[proofplan]
We construct $L(\lambda)$ for each dominant integral weight by taking the irreducible highest weight module of highest weight $\lambda$ and invoking the finite-dimensionality theorem for dominant integral highest weights. Surjectivity follows by starting with an arbitrary finite-dimensional irreducible module, finding a highest weight vector, proving its highest weight is dominant integral, and then identifying the module with $L(\lambda)$ by uniqueness at fixed highest weight. Injectivity follows by observing that an isomorphism preserves weights and highest weight vectors, while the PBW weight bound forces the two highest weights to be comparable in both directions in the dominance order.
[/proofplan]
[step:Fix the dominance order determined by the chosen simple roots]
Define a partial order $\le$ on $\mathfrak h^*$ by declaring that, for $\mu,\lambda\in\mathfrak h^*$,
\begin{align*}
\mu\le \lambda
\end{align*}
means
\begin{align*}
\lambda-\mu\in \sum_{\alpha\in\Delta}\mathbb Z_{\ge 0}\alpha .
\end{align*}
This is the dominance order associated to the chosen positive root system $\Phi^+$. A vector $v$ in a $\mathfrak g$-module $V$ is a highest weight vector of weight $\lambda\in\mathfrak h^*$ if $v\ne 0$,
\begin{align*}
\mathfrak n^+v=0,
\end{align*}
and
\begin{align*}
hv=\lambda(h)v
\end{align*}
for every $h\in\mathfrak h$.
[/step]
[step:Construct a finite-dimensional irreducible module for each dominant integral weight]
Let $\lambda\in P^+$. By the existence and uniqueness of simple highest weight modules, applied with highest weight $\lambda$, there exists an irreducible highest weight module $L(\lambda)$ with highest weight $\lambda$; equivalently, $L(\lambda)$ contains a nonzero vector $v_\lambda$ such that
\begin{align*}
\mathfrak n^+v_\lambda=0
\end{align*}
and
\begin{align*}
h v_\lambda=\lambda(h)v_\lambda
\end{align*}
for every $h\in\mathfrak h$. This is precisely the simple quotient supplied by [citetheorem:9379].
Since $\lambda\in P^+$, the weight $\lambda$ is dominant integral by definition. Therefore the finite-dimensionality theorem for irreducible highest weight modules with dominant integral highest weight, namely [citetheorem:9371], applies to $L(\lambda)$ and gives that $L(\lambda)$ is finite-dimensional.
[guided]
Fix $\lambda\in P^+$. The goal is not merely to name a module with highest weight $\lambda$, but to ensure that it is both irreducible and finite-dimensional. The construction begins with the general highest weight theory: by [citetheorem:9379], for every $\lambda\in\mathfrak h^*$ there exists a simple highest weight module $L(\lambda)$ of highest weight $\lambda$. Thus there is a vector $v_\lambda\in L(\lambda)$ with $v_\lambda\ne 0$ such that
\begin{align*}
\mathfrak n^+v_\lambda=0
\end{align*}
and
\begin{align*}
h v_\lambda=\lambda(h)v_\lambda
\end{align*}
for every $h\in\mathfrak h$.
This gives irreducibility, but finite-dimensionality is the extra point. The hypothesis $\lambda\in P^+$ says exactly that
\begin{align*}
\lambda(H_\alpha)\in\mathbb Z_{\ge 0}
\end{align*}
for every simple root $\alpha\in\Delta$. Hence $\lambda$ is dominant integral. Therefore the hypotheses of [citetheorem:9371] are satisfied: $L(\lambda)$ is an irreducible highest weight module whose highest weight is dominant integral. That theorem gives that $L(\lambda)$ is finite-dimensional. Thus every $\lambda\in P^+$ produces a finite-dimensional irreducible highest weight module with highest weight $\lambda$.
[/guided]
[/step]
[step:Show every finite-dimensional irreducible module has one of these highest weights]
Let $V$ be a finite-dimensional irreducible $\mathfrak g$-module. The [Lie algebra](/page/Lie%20Algebra) $\mathfrak g$, the Cartan subalgebra $\mathfrak h$, and the positive root system $\Phi^+$ are exactly the fixed data in the statement, and $V$ is nonzero and finite-dimensional. Therefore [citetheorem:9367] applies and gives a weight $\lambda\in\mathfrak h^*$ and a nonzero vector $v\in V$ such that
\begin{align*}
\mathfrak n^+v=0
\end{align*}
and
\begin{align*}
hv=\lambda(h)v
\end{align*}
for every $h\in\mathfrak h$.
Let $U(\mathfrak g)$ denote the universal enveloping algebra of $\mathfrak g$. The cyclic subspace $U(\mathfrak g)v\subset V$ is a nonzero $\mathfrak g$-submodule of $V$. Since $V$ is irreducible, this submodule equals $V$. Thus $V$ is an irreducible highest weight module of highest weight $\lambda$.
By [citetheorem:9368], the highest weight of a nonzero finite-dimensional irreducible $\mathfrak g$-module is dominant integral. Hence $\lambda\in P^+$. Both $V$ and $L(\lambda)$ are finite-dimensional irreducible highest weight modules with highest weight $\lambda$, so [citetheorem:9372] gives an isomorphism
\begin{align*}
V\cong L(\lambda)
\end{align*}
as $\mathfrak g$-modules. Therefore every finite-dimensional irreducible $\mathfrak g$-module lies in the image of $\lambda\mapsto [L(\lambda)]$.
[/step]
[step:Prove distinct dominant integral weights give non-isomorphic modules]
Let $\lambda,\mu\in P^+$ and suppose that
\begin{align*}
\varphi:L(\lambda)\to L(\mu)
\end{align*}
is a $\mathfrak g$-module isomorphism. Let $v_\lambda\in L(\lambda)$ and $v_\mu\in L(\mu)$ be nonzero highest weight vectors of weights $\lambda$ and $\mu$, respectively. Since $\varphi$ is injective, $\varphi(v_\lambda)\ne 0$. For every $h\in\mathfrak h$,
\begin{align*}
h\varphi(v_\lambda)=\varphi(hv_\lambda)=\lambda(h)\varphi(v_\lambda),
\end{align*}
and, for every $x\in\mathfrak n^+$,
\begin{align*}
x\varphi(v_\lambda)=\varphi(xv_\lambda)=0.
\end{align*}
Thus $\varphi(v_\lambda)$ is a nonzero highest weight vector in $L(\mu)$ of weight $\lambda$.
By the [PBW spanning theorem for highest weight modules](/theorems/9366), [citetheorem:9366] applied to the highest weight module $L(\mu)=U(\mathfrak g)v_\mu$ shows that every weight of $L(\mu)$ has the form
\begin{align*}
\mu-\sum_{\alpha\in\Delta}m_\alpha\alpha
\end{align*}
with $m_\alpha\in\mathbb Z_{\ge 0}$. Since $\lambda$ is a weight of $L(\mu)$, this gives $\lambda\le\mu$ in the dominance order. Applying the same argument to the inverse isomorphism
\begin{align*}
\varphi^{-1}:L(\mu)\to L(\lambda)
\end{align*}
gives $\mu\le\lambda$. Hence both
\begin{align*}
\mu-\lambda\in\sum_{\alpha\in\Delta}\mathbb Z_{\ge 0}\alpha
\end{align*}
and
\begin{align*}
\lambda-\mu\in\sum_{\alpha\in\Delta}\mathbb Z_{\ge 0}\alpha.
\end{align*}
Because the simple roots $\Delta$ are linearly independent, all coefficients in these two nonnegative simple-root expansions are zero. Therefore
\begin{align*}
\lambda=\mu.
\end{align*}
The assignment $\lambda\mapsto [L(\lambda)]$ is injective.
[/step]
[step:Combine existence, surjectivity, and injectivity]
The construction step shows that the assignment
\begin{align*}
\lambda\longmapsto [L(\lambda)]
\end{align*}
is defined for every $\lambda\in P^+$. The surjectivity step shows that every isomorphism class of finite-dimensional irreducible $\mathfrak g$-modules is represented by some $L(\lambda)$ with $\lambda\in P^+$. The injectivity step shows that two dominant integral weights represent the same isomorphism class only when the weights are equal. Therefore the assignment is a bijection from $P^+$ to the set of isomorphism classes of finite-dimensional irreducible $\mathfrak g$-modules.
[/step]
