Singular matrices have measure zero

Lemma. If $p\in\mathbb{R}[x_1,\dots,x_k]$ is not the zero polynomial, then $Z(p)=\{p=0\}$ has measure zero in $\mathbb{R}^k$.

*Proof by induction on $k$.* For $k=1$, a nonzero polynomial has finitely many roots. Let $k\ge2$ and write
$$p(x_1,\dots,x_k)=\sum_{j=0}^dq_j(x_1,\dots,x_{k-1})\,x_k^j,$$
with $q_d\not\equiv0$. By induction, $N=\{q_d=0\}\subset\mathbb{R}^{k-1}$ is null. For $x'=(x_1,\dots,x_{k-1})\notin N$, the polynomial $x_k\mapsto p(x',x_k)$ is nonzero, so its zero set in $\mathbb{R}$ is finite, hence null. By Fubini-Tonelli,
$$\lambda_k(Z(p))=\int_{\mathbb{R}^{k-1}}\lambda_1\big(\{x_k:p(x',x_k)=0\}\big)\,dx'=0,$$
because the integrand vanishes for $x'\notin N$ and $N$ is null. $\square$

Application. $\det A$ is a polynomial of degree $n$ in the $n^2$ entries, and it is not the zero polynomial since $\det I=1$. By the lemma $\{\det=0\}$ is null.

Remark. The same proof shows that $\{\det=0\}$ is also nowhere dense: it is closed with empty interior (a null set cannot contain a ball). So a "random" matrix is almost surely invertible, and generic statements about invertible matrices extend to all matrices by continuity, a technique used throughout this topic.