A simple case of translation invariance of Riemann integrals

[ttsp]

Homework Statement

Show that

$$\int_{A} 1 = \int_{T(A)} 1$$

given A is an arbitrary region in R^n (not necessarily a rectangle) and T is a translation in R^n.

Homework Equations

Normally we find Riemann integrals by creating a rectangle R that includes A and set the function to be zero when x is in R\A and x is in A. I can only use the definition.

The Attempt at a Solution

I tried to create a rectangle that includes both A and T(A). However that was not useful. I'm really at a lost here.

 

[fresh_42]

You can either use the definition of Riemann integrals, which might be a bit messy and long, or you could use Stoke's theorem which reduces the task to one line. Your rectangles are actually $n-$dimensional cubes, which makes the entire thing troublesome. So what else can you use? Unfortunately your section 2 of the template is a bit too thin.

If you can use nothing than the definition, which seems to be the case, then I'll have to ask you: What is $\int_A \mathbf{1}$ in exact terms?

 

[ttsp]

You can either use the definition of Riemann integrals, which might be a bit messy and long, or you could use Stoke's theorem which reduces the task to one line. Your rectangles are actually $n-$dimensional cubes, which makes the entire thing troublesome. So what else can you use? Unfortunately your section 2 of the template is a bit too thin.

If you can use nothing than the definition, which seems to be the case, then I'll have to ask you: What is $\int_A \mathbf{1}$ in exact terms?

Indeed I can only use the definition. I would first create a rectangle $R$ such that A is included in R and $\tilde{f}$ such that $\tilde{f}(x) = 1$ if x is in A and $\tilde{f}(x) = 0$ if x is in R\A

And by definition, I have: $\int_{A} 1 = \int_{R} \tilde{f} = sup \{L(\tilde{f},P) \}$.

How should I proceed? Can I get a hint?

 

[fresh_42]

I don't understand what you wrote. Your rectangles are cubes, n-dimensional cubes. If you know already, that it is sufficient to integrate over the boundary instead - the rectangles $R$ - of the whole space $A$ inside, which is Stoke's theorem, then you're done, because translation doesn't change the boundary. Otherwise you have to deal with a volume $A \subseteq \mathbb{R}^n$ and you need to describe this volume somehow and prove that the translation doesn't change the volume. It is obviously true, so the main task is to find out how to write it. As we start with $\int_A 1$ we need to get hold on $A$ somehow.

E.g. a shear mapping does change volume as well as the boundaries, so your proof needs to make use of the difference between a translation and a shear mapping. Both are linear. Do you have volume elements $\wedge_i dx^i\,$? How to describe $A$ and then $T(A)$ is the key.

 

[ttsp]

I don't understand what you wrote. Your rectangles are cubes, n-dimensional cubes. If you know already, that it is sufficient to integrate over the boundary instead - the rectangles $R$ - of the whole space $A$ inside, which is Stoke's theorem, then you're done, because translation doesn't change the boundary. Otherwise you have to deal with a volume $A \subseteq \mathbb{R}^n$ and you need to describe this volume somehow and prove that the translation doesn't change the volume. It is obviously true, so the main task is to find out how to write it. As we start with $\int_A 1$ we need to get hold on $A$ somehow.

E.g. a shear mapping does change volume as well as the boundaries, so your proof needs to make use of the difference between a translation and a shear mapping. Both are linear. Do you have volume elements $\wedge_i dx^i\,$? How to describe $A$ and then $T(A)$ is the key.

I don't know the Stokes' theorem. Is it possible to prove it without the Stokes' theorem? Can I do it straight from the definition?

Does the volume element matter? It can be any $dx^i$. $A$ can be any arbitrary bounded region, not necessarily a cube.

 

[fresh_42]

Yes, but for Riemann integrals, you enclose it by cubes, from the outside and / or from the inside and build the infimum / supremum of those cubes. So the kernel statement is, that cube and T(cube) have the same volume. The arbitrary case are simply more cubes of all sizes to gradually fill $A$. So if you can prove if for any cube, it holds for $A$ as well.

So the first line of a proof could be. W.l.o.g. let $A$ be a cube. Then we need a formalism to describe $T(cube)$. E.g. if $Vol(A)= \wedge_i dx^i$ then what is $Vol(T(A))$?

 

[ttsp]

Yes, but for Riemann integrals, you enclose it by cubes, from the outside and / or from the inside and build the infimum / supremum of those cubes. So the kernel statement is, that cube and T(cube) have the same volume. The arbitrary case are simply more cubes of all sizes to gradually fill $A$. So if you can prove if for any cube, it holds for $A$ as well.

So the first line of a proof could be. W.l.o.g. let $A$ be a cube. Then we need a formalism to describe $T(cube)$. E.g. if $Vol(A)= \wedge_i dx^i$ then what is $Vol(T(A))$?

That's a good hint. I know how to prove it when $A$ is a cube, and I can see intuitively why that would imply the same when $A$ is an arbitrary area. However, how do I show that mathematically? Suppose that I was able to show the statement for any cube $R$. How do I generalize to any arbitrary region $A$?

 

[fresh_42]

For Riemann integrals we approximate the volume $A$ by cubes but always finitely many, and if $T(\sum_n \operatorname{Vol}(A_n))=\sum_n \operatorname{Vol}(T(A_n))$, then the question is, whether this holds for the limits as well:
$$
\int_A 1 = \lim_{n \to \infty} \sum_n \operatorname{Vol}(A_n) = \lim_{n \to \infty} \sum_n \operatorname{Vol}(T(A_n)) = \lim_{n \to \infty} T( \sum_n \operatorname{Vol}(A_n) ) = \int_{T(A)} 1
$$
$T$ has all the nice properties needed: continuous, linear, smooth, and bounded. I'm not quite sure if we also have to swap the limit and $T$ but this shouldn't be a problem, too.

 

[ttsp]

For Riemann integrals we approximate the volume $A$ by cubes but always finitely many, and if $T(\sum_n \operatorname{Vol}(A_n))=\sum_n \operatorname{Vol}(T(A_n))$, then the question is, whether this holds for the limits as well:
$$
\int_A 1 = \lim_{n \to \infty} \sum_n \operatorname{Vol}(A_n) = \lim_{n \to \infty} \sum_n \operatorname{Vol}(T(A_n)) = \lim_{n \to \infty} T( \sum_n \operatorname{Vol}(A_n) ) = \int_{T(A)} 1
$$
$T$ has all the nice properties needed: continuous, linear, smooth, and bounded. I'm not quite sure if we also have to swap the limit and $T$ but this shouldn't be a problem, too.

What are the $A_n$? Are they cubes used to approximate $A$? How do you know that we can actually approximate $A$ arbitrarily close?

 

[fresh_42]

What are the $A_n$? Are they cubes used to approximate $A$?

Yes.

How do you know that we can actually approximate $A$ arbitrarily close?

The same as on the real line with intervals or a bit more visual in the plane with rectangles. At least for finite volumes. Don't ask me for the formal procedure, though. Maybe so: Open cubes are an open covering and such we can select a finite subcover for compact sets. Now refine the covering by taking smaller cubes and so on. We get more and more, but always finitely many and the procedure should result in an approximation of $\operatorname{Vol}(A)$.