Throughout this chapter $A$ will denote a complexified semi-simple Lie-algebra, $H$ a Cartan sub-algebra, $R\sbe H$ the root system and $B$ a base for $R$.
$\proof$
Let $F$ be the sub-space of $E$ generated by all weight vectors in $E$ whose weights lie in $\mu+\Z r$. These weights are shifted by $\psi(x_r)$ and $\psi(y_r)$ by $\pm r$ and therefore the restrictions $\psi|\lhull{r,x_r,y_r}$ gives a representation $\vp$ of $\sla(2,\C)$ in $F$ - with isomorphism $r^\vee\to H$, $x_r\to X$ and $y_r\to Y$. Since $R_r(\mu)=\mu-nr$ is also a weight for $\vp$ the operator $\vp(r^\vee)\in\Hom(F)$ has eigen-values $l\colon=\la r^\vee,\mu\ra$ and $\la r^\vee,\mu-nr\ra=l-2n$. By proposition $\vp(r^\vee)\in\Hom(F)$ must have the eigen-values $l,l-2,\ldots,l-2n$, which coincides with the set $\{\la r^\vee,\nu\ra: \nu\in\{\mu,\mu-r,\ldots,\mu-nr\}\}$. Thus for any such $\nu=\mu-kr$ there must be an eigen-vector $x\in F\sm\{0\}$ of $\vp(r^\vee)$ such that:
$$
\psi(r^\vee)x=\vp(r^\vee)x=\la r^\vee,\nu\ra x
$$
and this eigen-vector $x$ can be obtained by starting with a normed weight vector $x_0$ for the weight $\mu$ and applying $\psi(y_r)$ $k$ times; as $x\neq0$, it`s a weight vector for $\psi$.
$\eofproof$
$\proof$
Assume $\mu-\l=\sum_bn(b)b$, $n(b)\in\N_0$. Let $P\sbe H$ be the set
$$
P\colon=\mu+\sum_b\{0,1,\ldots,n(b)\}b~.
$$
and define for each $p=\mu+\sum k(b)b\in P$: $d(p)\colon=\sum k(b)$, which we think of as a sort of distance from $p$ to $\mu$. We claim that if $p\in P$ is a weight and $d(p) > 0$, then there is $b_0\in B$ such that $\la p,b_0\ra > 0$: As $p-\mu\neq0$ we have
$$
0 < \norm{p-\mu}^2
=\sum_{b\in B}k(b)\la p-\mu,b\ra
$$
and thus there is some $b_0\in B$ such that $k(b_0) > 0$ and $\la p-\mu,b_0\ra > 0$. Since $\mu$ is dominant we conclude that $\la p,b_0\ra > 0$.
Finally by lemma $p-b_0$ must be a weight. Hence for every weight $p\in P$ satisfying $d(p) > 0$ there is a weight $q\in P$ (e.g. $q\colon=p-b_0$) satisfying $d(q) < d(p)$, which shows that $\mu$ is a weight.
$\eofproof$
Finally we can state the generalization of theorem to semi-simple Lie-algebras:
$\proof$
As $\psi$ is equivalent to $\g_\l:A\rar\Hom(W_\l/V_\l)$, each weight must be integral and $\mu\in\l-\sum_{b\in B}\N_0b$. Finally for all $w$ in the Weyl group $W$: $w(\mu)\preceq\l$ and since $\l$ is dominant we conclude by theorem: $\mu\in\convex{W(\l)}$.
Conversely, assume the three conditions hold. By corollary we can find some $w\in W$ such that $w(\mu)$ is dominant. By proposition $w(\mu)$ is a weight for $\psi$ (and thus $\mu$ is a weight) if $\l-w(\mu)\in\sum_{\b\in B}\N_0b$. But as $\mu\in C$ this follows from theorem provided $\l-w(\mu)\in\sum_{b\in B}\Z b$.
Now for each $c\in B$: $R_c(\mu)-\mu=\la\mu,c^\vee\ra c\in\Z c$ is integral, which implies by proposition that for all $w\in W$: $w(\mu)-\mu\in\sum_{b\in B}\Z b$ and therefore $\l-w(\mu)=\l-\mu+\mu-w(\mu)$ and since both $\l-\mu$ and $\mu-w(\mu)$ lie in $\sum_{b\in B}\Z b$, we are done.
$\eofproof$weight space