Two questions about changing variables in multiple integral

Question

Suppose we want to evaluate the double integral $\mathbf F(x, y)$ under domain $\mathit D$. And suppose there is a function $\mathbf φ(u, v)$ which transform $\mathbf F(x, y)$ into $\mathbf F \circ \mathbf φ(u, v)$. Then the area of each small rectangle $dxdy$ can be approximated by $|det \mathbf J \mathbf φ|dudv$. And we have neglected some higher order infinitesimal area. And here comes my first question: If we restrict our domain $\mathit D$ to be a bounded region. This works obviously. But what if our region $\mathit D$ is an unbounded region? I suspect the summation of those neglected area may not be neglectable. Are there any examples that the Jacobian transformation fails to work in an unbounded region?

My second question is the case of two and three variables in Jacobian transformation can be interpreted geometrically by the area of a curve parallelogram. However, how can we prove that Jacobian transformation is still true in higher dimension since I think the geometric meaning fails in dimension higher than fourth.

PS. I am sorry for the possibility of unclear of my question. Since some partial derivative symbol are hard to type so I try my best to express in words.

Answer 1

tl;dr: The theorem is true for unbounded domains/functions, and also true in arbitrary dimensions. The level of generality with which you see the theorem depends very much on your background knowledge. If you want a rigorous proof (as mentioned in comments) read an appropriate textbook.

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For your first question, I think a big reason why one might typically encounter bounded domains is because the very definition of (proper) Riemann-integrability necessitates it:

Let $A\subset\Bbb{R}^n$ be a closed rectangle, and $f:A\to\Bbb{R}$ a bounded function. We say $f$ is (Darboux/Riemann) integrable on $A$ if the supremum of the lower (Darboux) sums equals the infimum of the upper (Darboux) sums. In this case, this common number is denoted as $\int_Af$.

From here, one can try to extend the definition to encompass integrability over bounded regions other than rectangles. For example, if $C\subset\Bbb{R}^n$ is a bounded subset so that it lies inside some rectangle $A$, and $f:A\to\Bbb{R}$ is a bounded function, we say $f$ is integrable over $C$ if $f\cdot \chi_C$ is Riemann-integrable over $A$ in the above sense, and in this case one defines $\int_Cf:=\int_Af\cdot \chi_C$. Now, going a step further, one might want to define the Riemann integral of a possibly unbounded function, over a possible unbounded region, and here is where a lot of care is needed to define things. There are several ways of presenting the material and several books spell this out in detail, for example, Spivak's Calculus on Manifolds, or Munkres' Analysis on Manifolds, or Hubbard and Hubbard's Vector Calculus text. Spivak and Munkres use a partition of unity to define these integrability of possibly unbounded functions on unbounded domains and prove an appropriate change of variables theorem.

Everything I've said so far only goes to indicate why (in the setting of Riemann integrals) unbounded things are always a little iffy and much care needs to be taken. So, I wouldn't be surprised if an introductory book on advanced calculus only presents the change of variables theorem in a manner with several restricted hypotheses. Anyway, if one allows for Lebesgue integrals, the theorems are much cleaner to state (partially because Lebesgue integrals deal with $\infty$ in a much simpler way):

Let $U\subset\Bbb{R}^n$ be open, and $\varphi:U\to\varphi[U]\subset\Bbb{R}^n$ a $C^1$ diffeomorphism onto its image. Then, for any Lebesgue measurable function $f:\varphi[U]\to [0,\infty]$ and any Lebesgue measurable subset $A\subset U$, we have \begin{align} \int_{\varphi[A]}f\,d\lambda&=\int_Af\circ \varphi \cdot |\det D\varphi|\,d\lambda \end{align} In particular, for a Lebesgue measurable function $F:\varphi[U]\to \Bbb{C}$, we have that $F$ is Lebesgue integrable over $\varphi[A]$ if and only if $F\circ \varphi \cdot |\det D\varphi|$ is Lebesgue integrable over $A$, in which case, \begin{align} \int_{\varphi[A]}F\,d\lambda&=\int_AF\circ \varphi \cdot |\det D\varphi|\,d\lambda \end{align}

There are several proofs of this theorem. The most logical proof (i.e the one which I can remember and always reconstruct from scratch if I'm given about an hour) of this which I know of course involves measure theory/Lebesgue integrals. The only non-trivial part is knowing that the Radon-Nikodym derivative $\frac{d(\varphi^*\lambda)}{d\lambda}$ exists (for which one has to invoke the Radon-Nikodym theorem, which is a useful albeit non-trivial theorem), and then proving that this is equal to $|\det D\varphi|$.

So, the boundedness/unboundedness of the domain of integration $A$ makes no difference here. The theorem is equally valid in the unbounded case. I should mention that the change of variables theorem is one of the most important theorems in advanced calculus, yet it is rarely proven properly/with sufficient generality for applications. The amount of work needed to make all the approximations valid is just too much if one doesn't have all the necessary background. So, to your statement

If we restrict our domain $D$ to be a bounded region. This works obviously.

I would say, even with boundedness, it's still not an easy thing to prove, and I'd say this is far from obvious. Anyway, long story short, the theorem is true in much greater generality than you may have first seen it (and definitely also in a greater generality than what I have stated above).

For your second question I'm not sure why you think there ought to be a restriction on the dimension. As I have stated the theorem above, there is no restriction on the dimenion $n$.

Answer 2

Actually to prove the change of variables theorem for nonnegative Lebesgue measurable functions it suffices (and is convenient) to assume $D$ is bounded. This is because any open subset of $\mathbb{R}^n$ is an increasing countable union of compact sets, so we can use the monotone convergence theorem to recover the unbounded case of the theorem from the bounded case.

As for your second question, the factor has geometric meaning in all dimensions. It stems from the equality $m(A(S)) = |\det(A)|m(S)$, true for all Lebesgue measurable $S \subset \mathbb{R}^n$, invertible linear maps $A \colon \mathbb{R}^n \to \mathbb{R}^n$. This can be proved heuristically by seeing how an individual row operation affects the volume of $S$, and then using the fact that $A$ is a product of row operation matrices. The common rigorous proof of this equality also uses this proof strategy.