Do proper Zariski closed sets of algebraic sets have measure zero 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 $\mu_V(S)=0$.
 A: 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.
A: 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.
