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This is a question related to another question I asked: here.

Say we induce a probability measure that is absolutely continuous with respect to to Lebesgue measure onto an irreducible real algebraic set $V$. This is done by considering the $w^*$-limit of the sequence of measures $\lim_{\epsilon \to 0} \mu_{V,\epsilon}=\mu_V$, where

$\mu_{V,\epsilon}(A) := \frac{\mu(A\cap V_\epsilon)}{\mu(V_\epsilon)}$, and, $V_\epsilon := \{x:dist(x,V)<\epsilon\}$.

We assume that the induced measure, $\mu_V$, is well defined and is supported on $V$.

I would like to prove (or better, find reference) that any proper zariski closed set of $V$ has measure zero. i.e. for any irreducible algebraic set $U$ such that $U\cap V \ne V$ we have $\mu_V(U)=0$.


This is what I have: If the Minkowski dimension of an algebraic set equals its krull dimension, because than $\mu(U_\epsilon)$ decays faster than $\mu(V_\epsilon)$ and then the result might follows.

If at any non-singular point, $V$ and $U$ are differential manifolds the former works at every non singular point. Still one needs to show that if $S$ are all the singular values then $\mu_V(S)=0$.

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I don't know anything about measure theory, but if this measure has any comparison to Lebesgue measure at all, then this should follow easily. If the Zariski closed subset is proper, then it is defined by some zero sets and hence has strictly smaller (integer) dimension (for any sense of the word dimension I can think of!). –  Matt Jul 29 '13 at 13:31
    
Yes, you're right. The proper set has smaller dimension - in Krull dimension sense (and maybe in another sense too). So, one only needs to show that any set of strictly smaller dimension than $V$ will have measure zero. –  Ron Jul 29 '13 at 14:24
    
Also, strictly smaller Hausdorff dimension and the smooth locus is open, dense and is strictly smaller in terms of dimension as a real manifold. –  Matt Jul 29 '13 at 15:56
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This was asked (and answered very well by @SergeiIvanov) before: mathoverflow.net/questions/25513/… –  Igor Rivin Aug 5 '13 at 17:22
    
The main difference between the questions is that here we consider the measure of algebraic sets with respect to a non-Lebesgue continuous measure, that is supported on an algebraic set (and roughly respresent the "conditional distribution"). Second, the question discusses the reals, but this is a minor difference as far as I can see. –  Ron Aug 6 '13 at 8:54

2 Answers 2

If you are fine with closed submanifolds having measure zero then you are done, since any subvariety can be stratified into finitely many locally closed submanifolds. You construct this stratification by noetherian induction: the smooth locus of the subvariety is dense and Zariski open, so let this be a stratum, and repeat the procedure with the singular locus, which is again a subvariety.

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why does closed sub-mainfolds have measure zero? Notice that a family of a.c. measures can converge to say atomic ones (think of Gaussians which have smaller and smaller variance). By using $\sigma$-compactness of $\mathbb{R}^{n}$ and say smooth partition of unity, it becomes a really local question about affine varieties, which essentially boils down to the following thing - you've got a polynomial mapping into $\mathbb{R}^{n}$, you look at the thickening of a small patch of it, why it must becomes Lebesgue. You probably need to address self-intersections and some "isolation" –  Asaf Aug 4 '13 at 18:38
    
of the patch from the other patches, this is probably done by the polynomial behaviour of it (as it is not true for analytical mappings, say $\sin{1/x}$). I have an hunch that maybe the Dani-Margulis lemma from dynamics can help here, because it deals with precisely the situation where polynomial mappings can't be "too small" for long period of times. –  Asaf Aug 4 '13 at 18:40
    
My apologies if I gave the impression that the case of a submanifold is easy -- I'm just going off what's in the final paragraph of the original question. –  Dan Petersen Aug 4 '13 at 19:41
    
The result should not follow for any set, or for the zeros of any function. In fact, the assumption $\mu_V(V)=1$, rules out some pathological cases. I think, that it should follow for differential manifolds and submanifold of strict lower dimension, but I'm not sure. In any case, it mustn't become Lebesgue. Thanks Dan. From what I understand, the singularities are indeed not an issue, so it remains only to show that a submanifold will have measure 0. –  Ron Aug 5 '13 at 18:43

In your comment, you exhibit two important features of “having measure zero”: it is a local property and it is invariant under diffeomorphisms. This is because 1. you are considering a measure continuous with respect to the Lebesgue measure, 2. the ambient space is sigma countable and 3. locally compact

Now let us simplify the problem a bit: we only need to prove that

PROPOSITION. Any algebraic hypersurface of ${\bf C}^N$ is a null set for the Lebesgue measure.

To justify this simplification, observe that $\mu$ is continuous with respect to the Lebesgue measure, so that it has more null sets (i.e. sets of measure zero) and it is then enough to show that your algebraic variety is a null set for the Lebesgue measure. An algebraic variety is cut out by hypersurfaces, so if any (real) hypersurface is a null set, any (real) algebraic variety must also be a null set. The last point to check is that if the proposition holds it implies the analog statement for real algebraic sets. I will not go in the details, but the point is that if V is the complex zero locus of a real equation, you can cover it by acountable set of self-conjugated measurable compact sets whose union has an arbitrary small measure: you conclude by observing that the Lebesgue measure of ${\bf R}^N$ can be recovered from the Lebesgue measure of ${\bf C}^N$ with an $\epsilon$-based construction similar to yours. (To deal with this gently, use the cahracterisation of the Lebesgue measure as the Haar-mesure for the topological group ${\bf R}^N$.)

It is not a very clean exposition, but I nevertheless assume I convinced you that the PROPOSITION implies that algebraic subvarieties of ${\bf R}^N$ are null sets for the Lebesgue measure.

Proof of the PROPOSITION. Let $f$ be a complex polynomial on ${\bf C}^N$ vanishing at some point $p$. Thanks to the Weierstrass preparation theorem, we can assume that in a coordinate system centered on $p$, the function $f$ has the form $$ f(z, \zeta) = z^k + z^{k-1} g_1(\zeta) + \cdots + g_k(\zeta) $$ where $(z,\eta)$ belongs to some neighbourhood of the origin in ${\bf C} \times {\bf C}^{N+1}$. Because the (graded) algebra of symmetric polynomials in $k$ variables spanned by the $k$ elementary symmetric functions $\sigma_j$ is isomorphic to the (graded) algebra of polynomials in $k$ variables via the map $$ \Phi: \phi(X_1,\ldots,X_k) \mapsto \phi(\sigma_1, \ldots, \sigma_k) $$ there is polynomials $\gamma_1, \ldots, \gamma_k$ such that $$ f(z, \zeta) = \prod_{j = 1}^k (z - \gamma_j(\zeta)) . $$ In other words we look at the morphism from ${\bf C}^k$ to the set $M$ of monic polynomials of degree $k$ given by $$ \omega \mapsto \prod_{j = 1}^k (z - \omega_k), $$ it induces an isomorphism of algebraic varieties ${\bf C}^k / {\mathfrak{S}_k} \simeq M$ which is precisely $\Phi$. Looking at $f$ as a function of $\zeta$ with values in $M$, we define the $\gamma_j$ (up to permutation!) by composing $f$ with the map $M \to {\bf C}^k / {\mathfrak{S}_k}$.

This is very good, because we now see that the zero locus of $f$ is near $p$ the union of the graphs $z = \gamma_j(\eta)$ of $k$ functions. But a graph is diffeomorphic to a hyperplane and is thus a null set.

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I may be confused, but my measure $\mu_V$ is not continuous with respect to Lebesgue measure. In fact by assumption $\mu_V(V)=1$ where $V$ is an algebraic set (hence null set by Lebesgue measure). –  Ron Aug 5 '13 at 17:53
    
Well one the one hand, you are right, if $\mu(V) = 1$ your measure is probably not continuous with repsect to Lebesgue measure, but on the other hand, the first sentance of your question is “Say we induce a probability measure that is absolutely continuous with respect to to Lebesgue measure” so I got tripped. :) –  Michael Grünewald Aug 5 '13 at 18:26

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