The category of representations of $\mathfrak{g}$ is not just abelian but is isomorphic to the category of modules over a certain (associative) $k$-algebra (where $k$ is the base field). Indeed, define $U(\mathfrak{g})$ to be the free associative $k$-algebra on the underlying $k$-vector space $\mathfrak{g}$, modulo relations that say that for each $x,y\in \mathfrak{g}$, $xy-yx=[x,y]$ (here the left-hand side is computed using the multiplication of our associative algebra, and the right-hand side is the bracket in $\mathfrak{g}$). That is, we "freely" construct an associative algebra from $\mathfrak{g}$ in which the bracket becomes the commutator operation on elements of $\mathfrak{g}$. Then it is straightforward to verify that a $U(\mathfrak{g})$-module is the same thing as a $\mathfrak{g}$-representation, giving an isomorphism of categories. The algebra $U(\mathfrak{g})$ is known as the universal enveloping algebra of $\mathfrak{g}$.
For any ring $R$, you can form an $Ab$-enriched category $BR$ with one object whose endomorphisms are $R$ (with addition in $R$ being the $Ab$-enrichment and multiplication in $R$ being composition of maps). An $R$-module is then the same thing as a functor $BR\to Ab$ which preserves the $Ab$-enrichment. In particular, taking $R=U(\mathfrak{g})$, this gives a description of the representation category of $\mathfrak{g}$ as a certain functor category.
None of this depends on finite-dimensionality, or even on $k$ being a field ($k$ could be any commutative ring).
