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Let $I$ be a special rectangle in $\mathbb{R}^n$, and denote $\lambda(A)$ the measure of $A$. Prove that the following conditions are equivalent:

a) $\lambda(I)=0$

b) $I^{\circ}=\emptyset$ (i.e., the interior of $I$ is empty)

c) $I$ is contained in an affine subspace of $\mathbb{R}^n$ having dimension smaller than $n$. (An affine subspace is any set of the form $\{x_0+x|x\in E\}$, where $x_0\in\mathbb{R}^n$ is fixed and $E$ is a subspace of the vector space $\mathbb{R}^n$. Its dimension is equal to the dimension of $E$.)

The implication $a\implies b$ isn't too bad (using the definition if $I=[a_1,b_1]\times...\times[a_n,b_n]$ then $\lambda(I)=(b_1-a_2)...(b_n-a_n)$. So we conclude $a_i=b_i$ for some $i$. And since $I^{\circ}$ is open, there can't be anything contained in it). I also see how c) makes sense (at least intuitively), but not sure how to show it formally. Thank you.

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1 Answer 1

$a\implies b$: If $a_i=b_i$ for some $i$, then $(a_i,b_i)=\emptyset$ and the cartesian product of empty set and any set is empty.

$b\implies c$: Evidently, for some $i$, $a_i=b_i$ because $I^{\circ}=[a_1,b_1]^{\circ}\times...\times [a_n,b_n]^{\circ}$. So $I$ is of the form $x_0\times...\times [a_n,b_n]$ for some $x_0$. This is our $x_0$ in the definition of affine space. Now take $E=\{0\}\times\Bbb{R}^{n-1}$.

$c\implies a$: Here we assume that for any $\textbf{x}\in I$, $(x_1,...,x_n)=x_0+\textbf{e}$ for some $\textbf{e}\in E$. If $\dim E=m<n$, then $\textbf{x}=x_0+\textbf{e}=x_0+(e_1,...,e_m,0,...,0)$, which shows that at least for one $i$, $a_i=b_i=0$.

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