Root space $L_\alpha$ is completely contained in simple ideal? I'm having trouble understanding a section in Humphrey's Lie algebras on page 74.

Suppose $L$ is a semisimple Lie algebra which decomposes as a direct sum of simple ideals $L_1\oplus\cdots\oplus L_t$. Let $H$ be a maximal toral subalgebra, so $H=H_1\oplus\cdots\oplus H_t$, where $H_i=L_i\cap H$.
If $\alpha\in\Phi$ is a root of $L$ to $H$, and $L_\alpha=\{x\in L:[hx]=\alpha(h)x,\ \forall h\in H\}$, then $[H_iL_\alpha]\neq 0$ for some $i$, and then $L_\alpha\subset L_i$.

I just don't understand why this forces to $L_\alpha$ to be completely in $L_i$?

We have $0\neq [H_i,L_\alpha] = \{\alpha(h)x\,\big|\, h\in H_i, x\in L_\alpha\}$. In particular there exists $h\in H_i$ such that $\alpha(h) \neq 0$. But this means $[H_i,L_\alpha]= L_\alpha$. Since $L_i\subseteq L$ is a simple ideal, we also have $[L_i,L_\alpha]\subseteq L_i$. Putting things together, we arrive at
$$L_\alpha = [H_i,L_\alpha] \subseteq [L_i,L_\alpha] \subseteq L_i.$$
