This is related to [another question](https://mathoverflow.net/questions/15569) 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.



Zariski closed sets in C^n are of measure 0
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](http://en.wikipedia.org/wiki/Lebesgue%2527s_density_theorem). 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 $f\equiv 0$.


