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My question is very direct: considering the parametric space $\Theta = \mathbb{R}^n$, n>3, I want to know if the set $ \{ (x_1, \ldots, x_n) \in \mathbb{R^n}: x_1 = x_2 = x_3\} $ has Lebesgue measure zero in $\Theta$?

My intuition say "Yes", for the same reason pointed here: $\mathbb{R}^n\times\{0\}$ has measure zero in $\mathbb{R}^{n+1}$

But, I don't know for sure. If my intuition is correct, I would really like that someone show me the path to rigorously demonstrate it.

Renato Fernandes
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  • Do you know Fubini's theorem? – GEdgar Oct 19 '23 at 20:06
  • @GEdgar I don't remember this theorem exactly, but if there is a solution using it, I will try to understand it. – Renato Fernandes Oct 19 '23 at 20:30
  • There is a solution using it, so you probably should go try to understand the theorem; then you'll be able to compose the solution yourself. – John Hughes Oct 19 '23 at 20:42
  • @GEdgar A version of this for probability: If $X, Y$ are independent continuous r.v.'s, then $P(X = Y) = 0$. One proof of this uses Fubini's theorem if $X, Y$ have densities, but it can be proven without it using some trickery. – Jakobian Oct 19 '23 at 20:51

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Apply Sard's theorem to the linear map $(x_1, x_2, x_3, x_4, \ldots, x_n) \mapsto (x_1, x_1, x_1, x_4, \ldots x_n)$, which has rank zero, and the claim follows immediately.

Of course, this is a bit like killing a fly with a sledgehammer --- chances are good that if you're stuck on this particular claim, you've not yet encountered Sard's theorem. But in the rare cases when I want to prove something's neglible, Sard is the go-to tool in my toolbox.

John Hughes
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