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The following lemma is from Lee's "Introduction to Smooth Manifolds."

Lemma 6.2. Suppose $A\subseteq\mathbb R^n$ is a compact subset whose intersection with $\{c\}\times\mathbb R^{n-1}$ has $(n-1)$-dimensional measure zero for every $c\in\mathbb R$. Then $A$ has $n$-dimensional measure zero.

Do we need to hypothesize that $A$ is compact? I think the author included the compactness hypothesis so that the elementary proof (no measure theory) in the book will work. But if I prove the lemma using Tonelli's theorem, I think $A$ does not need to be compact:

Let $\chi_A$ be the characteristic function of $A$. We have to prove that $\int\chi_Adm^n=0$. By Tonelli's theorem, $\int\chi_Adm^n=\int\left(\int(\chi_A)_cdm^{n-1}\right)dm(c)$, where $(\chi_A)_c(x_2,\ldots,x_n)=(c,x_2,\ldots,x_n)$. Since we are given that $\{c\}\times\mathbb R^{n-1}$ has $(n-1)$-dimensional Lebesgue measure zero, it follows that $\int(\chi_A)_cdm^{n-1}=0$ for all $c\in\mathbb R$. Thus $\int\chi_Adm^n=\int 0dm(c)=0$.

Am I right, or did I miss something?

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    $\begingroup$ Compactness is not needed. $\endgroup$ – daw Nov 30 '20 at 6:26
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You're absolutely right -- the lemma is true without assuming compactness, with an easy proof based on Fubini/Torelli. And you're also right that the reason I assumed compactness was so I could give an elementary proof that doesn't use any measure theory.

My intention for my series of manifolds books (Introduction to Topological/Smooth/Riemannian Manifolds) was to assume only a decent undergraduate math degree as prerequisite, and to make the series otherwise self-contained. Since many undergraduate math programs don't cover Lebesgue integration (or at least that was the case when I started writing the books), I decided to use only Riemann integration and some very basic facts about sets of measure zero.

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