Finite-dimensional highest weights are dominant integral

Fix a simple root ${\alpha }_i$, its triple $(e_i,f_i,h_{{\alpha }_i})$ from [L1], and a highest weight vector $v$ of weight $\lambda$ generating $V$ as in [L2]; then $e_i\cdot v=0$ and $h_{{\alpha }_i}\cdot v=m_{i}v$ with $m_i=\lambda (h_{{\alpha }_i})$.

Let $W=U(\mathrm{span}\{e_i,f_i,h_{{\alpha }_i}\})v$ be the ${\mathfrak{s}\mathfrak{l}}_2$-submodule generated by $v$; it is a subspace of the finite-dimensional space $V$, so $W$ is finite dimensional, and by [L3] it is a direct sum of irreducible ${\mathfrak{s}\mathfrak{l}}_2$-submodules.

Write the direct-sum decomposition from step 2.1 as $W={⨁}_{a=1}^s{W}_a$ and decompose $v={\sum }_a{v}_a$ accordingly. Each $W_a$ is stable under $e_i$ and $h_{{\alpha }_i}$, so uniqueness of the direct sum and step 1.1 give $e_i\cdot v_a=0$ and $h_{{\alpha }_i}\cdot v_a=m_i{v}_a$ for every $a$. Since $v\mathrm{\ne }0$, some component $v_a$ is nonzero. That component is a highest weight vector of the irreducible module $W_a$ with highest weight $m_i$, so [L3] implies $m_i\in {\mathbb{Z}}_{\ge 0}$.

The argument of steps 1.1–3.1 applies to every simple root, so $⟨\lambda ,{\alpha }_i^{\vee }⟩=m_i\in {\mathbb{Z}}_{\ge 0}$ for every $i$, which by definition means that the highest weight $\lambda$ is dominant integral.