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?

  • 3
    $\begingroup$ Yes. Induction and Fubini. $\endgroup$ – Daniel Fischer Dec 17 '16 at 14:12

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 $P(y_1,\dotsc, y_{n-1},t) = 0$.

  • $\begingroup$ I don't quite understand this proof. You say that "since for all $y \in \mathbb{R}^{n-1}\setminus Q_m^{-1}(\{0\})$ there are only finitely many $t$ such that $P(y_1,\dots,y_{n-1},t) = 0$". That's certainly true if all the other $Q_k$ except $Q_m$ are $0$, but if they aren't it is not necessarily true imho. $\endgroup$ – user159517 Dec 25 '16 at 12:31
  • $\begingroup$ @user159517 For fixed $y_1,\dotsc, y_{n-1}$, we have a univariate polynomial in $t$, namely $$\sum_{k = 0}^m Q_k(y_1,\dotsc, y_{n-1})\cdot t^k,$$ and such a polynomial has only finitely many zeros (perhaps none at all), unless it is the zero polynomial. It suffices that any of the $Q_k$ for $0 \leqslant k \leqslant m$ is nonzero at $(y_1,\dotsc, y_{n-1})$. I chose $m$ so that $Q_m \not\equiv 0$, thus its zero set is by the induction hypothesis a Lebesgue null set in $\mathbb{R}^{n-1}$, and for all $y$ outside that, the section $t\mapsto \sum_{k = 0}^m Q_k(y)t^k$ is a polynomial of degree $m$. $\endgroup$ – Daniel Fischer Dec 26 '16 at 16:49
  • $\begingroup$ And therefore that section has at most $m$ zeros in $\mathbb{R}$ (typically fewer, since most polynomials have some nonreal zeros). $\endgroup$ – Daniel Fischer Dec 26 '16 at 16:51
  • $\begingroup$ That certainly clarifies things for me. Thank you and happy holidays! $\endgroup$ – user159517 Dec 27 '16 at 1:25

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