Zariski closed sets in C^n are of measure 0 - MathOverflow

Zariski closed sets in C^n are of measure 0

2010-05-21T17:48:31Z

This is related to another question in which it is proved that Zariski open sets are dense in analytic topology.

But it is intuitive that something more is true. Namely, that they are the sets where some polynomials vanish, and consideration of a few examples in $\mathbb R^n$ where they are of Lebesgue measure $0$, suggest strongly that the Zariski-closed sets(except the whole affine space) are of measure $0$ in $\mathbb C^n$ as well. This should be quite simple; but I am unable to prove it due to inexperience in measure theory.

The nice thing about proving this is that once this is done, then we are able to claim safely that so-and-so statement is true almost everywhere, if it is true on a Zariski-open set.

So, in a more measure theoretic formulation:

Let $X$ be a set in $\mathbb C^n$ contained in the zero locus of some collection of polynomials. How to show that $X$ is of measure $0$?

In fact my feeling is that more should be true, ie, we can replace polynomials by analytic functions at least, and get the same result.

Answer by Henri for Zariski closed sets in C^n are of measure 0

2010-05-21T18:04:41Z

Clearly, it is sufficient too show it for a closed set given by $f=0$ where $f$ is analytic. (write your set as included in a countable union of such described sets). Then, using the normal form of analytic germs as finite ramified coverings, you're done.

Answer by Sergei Ivanov for Zariski closed sets in C^n are of measure 0

2010-05-21T18:16:55Z

If a real analytic function $f:U\subset\mathbb R^n\to\mathbb R^m$ is zero on a set $Z$ of positive measure (and $U$ is connected), then $f\equiv 0$.

Indeed, almost every point of $Z$ is a density point. It is easy to see that the derivative at a density point is zero. Therefore $df=0$ a.e. on $Z$. Applying the same argument to $df$, conclude that the second derivative vanishes a.e. on $Z$ too. And so on. Thus $f$ has zero Taylor expansion at some point, hence $f\equiv 0$.

Answer by Robin Chapman for Zariski closed sets in C^n are of measure 0

2010-05-22T15:19:11Z

There is a very naive argument for this. As Henri says, it reduces to a zero set of a polynomial $f$. Write $$f(z_1,\ldots,z_n)=\sum_{j=0}^d g_j(z_1,\ldots,z_{n-1})z_n^j.$$ where the polynomial $g_d$ is not identically zero. For each $(z_1,\ldots,z_{n-1})\in\mathbb{C}^n$, there are only finitely many $z_n$ with $f(z_1,\ldots,z_n)=0$ unless $g_d(z_1,\ldots,z_{n-1})=0$. Inductively these exceptional $(n-1)$-tuples form a set of measure zero in $\mathbb{C}^{n-1}$ and now the result follows from Fubini's theorem (regarding $\mathbb{C}^n$ as $\mathbb{R}^{2n}$ and going down two real dimensions).

Answer by Michael Greenblatt for Zariski closed sets in C^n are of measure 0

2010-05-22T18:01:54Z

And if you want to take this to an extreme... for a function on a domain in ${\bf R}^n$, it's enough to assume that at every $x$ there's a ball $B_x$ centered at $x$ and a multiindex $\alpha$ for which $\partial^{\alpha} f$ is nonzero and continuous on $B_x$.

To see this, first note that it suffices to show that the zeroes of $f$ in a given $B_x$ have measure zero. This is proven by induction on $|\alpha|$. If $\alpha = 0$ it's trivial, and if $\partial^{\alpha'}f(x) \neq 0$ for any $\alpha '$ with $|\alpha '| < |\alpha|$ it follows by the induction hypothesis, shrinking $B_x$ if necessary. Otherwise we can write $\partial^{\alpha} f = \partial_{x_i}\partial^{\beta} f$ for some $\beta$, where we can assume $\partial^{\beta} f(x) = 0$. By the implicit function theorem, if $B_x$ is small enough the zeroes of $\partial^{\beta} f$ in $B_x$ form a $C^1$ hypersurface with measure zero. For each $y$ off this surface, $\partial^{\beta} f$ is nonzero and then you can apply the inductive hypothesis to find an appropriate $B_y$. A simple compactness argument shows you need only countably many $B_y$.. so you're done.