstanding?
Best Answer
Given the unbolded line in your quote, we know $W$ acts on the set of $\Phi$-extreme elements. What the bolded line shows is then that there is only one $W$-orbit of $\Phi$-extreme elements intersecting $P_{++}(\mathfrak{g}) \cap \chi(V)$—the orbit of $\lambda$: $$ \{\text{$\Phi$-extreme}\} \cap P_{++}(\mathfrak{g}) \cap \chi(V) = \{\lambda\}, $$ so $$ W \cdot (\{\text{$\Phi$-extreme}\} \cap P_{++}(\mathfrak{g}) \cap \chi(V)) = W \cdot \lambda $$ Then recall that every weight's $W$-orbit intersects $P_{++}(\mathfrak{g})$. So, given that the the 2 of the sets on the LHS are $W$-invariant and the other contains representatives of every $W$-orbit, taking $W$-orbits is equivalent to removing the intersection with $P_{++}(\mathfrak{g})$ on the left-hand side: $$ W \cdot P_{++}(\mathfrak{g}) = P(\mathfrak{g}) \ ; \ W \cdot \{\text{$\Phi$-extreme}\} = \{\text{$\Phi$-extreme}\} \ ; \ W \cdot \chi(V) = \chi(V) $$ implies $$ W \cdot (\{\text{$\Phi$-extreme}\} \cap P_{++}(\mathfrak{g}) \cap \chi(V)) = \{\text{$\Phi$-extreme}\} \cap \chi(V) = W \cdot \lambda $$
Edit: To address OP's last paragraph, it sounds like there's some confusion about the meaning and consequences of integrality for a weight. For me, and it seems like this is the way it's used in the proof, an integral weight $\lambda$ is an element of the vector space spanned by $\Phi$ whose pairing with all roots is an integer. Equivalently, it is an integral combination of fundamental weights. So it's just equivalent to being in $P(\mathfrak{g})$, as opposed to being in the real vector space spanned by $P(\mathfrak{g})$.
So one consequence is that $W$ does preserve integrality, and another is that you typically do have several (integral) weights in the intersection of $\chi(\lambda)$ and $P_{++}(\mathfrak{g})$.
For instance, in the simplest case of $\mathfrak{sl}_2$, we have $\Phi = \pm{\alpha}$ and $P(\mathfrak{sl}_2) = \mathbb{Z}\lambda_{\alpha}$ where $2 \lambda_{\alpha} = \alpha$. Then as long as $V$ is a highest weight representation of a weight $\mu = n \lambda_{\alpha}$ where $n \ge 2$, then $\chi(V) \ni (n - 2)\lambda_{\alpha}$, which is dominant and integral. Almost every highest weight representation $V(n \lambda_{\alpha})$ is included, since $n \lambda_{\alpha}$ is dominant iff $n \ge 0$. However, for all $n$, the only $\Phi$-extreme weights of $V(n \lambda_{\alpha})$ will be $\pm n \lambda_{\alpha}$, and of those only $+n \lambda_{\alpha}$ is dominant.