# The Lebesgue measure of zero set of a polynomial function is zero

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.

• It does not hold for a smooth function. Analytic, on the other hand... Commented Sep 9, 2016 at 11:00
• Thanks i dint know that.Is there any simple counter example? Commented Sep 9, 2016 at 11:01
• $$f(x)=\cases{0&if x\leq0\\e^{-1/x^2}&if x>0}$$ Commented Sep 9, 2016 at 11:04
• @Arthur: Ah "the" famous counterexample.Thanks! Commented Sep 9, 2016 at 11:06
• Do you really need an inductive step? Is it ok if I suggest an alternate proof? Commented Sep 9, 2016 at 11:33

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.)

• 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. Commented Jul 22, 2021 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". Commented Jul 22, 2021 at 14:40

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 $f$.