1 Fully prove steps 1 through 4 of the Change of Variables Theorem

Proof: The proof begins with several reductions which allow us to assume that f ≡ 1, that A is a small open set about a point a, and that g′(a) is the identity matrix. Then the argument is completed by induction on n with the use of Fubini’s Theorem.

(a) Step 1

∈

Then the theorem is true for all A.

Proof.
The collection of all g(U) is an open cover of g(A). Let Φ be a partition of unity subordinate to this cover. For any Riemann integrable f : g(A) → R, if φ = 0 outside of g(U), then, since g is one-to-one we have that (φf) ∘g = 0 outside of U. Hence φf is integrable and the equation

can be written as

Summing over all φ ∈ Φ yields

∑ φ∈Φ ∫ g(A)φf	= ∑ φ∈Φ ∫ A((φf) ∘ g)|detg′|
∫ g(A)f	= ∫ A|detg′|
∫ g(A)f	= ∫ A|detg′|
∫ g(A)f	= ∫ A(f ∘ g)|detg′|

(b) Step 2

Proof.
If the theorem holds for f = 1 then it holds for f = constant. Let V be a rectangle in g(A) and P a partition of V . For each subrectangle S of P, let f_S be the constant function m_S(f). Then

Likewise, letting f_S = M_S(f), we get the opposite inequality, and so that conclude that

Then as in Step 1, it follows that

(c) Step 3

Proof.
To ease the proof slightly, define X = g(A) and f′ = (f ∘ h)|deth′|. Then we have

∫ h∘g(A)f	= ∫ h(g(A))f
	= ∫ h(X)f
	= ∫ X(f ∘ h)|deth′|
	= ∫ Xf′
	= ∫ g(A)f′
	= ∫ A(f′∘ g)|detg′|
	= ∫ A|detg′|
	= ∫ A((f ∘ h) ∘ g)|detg′|
	= ∫ A(f ∘ (h ∘ g))|detg′|
	= ∫ A(f ∘ (h ∘ g))|det(h ∘ g)′|

(d) Step 4

Proof.
By steps 1 and 2, it suffices to show for any open rectangle U that

Note that for a linear transformation g, we have g′ = g. Then this is just the fact from linear algebra that a linear tranformation g : Rⁿ → Rⁿ multiplies volumes by |detg|.

2 Fully prove the Fundamental Theorem

Then f is Riemann integrable on A if and only if B has measure zero.

Proof.
Suppose first that B has measure zero.

Let ε > 0. Define B_ε = {x ∈ A : o(f,x) ≥ ε}. Now B_ε ⊂ B, hence B_ε has measure zero. By problem 13 of our previous Chapter 2, the set B_ε is closed. Since B_ε is also bounded, it is compact, and so has content zero. Hence there is a finite collection U₁,…,U_n of closed rectangles, whose interiors cover B_ε, with total volume less than ε.

Now let P be a partition of the original rectangle A which “refines” the collection of rectangles U_i in the following sense. Each rectangle S ∈ P is in one of the following two groups:

1. Group 1 (G₁): S ⊂ U_i for some i
2. Group 2 (G₂): otherwise; i.e. S is disjoint from B_ε

Since the function f is, by hypothesis, bounded on A, choose M so that |f(x)| < M for all x ∈ A. Then

for all S ∈ P. Thus since

we can divide the above difference into two parts, the first corresponding to G₁ and the other to G₂. We have the following for the first part

(2.1)

As for the second part, since each point x ∈ S ∈ G₂ has o(f,x) < ε, then any S ∈ G₂ can be further partitioned into rectangles S′ so that

Thus replacing the partitions in G₂ with these refined partitions implies the following bound

(2.2)

Putting together the partial sums from G₁ and G₂ of equations 2.1 and 2.2 yields

The value on the right-hand side can be made arbitrarily small by appropriate choice of ε and so we conclude that f is Riemann integrable.

Conversely, suppose that f is Riemann integrable. We must show that the set B has measure zero. Since B = B₁ ∪ B_1∕2 ∪ B_1∕3 ∪, it is enough to show that each B_1∕n has measure zero.

Since B_1∕n is compact, that is the same as having content zero. Since f is Riemann integrable, then given any ε > 0 we can find a partition P of A such that

Let G be the subfamily of P consisting of rectangles which meet B_1∕n. Then the rectangles S in G cover B_1∕n. Expand slightly each of these rectangles S to a rectangles S′, so that the interiors of the S′ now cover B_1∕n. Then each of the rectangles S′ contains in its interior a point x ∈ B_1∕n where the oscillation o(f,x) ≥ 1∕n. It follows from this that

Hence