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This is an exercise on my homework. The professor gave us a hint that we just need to prove that one dimension of the boundary has measure zero, and then we can take the countable union to have measure zero as well.

How would I define just one dimension of the boundary of the rectangle? I'm not quite sure what that even means.

Ignoring the hint, here are some preliminary thoughts that I had: So I have $R$, a rectangle in $\mathbb{R}^n$ with $R=I_1\times ...\times I_n$, where $I_i,i=1,...,n,$ are bounded intervals in $\mathbb{R}^n$ with lower bound $a_i$ and upper bound $b_i$. I know that the union of the boundaries of $I_i$ have measure zero, and that the boundary of the unions of $I_i$ has measure zero (although I do not yet know how to prove that second part). Can I relate the boundaries of $I_i$ to the boundary of $R$ in a meaningful way?

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The boundaries of the $I_i$ are in different space so you cannot really consider them. Think about the rectangle $R$. It's boundary consists of the "walls" where one coordinate is fixed. You only need to consider one of these walls. The wall is contained in a hyperspace (a plane in $3d$) so you can actually consider the whole hyperspace and prove that it has measure zero. You can do that by covering it with rectangles that have appropriately small (something like $\epsilon * 2^{-k}$) width in the fixed dimension.

On the other hand, it might be easier to consider just the "wall" if it is bounded, since then one covering rectangle would suffice, but you can do it more generally with proving that a hyperspace has measure zero.

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  • $\begingroup$ Why is it sufficient to consider only one wall, even if two of the total four walls would have a different length? How can one generalise a "measure" done on e.g. that smaller wall to the larger wall without considering their length differences? $\endgroup$ – mavavilj Jan 29 '16 at 1:48
  • $\begingroup$ @mavavilj You do it for an arbitrary wall. A wall is always contained in a hypersurface. $\endgroup$ – ploosu2 Feb 3 '16 at 14:13

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