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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.

Math Lover
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2 Answers2

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Suppose the theorem is established for polynomials in $n-1$ variables. Let $p$ be a nontrivial polynomial in $n$ variables, say of degree $k \ge 1$ in $x_n$. We can then write $$p(\mathbf{x}, x_n) = \sum_{j=0}^k p_j(\mathbf{x}) x_n^j$$ where $\mathbf{x} = (x_1, \dots, x_{n-1})$ and $p_0, \dots, p_k$ are polynomials in $n-1$ variables, where at least $p_k$ is nontrivial.

Let us note that since $p$ is continuous, the zero set $Z(p)$ is a measurable subset of $\mathbb{R}^n$.

Now if $(\mathbf{x}, x_n)$ is such that $p(\mathbf{x}, x_n) = 0$ then there are two possibilities:

  1. $p_0(\mathbf{x}) = \dots = p_k(\mathbf{x}) = 0$, or

  2. $x_n$ is a root of the (nontrivial) one-variable polynomial $p_{\mathbf{x}}(t) =\sum_{j=0}^k p_j(\mathbf{x}) t^j$.

Let $A,B$ be the subsets of $\mathbb{R}^n$ where these respective conditions hold, so that $Z(p) = A \cup B$.

Use the inductive hypothesis to show $A$ has measure zero.

Use the fundamental theorem of algebra (its easy direction) to show that for each fixed $\mathbf{x}$, there are finitely many $t$ such that $(\mathbf{x},t) \in B$. (Indeed, there are at most $k$.) A finite set has measure zero in $\mathbb{R}$. Now apply Fubini's theorem to conclude that $B$ has measure zero. (Note that $B = Z(p) \setminus A$ is measurable.)

Nate Eldredge
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  • Technically, you're not using the fundamental theorem of algebra in the last paragraph, are you? A degree-$n$ polynomial having no more than $n$ roots only rests on the assertion that $\mathbb{R}$ or $\mathbb{C}$ are fields. – Thibaut Demaerel Jul 22 '21 at 11:18
  • @ThibautDemaerel: True, I tend to think of FTA as "a degree $n$ polynomial has exactly $n$ complex roots", but you're right that we're using the easy direction here - the hard part is "at least one root". – Nate Eldredge Jul 22 '21 at 14:40
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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 $f$.

anomaly
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