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Let $f : [0,1]^d \to [0,1]$ be a "nice" function and consider the uniform distribution on (i.e., Lebesgue measure on $\mathbb R^{d+1}$ restricted to) the set $$\mathcal X = \{(x,y): x \in [0,1]^d,\; y \in[0, f(x)]\}.$$ Let us construct a subset $\mathcal X_1$ of $\mathcal X$ by picking at most $n$ $y$-points for any given $x$. That is, $|\{ y: (x,y) \in \mathcal X_1\}| \le n$ for all $x \in [0,1]^d$. Can we use Fubini theorem to conclude that $\mathcal{X}_1$ has measure zero? Intuitively, this should be true, on the other hand $\mathcal X_1$ is an uncountable union of finite sets and potentially uncountable which makes the conclusion a bit uncomfortable.

EDIT: I guess part of the problem might be even the measurability of such set. For concreteness, for every $x$ the $y$-points chosen are zeros of a polynomial of degree $n$ whose coefficients are given by "nice" fixed functions of $x$.

EDIT: In anticipation of a counter-example, what is a general regularity condition for this selection process to make the resulting set a measurable set with measure zero.

passerby51
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    "intuitively, this should be true" -- why? – Chill2Macht Nov 22 '16 at 06:33
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    @William, usually intuition does not need a proof (!) ...and because if $x$-domain was countable it would be true, and maybe hopefully in my case the $x$-domain can be approximated by something countable (the polynomial example). Another case is a smooth surface/curve $\mathcal{X}_1 = { (x,c(x)): x \in [0,1]^d }$. It is just hard for me to think of a counter-example, but I guess someone has one. My intuition is that if the process is sufficiently "regular" this should be true. Not sure what regular means in this context. – passerby51 Nov 22 '16 at 06:44
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    I get what you're saying -- a curve is an uncountable point set, but still has Lebesgue measure zero in $\mathbb{R}^d$ for $d > 1$. And your intuition is that this set should somehow be similar to "$n$ curves" in $\mathbb{R}^{d+1}$. I don't think I know the answer to your question -- in any case you might want to provide more details about the specific context you have in mind because it seems like they would be necessary to determine a yes/no answer -- e.g. which polynomials, etc. – Chill2Macht Nov 22 '16 at 06:56
  • @William, yes (more or less). – passerby51 Nov 22 '16 at 06:58
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    @William In fact there are simple curves in $\mathbb R^2$ (homeomorphs of the closed unit interval) which have positive Lebesgue measure. – bof Nov 22 '16 at 07:02
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    @William, thanks. Yes, probably more details are needed. It is related to this other questions I asked http://math.stackexchange.com/questions/2025056/total-number-of-roots-of-polynomials-obtained-by-perturbation. I will try to provide more details when I get a chance. But if the answer to that question is positive (with uniform $\delta$), then it seems to me that we can approximate the $x$-domain in my polynomial example with a countable dense subset of $[0,1]^d$ and the result should follow. – passerby51 Nov 22 '16 at 07:04
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    @passerby51 What is the reason for stating your question in such tedious generality? Would it satisfy you to see a counterexample in the case where $n=d=1$ and $f(x)=1?$ Of course the counterexample will have to be a nonmeasurable set. – bof Nov 22 '16 at 07:05
  • @bof, OK.. thanks. I wouldn't mind the counter-example (I remembered space-filling curves, etc. after your comment.) I would change the question if I could. One question: why would it has to be nonmeasruable? I assume if I just add that $\mathcal{X}_1$ is measurable, then the application of Fubini is correct? So, it seems the most general regularity condition I need is measruability of $\mathcal{X}_1$? – passerby51 Nov 22 '16 at 07:14
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    Space-filling curves are easy, but they are not simple curves, they have self-intersections. It is harder but possible to construct a simple continuous curve (homeomorphic image of the unit interval) with positive Lebesgue measure. – bof Nov 22 '16 at 07:19
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    I believe there is a function $f:\mathbb R\to\mathbb R$ whose graph is a non-measurable set for planar Lebesgue measure. But I'd be hard pressed to prove that with what little I recall from that class in measure theory I took a long time ago. It's just an intuition or a hunch or a vague memory I have. I wouldn't try to write an answer, but maybe I can find a reference. – bof Nov 22 '16 at 07:30
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    The answer to this question should be relevant to your concerns. – bof Nov 22 '16 at 07:34
  • @bof, OK, thanks for the clarification and the pointer. The reference would be appreciated if you can find it. I am more interested in a positive answer though, so I will try to rephrase my specific problem in a new question. This has been an enlightening discussion. (I too have vague memories of measure theory, hence my questions.) – passerby51 Nov 22 '16 at 16:51

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