For semi-simple Lie group, there exist a bi-invariant non-degenerated metric $g$ on Lie group and the metric in the Tangent space of indentiy is Killing form. Certainly for manifold with non-degenerated metric, we can construct the Haar measure $\sqrt{|g|}dx^1\wedge\cdots\wedge dx^n$.
My question is how to construct the Haar measure in a non-semisimple Lie group? Because the bi-invariant metric in these Lie group is degenerate.
Haar measure by definition is only left invariant, and it is not bi-invariant in general (this is measured by the so-called modular function). Hence you can simply choose a non-zero element in $\Lambda^n\mathfrak g^*$, where $\mathfrak g$ is the Lie algebra of $G$ and $n$ is its dimension. This gives rise to a left invariant form of top degree on $G$, which then induces the Haar measure. As inidcated in the comment by @Cronus, there is a general constuction for locally compact topological groups, but this is significantly more complicated.