[proofplan]
We verify the defining conditions of BGG category $\mathcal O$ directly. Finite-dimensionality gives finite generation and local $U(\mathfrak n_+)$-finiteness, while the $\mathfrak h$-weight decomposition follows by Weyl complete reducibility together with the highest-weight theory of finite-dimensional simple modules. For a simple finite-dimensional module, the finite-dimensional highest-weight classification gives a dominant integral highest weight $\lambda$ and identifies the module with $L(\lambda)$. Finally, the universal property of the Verma module gives the quotient map from $M(\lambda)$, and the uniqueness of the simple highest-weight quotient gives the asserted uniqueness.
[/proofplan]
[step:Check finite generation from finite-dimensionality]
Let
\begin{align*}
\rho:\mathfrak g\to \mathfrak{gl}(V)
\end{align*}
denote the [Lie algebra](/page/Lie%20Algebra) representation defining the $\mathfrak g$-module structure on $V$. By the [universal property of the universal enveloping algebra](/theorems/9355), the representation extends uniquely to a unital algebra homomorphism
\begin{align*}
\widetilde{\rho}:U(\mathfrak g)\to \operatorname{End}_{\mathbb C}(V).
\end{align*}
Since $V$ is finite-dimensional over $\mathbb C$, choose a finite $\mathbb C$-basis $(v_1,\dots,v_m)$ of $V$. Then every $v\in V$ has the form
\begin{align*}
v=\sum_{i=1}^{m} c_i v_i
\end{align*}
with $c_i\in\mathbb C$. Because $\mathbb C\subset U(\mathfrak g)$ through scalar multiples of the unit, this expression also shows that $V$ is generated as a $U(\mathfrak g)$-module by the finite set $\{v_1,\dots,v_m\}$. Hence $V$ is finitely generated as a $U(\mathfrak g)$-module.
[/step]
[step:Decompose the module into finite-dimensional $\mathfrak h$-weight spaces]
By the [Weyl complete reducibility theorem](/theorems/3817) for finite-dimensional modules over a complex semisimple Lie algebra, the finite-dimensional $\mathfrak g$-module $V$ is a [direct sum](/page/Direct%20Sum) of simple finite-dimensional $\mathfrak g$-submodules. Thus there exist simple finite-dimensional $\mathfrak g$-submodules $V_1,\dots,V_r\subseteq V$ such that
\begin{align*}
V=V_1\oplus \cdots \oplus V_r.
\end{align*}
For each $a\in\{1,\dots,r\}$, the module $V_a$ is simple and finite-dimensional. By [citetheorem:9367] and [citetheorem:9368], there is a dominant integral weight $\lambda_a\in\mathfrak h^*$ such that $V_a$ is a finite-dimensional highest-weight module of highest weight $\lambda_a$. The finite-dimensional highest-weight theory includes the corresponding $\mathfrak h$-weight-space decomposition, so $V_a$ decomposes as a direct sum of its $\mathfrak h$-weight spaces:
\begin{align*}
V_a=\bigoplus_{\mu\in\mathfrak h^*}(V_a)_\mu,
\end{align*}
where
\begin{align*}
(V_a)_\mu=\{v\in V_a:h\cdot v=\mu(h)v\text{ for every }h\in\mathfrak h\}.
\end{align*}
Because each $V_a$ is finite-dimensional, only finitely many of the spaces $(V_a)_\mu$ are nonzero, and each nonzero $(V_a)_\mu$ is finite-dimensional over $\mathbb C$.
Define, for each $\mu\in\mathfrak h^*$,
\begin{align*}
V_\mu=\{v\in V:h\cdot v=\mu(h)v\text{ for every }h\in\mathfrak h\}.
\end{align*}
Using the direct sum decomposition $V=V_1\oplus\cdots\oplus V_r$ and the fact that each $V_a$ is stable under $\mathfrak h$, we obtain
\begin{align*}
V_\mu=(V_1)_\mu\oplus\cdots\oplus (V_r)_\mu.
\end{align*}
Therefore
\begin{align*}
V=\bigoplus_{\mu\in\mathfrak h^*}V_\mu,
\end{align*}
and every $V_\mu$ is finite-dimensional.
[guided]
The category $\mathcal O$ condition is not merely that every vector sits in some finite-dimensional space; it requires a genuine decomposition into simultaneous eigenspaces for $\mathfrak h$. For an arbitrary finite-dimensional representation, this is the point where semisimplicity of $\mathfrak g$ matters.
By the Weyl complete reducibility theorem for finite-dimensional modules over a complex semisimple Lie algebra, the module $V$ splits as a direct sum of simple $\mathfrak g$-submodules. Thus there are simple finite-dimensional $\mathfrak g$-submodules $V_1,\dots,V_r\subseteq V$ such that
\begin{align*}
V=V_1\oplus \cdots \oplus V_r.
\end{align*}
Now fix an index $a\in\{1,\dots,r\}$. Since $V_a$ is simple and finite-dimensional, [citetheorem:9367] gives a highest weight for $V_a$, and [citetheorem:9368] says that this highest weight is dominant integral. Let this highest weight be denoted by $\lambda_a\in\mathfrak h^*$. The finite-dimensional highest-weight theory also supplies the decomposition into simultaneous $\mathfrak h$-eigenspaces, namely
\begin{align*}
V_a=\bigoplus_{\mu\in\mathfrak h^*}(V_a)_\mu,
\end{align*}
where each weight space is explicitly
\begin{align*}
(V_a)_\mu=\{v\in V_a:h\cdot v=\mu(h)v\text{ for every }h\in\mathfrak h\}.
\end{align*}
Because $V_a$ is finite-dimensional as a complex [vector space](/page/Vector%20Space), only finitely many of these subspaces can be nonzero, and every nonzero one is finite-dimensional.
We now pass from the simple summands back to $V$. For each $\mu\in\mathfrak h^*$, define the $\mu$-weight space of $V$ by
\begin{align*}
V_\mu=\{v\in V:h\cdot v=\mu(h)v\text{ for every }h\in\mathfrak h\}.
\end{align*}
Since each $V_a$ is a $\mathfrak g$-submodule, it is stable under the action of $\mathfrak h$. Therefore a vector $v=v_1+\cdots+v_r$ with $v_a\in V_a$ is a $\mu$-weight vector exactly when each component $v_a$ is a $\mu$-weight vector. This gives
\begin{align*}
V_\mu=(V_1)_\mu\oplus\cdots\oplus (V_r)_\mu.
\end{align*}
Taking the direct sum over all weights gives
\begin{align*}
V=\bigoplus_{\mu\in\mathfrak h^*}V_\mu.
\end{align*}
Finally, each $V_\mu$ is a finite direct sum of finite-dimensional vector spaces, so each $V_\mu$ is finite-dimensional. This proves the weight-space condition in the definition of $\mathcal O$.
[/guided]
[/step]
[step:Verify local $U(\mathfrak n_+)$-finiteness]
Let $v\in V$. Define the $U(\mathfrak n_+)$-submodule generated by $v$ by
\begin{align*}
U(\mathfrak n_+)\cdot v=\{u\cdot v:u\in U(\mathfrak n_+)\}\subseteq V.
\end{align*}
This is a complex vector subspace of $V$. Since $V$ is finite-dimensional over $\mathbb C$, every subspace of $V$ is finite-dimensional. Hence $U(\mathfrak n_+)\cdot v$ is finite-dimensional for every $v\in V$, so $V$ is locally finite under the action of $U(\mathfrak n_+)$.
[/step]
[step:Conclude that the finite-dimensional module lies in category $\mathcal O$]
By [citetheorem:9397], the category $\mathcal O$ associated to the chosen triangular decomposition consists precisely of those left $U(\mathfrak g)$-modules that are finitely generated over $U(\mathfrak g)$, decompose into finite-dimensional $\mathfrak h$-weight spaces, and are locally finite for $U(\mathfrak n_+)$. The preceding steps verify these three conditions for $V$. Therefore $V\in\mathcal O$.
[/step]
[step:Identify a simple finite-dimensional module as a dominant integral highest-weight module]
Assume now that $V$ is simple. By [citetheorem:9367], the simple finite-dimensional $\mathfrak g$-module $V$ has a highest weight $\lambda\in\mathfrak h^*$. By [citetheorem:9368], this highest weight is dominant integral. By the finite-dimensional highest-weight classification [citetheorem:9373], the simple finite-dimensional $\mathfrak g$-module with highest weight $\lambda$ is isomorphic to the irreducible highest-weight module $L(\lambda)$. Hence
\begin{align*}
V\cong L(\lambda)
\end{align*}
as $\mathfrak g$-modules.
[/step]
[step:Use the Verma module quotient to prove uniqueness]
Let $v_\lambda\in V$ be a nonzero highest-weight vector of weight $\lambda$. By the universal property of the Verma module [citetheorem:9375], there exists a unique $\mathfrak g$-[module homomorphism](/page/Module%20Homomorphism)
\begin{align*}
\pi:M(\lambda)\to V
\end{align*}
sending the canonical highest-weight vector of $M(\lambda)$ to $v_\lambda$. Since $V$ is simple and $v_\lambda\neq 0$, the image $\operatorname{im}\pi$ is a nonzero $\mathfrak g$-submodule of $V$, so $\operatorname{im}\pi=V$. Thus $\pi$ is surjective.
By [citetheorem:9379], the Verma module $M(\lambda)$ has a unique simple quotient, namely
\begin{align*}
L(\lambda)=M(\lambda)/J(\lambda),
\end{align*}
where $J(\lambda)$ is the unique maximal proper submodule of $M(\lambda)$. Therefore any simple quotient of $M(\lambda)$ is isomorphic to $L(\lambda)$. Since every finite-dimensional simple quotient is, in particular, a simple quotient, every finite-dimensional simple quotient of $M(\lambda)$ is isomorphic to $L(\lambda)$. Combining this with $V\cong L(\lambda)$ proves the asserted uniqueness up to isomorphism and completes the proof.
[/step]
