# When does a proper Zariski closed set have measure zero with respect to a conditional measure?

Assume we have a probability measure $\mu$ over $\mathbb{R}^d$ that is absolutely continuous with respect to Lebesgue measure.

Given $m$ polynomials $p_1,\ldots,p_{m}\in \mathbb{R}[x_1,\ldots,x_d]$ we define a mapping from $m$ points in $\mathbb{R}^d$ to $\mathbb{R}^{m}$ $F(\bar{x}_1,\ldots,\bar{x}_m):\mathbb{R}^{d\times m} \to \mathbb{R}^{m}$ in the following way:

$F(\bar{x}_1,\ldots,\bar{x}_m)=\alpha_1,\ldots,\alpha_{m}$ if and only if for every $\bar{x}_i$ we have $\sum_{n=1}^{m} \alpha_n p_n(\bar{x}_i)+1=0$. (This is almost everywhere well defined.)

The mapping can be thought of as a mapping from $m$ points to an algebraic set: The algebraic set which is the common real zeros of a polynomial $\sum_{n=1}^m\alpha_np_n(x)+1$. For each $\bar{\alpha}$ we will denote the corresponding algebraic set by $V_\bar{\alpha}$.

Now assume that we chose our original polynomials in such a way that almost surely each $V_\bar{\alpha}$ is an irreducible algebraic set, and that we can apply the disintegration theorem and obtain a family of probability measures $\{\mu(\cdot | \alpha)\}_{\alpha}$. Each is almost surely supported on $V_\bar{\alpha}$ (on $V_\bar{\alpha}^m$ actually).

It looks kind of obvious that almost all measures $\mu(\cdot|\alpha)$ are probability measure and if $V$ is another algebraic set such that $V_\bar{\alpha} \nsubseteq V$ then $\mu(V|\bar{\alpha})=0$. (i.e. that proper Zariski closed sets of $V_\bar{\alpha}$ have measure zero). Is that true and how can it be proved?

-