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Let $A$ be a non atomic measure on $\mathbb R$. Consider the product measure $\mu := A \times \dots \times A$ on $\mathbb R^n$.

Question: Let $M$ be a $n-1$ dimensional smooth submanifold of $\mathbb R^n$. Is it true that $M$ has measure $0$ under $\mu$?

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  • $\begingroup$ If there's a projection $p_i$ whose fibers are countable, you can use Cavalieri's principle. $\endgroup$
    – user130903
    Commented Dec 25, 2021 at 6:06

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A submanifold $M$ of ${\mathbb R}^{n}$ (with codimension 1) can always (after permuting the coordinates) be locally represented in the form $G_f=\{(x,y) : y=f(x)\}$ where $x$ runs over $V \subset {\mathbb R}^{n-1}$ and $f:V \to {\mathbb R}$. That implies that $M$ is contained in a countable union of such graphs. Each graph has measure zero under the given product measure by Fubini since $A$ is nonatomic. Thus $M$ indeed has measure zero under the product measure by countable additivity.

A submanifold $M$ of ${\mathbb R}^{n}$ with positive codimension is always a subset of a manifold of codimension 1.

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  • $\begingroup$ Hm, the comment on the graph having measure $0$ by Fubini - this implicitly assumes that the induced measure after permuting the coordinates is still a product of nonatomic measures right? How does one see this? $\endgroup$
    – Nate River
    Commented Dec 26, 2021 at 0:33
  • $\begingroup$ @Nate River: You just permute the coordinates, relative to which it is a product of measures. So you really only permute these non-atomic measures. $\endgroup$
    – user130903
    Commented Dec 26, 2021 at 5:54
  • $\begingroup$ Permuting the coordinates preserves the product measure $A \times\ldots \times A$. And this is just a notational device so the map $f$ takes the first $n-1$ coordinates to the last one. $\endgroup$
    – Yuval Peres
    Commented Dec 26, 2021 at 10:07
  • $\begingroup$ Oh, its literally only a permutation of coordinates, instead of a more general transformation. That makes sense, thanks! I actually did not know you could always write it this way as a graph with only a permutation of coordinates. @Yuval Peres $\endgroup$
    – Nate River
    Commented Dec 26, 2021 at 10:22

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