Lemma 5

Let $z\in \mathfrak{a}$. Then $z\in \mathfrak{a}^\mu$ if and only if $[h_{\alpha}, z] =\langle \mu, \alpha^{\vee}\rangle z$ for all $\alpha \in B$.

For $\mu\in Q$, let $\mathfrak{a}^{(\mu)}$ be the set of $x\in \mathfrak{a}$ such that $[h_{\alpha}, x] =\langle \mu, \alpha^{\vee}\rangle x$ for all $\alpha \in B$. The sum of the $\mathfrak{a}^{(\mu)}$ is direct. To prove the lemma, it therefore suffices to show that $\mathfrak{a}^\mu\subset \mathfrak{a}^{(\mu)}$. Let $\alpha \in B$. The endomorphism $u$ of the vector space $\mathfrak{a}$ such that $u|\mathfrak{a}^\mu=\langle \mu, \alpha^{\vee}\rangle .1$ is a derivation of $\mathfrak{a}$ such that $ux=$ (ad $h_{\alpha}$)$.x$ for $x=x_{\beta}, x=h_{\beta}, x=x_{-\beta}$; hence $u=$ ad $h_{\alpha}$, which proves our assertion.