I know that if some specific statement $S$ holds for $x\in\mathbb C^n$, then $x$ belongs to a constructible set $X\subset C^n$ (in Zariski topology) of dimension strictly lees than $n$.

${\bf Question\ 1:}$ Can I say that

the statement $S$ does not hold for almost all $x\in \mathbb C^n$, that is $mes\{x\in\mathbb C^n:\ S \text{ holds}\}=0$?

I understand that the original formulation is stronger than the reformulation in terms of measure. On the other hand, the second statement is more usual in engineering community (I am preparing a paper on engineering).

My proof is: since $X$ is constructible, it follows that there exist nonzero polynomials $p_1,..,p_k$, $q_1,\dots, q_l$ such that $$ X=\{x\in\mathbb C^n:\ p_1(x)=0,\dots,p_k(x)=0, q_1(x)\ne 0,\dots, q_l(x)\ne 0\}. $$ Hence, $$ X\subset Y:=\{x\in\mathbb C^n:\ p_1(x)=0\}. $$ It is well known that the measure of $Y$ is zero.

${\bf Question\ 2:}$ Can I immediately conclude that

the statement $S$ does not hold for almost all $x\in \mathbb R^n$, that is $mes\{x\in\mathbb R^n:\ S \text{ holds}\}=0$?

Proof: assume that $mes\{x\in \mathbb R^n:\ S \text{ holds}\}>0$, then $mes\{x\in \mathbb R^n:\ S \text{ holds and } q_1(x)\ne 0,\dots, q_l(x)\ne 0\}>0$. This means that each polynomial $p_i$ vanishes on the set of nonzero measure and hence $p_i$ is identical zero.

  • $\begingroup$ In question 1, do you mean to write that the set of points for which $S$ does hold has measure $0$? Or do you want to change the role of $X$ above? $\endgroup$
    – Stephen
    Jun 11, 2013 at 15:05
  • $\begingroup$ @Steve I have corrected questions (there were typos with negations) $\endgroup$ Jun 11, 2013 at 15:15
  • 1
    $\begingroup$ I certainly agree with the conclusions: given a constructible set $X$ of dimension one less than the ambient space, its measure in the ambient space (and that of its set of real points in the ambient real space) is zero. The proofs at first read seem basically right as well. $\endgroup$
    – Stephen
    Jun 11, 2013 at 15:24


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.