TITLE: The Lebesgue measure of zero set of a polynomial function is zero
QUESTION [14 upvotes]: Suppose $f :\mathbb R^n \to \mathbb R$ be a non zero polynomial(more generally smooth) function.Suppose $Z(f)=\{ x \in \mathbb R^n \mid f(x)=0 \}$. Show that  Lebesgue measure of $Z(f)$ is zero.

I am trying to use induction on $n$.The result holds obviously if $n=1$.Could someone give me some idea to prove the inductive step.The proof without induction on $n$ is also appreciated.

REPLY [3 votes]: As was mentioned in a comment above, the result doesn't hold for a general smooth function. Suppose $f(X_1, \dots, X_n)$ is a polynomial, and assume without loss of generality that all $\partial f/\partial X_i\not\equiv 0$. By the constant rank theorem, the result holds off the $Z(\partial f/\partial X_i)$. Now induct on the degree of $f$.