= Solution
First consider finite-dimensional <Lie algebra representations>. The operators representing an <abelian Lie algebra> commute. They have a common <eigenvector>: take an <eigenspace> for the first operator of a <basis> of the algebra, observe that every remaining operator preserves it, and repeat inside this nonzero space. The resulting line is an <invariant subspace>, so an <irreducible representation> must be one-dimensional. Its action has the form $xv=\lambda(x)v$ for a <linear functional> $\lambda$. Conversely, every such functional defines a <Lie algebra representation>, since both the <Lie brackets> in the algebra and the <commutators> of its scalar operators vanish. Distinct functionals give nonisomorphic representations. Thus \b[the irreducibles are precisely the characters $\lambda\in\mathfrak g^*$].
Even without a finite-dimensional assumption on the module, this conclusion for a finite-dimensional <abelian Lie algebra> follows from the <Weak Hilbert Nullstellensatz>: its <universal enveloping algebra> is the <polynomial> algebra $\operatorname{Sym}(\mathfrak g)$. A simple module over a commutative algebra is its quotient by a <maximal ideal>, and every <maximal ideal> of this <polynomial> algebra over $\mathbb C$ has residue field $\mathbb C$.
\b[Indecomposable does not imply irreducible.] For the one-dimensional <abelian Lie algebra> $\mathbb Cx$, let $x$ act on $\mathbb C^2$ by the <Nilpotent Jordan block>
$$
N=\begin{pmatrix}0&1\\0&0\end{pmatrix}.
$$
The line spanned by the first coordinate vector is an <invariant subspace>, so the module is reducible. If it were a <direct sum> of two nonzero submodules, both would have <dimension> one; nilpotence would force $N$ to act as zero on both, contrary to $N\ne0$. Thus it is an <indecomposable representation>.
For a finite-dimensional module $V$, put $d=\dim V$ and define its <generalized weight spaces> intrinsically by
$$
V^\lambda=\{v\in V:(\rho(x)-\lambda(x)I)^d v=0\text{ for every }x\in\mathfrak g\}.
$$
To obtain the decomposition, choose a <basis> $x_1,\ldots,x_r$ of $\mathfrak g$, split into <generalized eigenspaces> of $\rho(x_1)$, and successively split each piece by $\rho(x_2),\ldots,\rho(x_r)$. All the pieces are submodules because the acting operators commute. On one resulting piece, $\rho(x_j)$ has a unique <eigenvalue> $\lambda_j$. The commuting operators can be simultaneously upper triangularized by repeating the common-eigenvector argument on quotients. Hence on this piece every $\rho(x_j)-\lambda_j I$ is strictly upper triangular. Any <linear combination> is also strictly upper triangular, and its $d$th power vanishes. Extending $\lambda_j$ linearly therefore identifies the piece with $V^\lambda$. Different tuples give disjoint pieces; if a component has a different tuple from $\lambda$, at least one $\rho(x_j)-\lambda(x_j)I$ is invertible there. Consequently
$$
\boxed{V=\bigoplus_{\lambda\in\mathfrak g^*}V^\lambda,}
$$
with only finitely many nonzero summands. Since the definition uses every $x$ rather than a chosen <basis>, this <generalized-weight decomposition for a nilpotent Lie algebra> is canonical. Its summands need not be irreducible, as the <Jordan block> example already shows.
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