Proper algebraic subsets of M_n(R) have measure zero

Reduction. A proper algebraic subset $Z=\{p_1=\dots=p_m=0\}\subsetneq M_n$ must have some $p_i\not\equiv0$ (if all were $0$, $Z$ would be everything). Then $Z\subseteq\{p_i=0\}$, so it suffices to treat one nonzero polynomial $p$ on $\mathbb{R}^k$, $k=n^2$.

Claim. If $p\not\equiv0$ is a polynomial on $\mathbb{R}^k$, then $\{p=0\}$ has Lebesgue measure $0$.

*$k=1$:* a nonzero polynomial has finitely many roots.

*Induction step.* Assume the claim in dimension $k-1$. Write
$$p(x',x_k)=\sum_{j=0}^dq_j(x')x_k^j,\qquad x'=(x_1,\dots,x_{k-1}),$$
with $q_d\not\equiv0$ (the top coefficient is a nonzero polynomial in $x'$). Let $Z'=\{q_d=0\}\subset\mathbb{R}^{k-1}$, null by the induction hypothesis.
- For $x'\notin Z'$, $x_k\mapsto p(x',x_k)$ is a polynomial of degree exactly $d$, so it has at most $d$ roots: the slice $\{x_k:p(x',x_k)=0\}$ is finite, hence null.
- The set $\{p=0\}$ is closed, hence measurable, so by Tonelli
$$\lambda_k(\{p=0\})=\int_{\mathbb{R}^{k-1}}\lambda_1\bigl(\{x_k:p(x',x_k)=0\}\bigr)dx'=\int_{Z'}(\cdots)dx'+\int_{\mathbb{R}^{k-1}\setminus Z'}0\,dx'=0,$$
because $Z'$ is null.

Remark. So "generic" in the sense of Zariski (complement of an algebraic set) implies "almost every" in the sense of Lebesgue. For example, almost every matrix is invertible, has distinct eigenvalues and is diagonalisable over $\mathbb{C}$. Lesson. Induction on the number of variables plus Fubini is the standard way to handle null sets of polynomials.