[proofplan]
The Harish-Chandra parametrization identifies equality of central characters for highest-weight modules with equality of the Weyl group orbits of the shifted weights $\lambda+\rho$ and $\mu+\rho$. Since $\lambda$ and $\mu$ are dominant integral, the shifted weights are strictly dominant: they pair positively with every simple coroot. A Weyl group orbit contains at most one strictly dominant element, so the equality of shifted orbits forces $\lambda+\rho=\mu+\rho$. Subtracting $\rho$ gives $\lambda=\mu$, and the equivalent module formulation follows from the highest-weight classification of finite-dimensional simple modules.
[/proofplan]
[step:Use the Harish-Chandra parametrization to put the shifted weights in one Weyl orbit]
Let $\lambda,\mu\in P^+$ and assume $\chi_\lambda=\chi_\mu$. Since $L(\lambda)$ and $L(\mu)$ are simple highest-weight modules with highest weights $\lambda$ and $\mu$, the [Harish-Chandra parametrization of central characters](/theorems/9398), in the $\rho$-shift convention, applies to these two modules. By [citetheorem:9398], equality of central characters gives an element $w\in W$ such that
\begin{align*}
\mu+\rho = w(\lambda+\rho).
\end{align*}
[guided]
Let us isolate exactly where the center enters. The modules $L(\lambda)$ and $L(\mu)$ are simple highest-weight $\mathfrak g$-modules, so the Harish-Chandra parametrization of central characters applies to them. In the convention used here, the central character of a highest-weight module of highest weight $\eta$ is determined by the Weyl orbit of the shifted weight $\eta+\rho$, not by the orbit of $\eta$ itself.
The hypothesis is $\chi_\lambda=\chi_\mu$. Applying [citetheorem:9398] to the two simple highest-weight modules $L(\lambda)$ and $L(\mu)$ gives the existence of an element $w\in W$ satisfying
\begin{align*}
\mu+\rho = w(\lambda+\rho).
\end{align*}
This is the only point in the proof where the structure of $Z(U(\mathfrak g))$ is used. The remaining argument is a statement about Weyl chambers.
[/guided]
[/step]
[step:Show that the shifted dominant weights are strictly dominant]
Let $\Delta\subset\Phi^+$ denote the set of simple roots. Since $\lambda,\mu\in P^+$, for every $\alpha\in\Delta$ one has
\begin{align*}
\lambda(\alpha^\vee)\in\mathbb Z_{\ge 0}
\end{align*}
and
\begin{align*}
\mu(\alpha^\vee)\in\mathbb Z_{\ge 0}.
\end{align*}
By the definition of $\rho$, one has $\rho(\alpha^\vee)=1$ for every simple root $\alpha\in\Delta$. Hence
\begin{align*}
(\lambda+\rho)(\alpha^\vee)=\lambda(\alpha^\vee)+1>0
\end{align*}
and
\begin{align*}
(\mu+\rho)(\alpha^\vee)=\mu(\alpha^\vee)+1>0.
\end{align*}
Thus $\lambda+\rho$ and $\mu+\rho$ both lie in the interior of the dominant Weyl chamber.
[/step]
[step:Prove that a Weyl orbit has at most one strictly dominant representative]
We prove the root-system fact needed in this case. Let $\eta\in\mathfrak h^*$ satisfy $\eta(\alpha^\vee)>0$ for every $\alpha\in\Delta$. Suppose $w\in W$ and $w\eta$ is also strictly dominant. We show that $w=e$.
If $w\ne e$, then there exists a simple root $\alpha\in\Delta$ such that $w^{-1}\alpha\in-\Phi^+$. Write $w^{-1}\alpha=-\beta$ with $\beta\in\Phi^+$. Then
\begin{align*}
(w\eta)(\alpha^\vee)=\eta(w^{-1}\alpha^\vee)=\eta(-\beta^\vee)=-\eta(\beta^\vee).
\end{align*}
The coroot $\beta^\vee$ is a positive coroot, hence it is a nonnegative integral linear combination of the simple coroots, not all coefficients being zero. Since $\eta$ is positive on every simple coroot, it follows that $\eta(\beta^\vee)>0$. Therefore
\begin{align*}
(w\eta)(\alpha^\vee)<0,
\end{align*}
contradicting the strict dominance of $w\eta$. Hence $w=e$.
Applying this to $\eta=\lambda+\rho$ and using $\mu+\rho=w(\lambda+\rho)$, we obtain
\begin{align*}
\mu+\rho=\lambda+\rho.
\end{align*}
Therefore $\lambda=\mu$.
[/step]
[step:Deduce the equivalent statement for arbitrary finite-dimensional simple modules]
Let $V_1$ and $V_2$ be finite-dimensional simple $\mathfrak g$-modules with the same central character. By the classification of finite-dimensional irreducible highest-weight modules, [citetheorem:9373], there exist dominant integral weights $\lambda,\mu\in P^+$ such that
\begin{align*}
V_1\cong L(\lambda)
\end{align*}
and
\begin{align*}
V_2\cong L(\mu).
\end{align*}
The central character is invariant under isomorphism of $\mathfrak g$-modules, so the common central character of $V_1$ and $V_2$ gives $\chi_\lambda=\chi_\mu$. The first part of the proof gives $\lambda=\mu$, and therefore
\begin{align*}
V_1\cong L(\lambda)=L(\mu)\cong V_2.
\end{align*}
This proves the equivalent formulation and completes the proof.
[/step]
