Parallel rectangles contained in oblique rectangle

[psie]

TL;DR

In my lecture notes, there is a geometrically intuitive result that seems to require more advanced tools, but the author has omitted a proof. I'm looking for a proof sketch of the result, or perhaps even a reference to where this is proved in more detail.

I am reading these notes on measure theory. On page 27, in chapter 2 on the construction of the Lebesgue measure, in section 2.8 on linear transformations, the author presents a lemma which is not proved. I wonder, how can one prove this?

The author uses the following terminology; a (closed) rectangle $R \in \mathbb R^n$ is parallel if it is parallel to the coordinate axes, and a (closed) rectangle $\tilde{R}\in\mathbb R^n$ is oblique if it isn't parallel (a parallel rectangle is an $n$-fold Cartesian product of compact intervals, and an oblique rectangle is the image under an orthogonal transformation of a parallel rectangle). An almost disjoint collection of sets means that the sets intersect each other at most along their boundary. $v(\cdot)$ is the volume of a rectangle, i.e. the product of the length of its sides (where length of say $[a,b]$ is simply $b-a$). Note, that since an orthogonal transformation preserves lengths and angles, we have $v(\tilde{R})=v(R)$.

Here's the lemma:

Lemma 2.29. If an oblique rectangle $\tilde{R}$ contains a finite almost disjoint collection of parallel rectangles $\{R_1, R_2 \dots, R_N\}$ then
$$\sum_{i=1}^N v(R_i) \leq v(\tilde{R}).$$

And the motivation that follows the Lemma is the following:

This result is geometrically obvious, but a formal proof seems to require a fuller discussion of the volume function on elementary geometrical sets, which is included in the theory of valuations in convex geometry. We omit the details.

Do you know what the author could mean by "fuller discussion of the volume function"? If you do know a proof, I'd be grateful for a sketch.

 

[BvU]

I'd be grateful for a sketch.

I think the least you can do is take $n=2$ and post a simple sketch

$\$

 

[psie]

If by sketch you mean a drawing, then I don't see how this could prove the lemma, even if one were to only consider $n=2$. What I meant by sketch is an outline of how one would go about proving the lemma.

Hence, I want to append some background material to my above post.

The author notes prior to this lemma that $\mu(R)=v(R)$ for parallel rectangles $R$, where $\mu$ is Lebesgue measure, but it has not yet been shown that $\mu(\tilde{R})=v(\tilde{R})$ for oblique rectangles $\tilde{R}$. Here's a proposition that has been proved earlier, I don't know if it's useful:

Proposition 2.6. If a rectangle $R$ is an almost disjoint, finite union of rectangles $\left\{R_{1}, R_{2}, \ldots, R_{N}\right\}$, then $$v(R)=\sum_{i=1}^{N} v\left(R_{i}\right).$$ If $R$ is covered by rectangles $\left\{R_{1}, R_{2}, \ldots, R_{N}\right\}$, which need not be disjoint, then $$v(R) \leq \sum_{i=1}^{N} v\left(R_{i}\right).$$

 

[Office_Shredder]

This seems surprisingly annoying. I guess the thing they're referring to is that volume is a valuation on convex bodies. This is not really that different from just observing that volume is the Riemann integral of the indicator function, which is always going to be Riemann integrable for convex sets, and since the indicator function is non negative integrating over the oblique rectangle minus the parallel interior rectangle gives a non negative number. (Since volume on convex bodies is probably defined by the Riemann integral, I'm not sure how else you would do it)

 

[Erland]

I think the simplest way to do this is to prove that an affine transformation (linear tranformation + translation) changes the measure of by constant factor, namely the absolute value of the determinant of the matrix that gives the linear transformation. Prove this for elementary linear transformations first (permutations of coordinates, dilations/contractions along one axis, reflections in an axis, shear transformations). Then, it must hold for products of such transformations, hence for all linear transformations (and affine transformations by applying translations). Since the determinant of an orthogonal matrix is +1 or -1, an orthogonal transformation (plus a translation) preserves measure, and since it also preserves length, it preserves volume, so the volume of an oblique rectangle must be the same as a the volume of a congruent parallell rectangle.

 

[jbergman]

TL;DR Summary: In my lecture notes, there is a geometrically intuitive result that seems to require more advanced tools, but the author has omitted a proof. I'm looking for a proof sketch of the result, or perhaps even a reference to where this is proved in more detail.

I am reading these notes on measure theory. On page 27, in chapter 2 on the construction of the Lebesgue measure, in section 2.8 on linear transformations, the author presents a lemma which is not proved. I wonder, how can one prove this?

The author uses the following terminology; a (closed) rectangle $R \in \mathbb R^n$ is parallel if it is parallel to the coordinate axes, and a (closed) rectangle $\tilde{R}\in\mathbb R^n$ is oblique if it isn't parallel (a parallel rectangle is an $n$-fold Cartesian product of compact intervals, and an oblique rectangle is the image under an orthogonal transformation of a parallel rectangle). An almost disjoint collection of sets means that the sets intersect each other at most along their boundary. $v(\cdot)$ is the volume of a rectangle, i.e. the product of the length of its sides (where length of say $[a,b]$ is simply $b-a$). Note, that since an orthogonal transformation preserves lengths and angles, we have $v(\tilde{R})=v(R)$.

Here's the lemma:


And the motivation that follows the Lemma is the following:

Do you know what the author could mean by "fuller discussion of the volume function"? If you do know a proof, I'd be grateful for a sketch.

I thought one could define the volume of the oblique rectangle as the limit of smaller and smaller parallel rectangles. In particular for almost disjoint rectangles the volume of them is the sum of their volumes. Then it seems somewhat trivial to show that the volume of the oblique triangle is at least as big as the volume of the parallel rectangles with a bit of care by choosing a grid such that the parallel rectangles volumes don't change under the limiting process.