We know that (in the measure-theoretic sense) almost all square matrices $M$ over $\mathbb R$ admit an LU decomposition: $M = LDU$. We are using the Lebesgue measure, and we may treat for each fixed $n$ the space of $n \times n$ $\mathbb R$-matrices as a measurable space.
One proof of this fact uses the fact that a necessary condition for a matrix not to have an LU decomposition is that:
The matrix is singular
Or at least one of the leading principal minors is zero.
Both conditions can be expressed as polynomial equations. Finally, we know that the zero-set of a polynomial has measure zero iff the polynomial is not the zero polynomial (as a formal expression).
Anyway, I'm asking if there are other ways of showing this.