Is the set $P^{-1}(\{0\})$ a set of measure zero for any multivariate polynomial?

Question

Let $P \in \mathbb{R}[X_1,\dots,X_n]$ be a non-constant multivariate polynomial with real coefficients. Let's consider it as a map $P: \mathbb{R}^n \to \mathbb{R}$. Is it true that the Lebesgue measure of the set $$P^{-1}(\{0\})$$ is necessarily of measure zero?

Answer 1

Polynomials are continuous, so $P^{-1}(\{0\})$ is closed, and therefore Borel measurable, a fortiori Lebesgue measurable.

For $n = 1$, we know the set is finite (since we assumed $P$ non-constant), and hence has Lebesgue measure $0$.

Next assume $n > 1$, and the assertion holds for all $Q \in \mathbb{R}[X_1,\dotsc,X_{n-1}]\setminus \{0\}$. Expanding $P$ by powers of $X_n$, we can write

$$P(X_1,\dotsc,X_n) = \sum_{k = 0}^m Q_k(X_1,\dotsc,X_{n-1})\cdot X_n^k.$$

Since $P$ isn't the zero polynomial, not all $Q_k$ can be the zero polynomial, and we can choose $m$ so that $Q_m \not\equiv 0$. By the induction hypothesis, $Q_m^{-1}(\{0\})$ is a null set in $\mathbb{R}^{n-1}$, and hence

\begin{align} \lambda^n(P^{-1}(\{0\}) &= \int_{\mathbb{R}^n} \chi_{P^{-1}(\{0\})}(x)\,d\lambda^n(x) \\ &= \int_{\mathbb{R}^{n-1}} \int_{\mathbb{R}} \chi_{P^{-1}(\{0\})}(y,t)\,d\lambda(t)\,d\lambda^{n-1}(y) \\ &= \int_{\mathbb{R}^{n-1}\setminus Q_m^{-1}(\{0\})} \int_{\mathbb{R}} \chi_{P^{-1}(\{0\})}(y,t)\,d\lambda(t)\,d\lambda^{n-1}(y) \\ &= \int_{\mathbb{R}^{n-1}\setminus Q_m^{-1}(\{0\})} 0\,d\lambda^{n-1}(y) \\ &= 0, \end{align}

since for all $y \in \mathbb{R}^{n-1}\setminus Q_m^{-1}(\{0\})$, there are only finitely many $t$ such that $P(y_1,\dotsc, y_{n-1},t) = 0$.