Change of variables · Selected prerequisite proofs

L10.7.2: Full compact Jordan change-of-variables theorem for Riemann-integrable amplitudes. The neighbourhood restriction and every used exercise are supplied by P19–P21.2.

Theorem 10.7.2.

Suppose

U \subset \R^n

is open,

S \subset U

is a compact Jordan measurable set, and

g \colon U \to \R^n

is a one-to-one continuously differentiable mapping, such that

J_g

is never zero on

S\text{.}

Suppose

f \colon g(S) \to \R

is Riemann integrable. Then

f \circ g

is Riemann integrable on

S

and

\begin{equation*} \int_S f\bigl(g(x)\bigr) \, \babs{J_g(x)} \, dx = \int_{g(S)} f(u) \, du . \end{equation*}

Proof.

The set

S

can be covered by finitely many closed rectangles

P_1,P_2,\ldots,P_k\text{,}

whose interiors do not overlap such that each

P_j \subset U

(Exercise 10.7.2). Proving the theorem for

P_j \cap S

instead of

S

is enough as we can simply add up the integrals. Define

f(y) \coloneqq 0

for all

y \notin g(S)\text{.}

The new

f

is Riemann integrable as

g(S)

is Jordan measurable. We can now replace the integrals over

S

with integrals over the whole rectangle. We therefore assume without loss of generality that

S

is a closed rectangle.

The matrix

g'(x)

is invertible for every

x \in S\text{,}

and it is continuous. Therefore,

{\bigl(g'(x)\bigr)}^{-1}

and consequently

\bnorm{{\bigl(g'(x)\bigr)}^{-1}}

is continuous and never zero. As

S

is compact, then there exists an

M > 1

so that

\bnorm{{\bigl(g'(x)\bigr)}^{-1}} \leq M

for all

x \in S\text{.}

Let

\epsilon > 0

be given. For every

x \in S\text{,}

let

\begin{equation*} W_x \coloneqq \left\{ y \in U : \bnorm{g'(x)-g'(y)} < \frac{\epsilon}{2M} \right\} . \end{equation*}

By Exercise 10.7.3,

W_x

is open. As

x \in W_x

for every

x\text{,}

we have an open cover. By the Lebesgue covering lemma (Lemma 7.4.10), there exists a

\delta > 0

such that for every

y \in S\text{,}

there is an

x

such that

B(y,\delta) \subset W_x\text{.}

In other words, if

Q

is a rectangle of maximum side length less than

\frac{\delta}{\sqrt{n}}

and

y \in Q\text{,}

then

Q \subset B(y,\delta) \subset W_x\text{.}

By the triangle inequality,

\bnorm{g'(\xi)-g'(\eta)} < \nicefrac{\epsilon}{M}

for all

\xi, \eta \in Q\text{.}

Let

\varphi(x) \coloneqq f\bigl(g(x)\bigr) \babs{J_g(x)}\text{.}

There exists a partition

P

S

such that

\epsilon + \int_S \varphi \geq U(P,\varphi)\text{.}

We can assume

\delta

is sufficiently small relative to the side of each subrectangle of the partition

P

so that we can cut each such subrectangle into further subrectangles each of whose sides

s

satisfies

\frac{\delta}{2\sqrt{n}} \leq s \leq \frac{\delta}{\sqrt{n}}

(Exercise 10.7.4). Denote these subrectangles by

R_1,R_2,\ldots,R_N\text{.}

For each

j=1,2,\dots,N\text{,}

find

x_j \in R_j

so that

\sabs{J_g(x_j)} \leq \sabs{J_g(x)}

for all

x \in R_j\text{,}

which is possible as

\sabs{J_g(x)}

is continuous and

R_j

is compact.

Consider some

R_j\text{.}

First suppose

x_j=0\text{,}

g(0) = 0\text{,}

and

g'(0) = I\text{.}

We claim that

g(R_j)

is contained in a rectangle of volume at most

V(R_j) \, {\bigl(1+4\sqrt{n} \, \epsilon\bigr)}^n\text{.}

Let us prove this claim. For any given

y \in R_j\text{,}

apply the fundamental theorem of calculus to the function

t \mapsto g(ty)

to find

g(y) = \int_0^1 g'(ty)y \,dt\text{.}

As the side of

R_j

is at most

\frac{\delta}{\sqrt{n}}\text{,}

we have

\snorm{y} \leq \delta\text{.}

We note that

\bnorm{g'(x)-I} < \epsilon

M > 1

and so

\begin{equation*} \begin{aligned} \bnorm{g(y)-y} = \norm{\int_0^1 \bigl(g'(ty) y - y\bigr) \,dt} & \leq \int_0^1 \bnorm{g'(ty) y - y} \,dt \\ & \leq \snorm{y} \int_0^1 \bnorm{g'(ty) - I} \,dt \leq \delta \epsilon . \end{aligned} \end{equation*}

Therefore,

g(R_j) \subset \widetilde{R}_j\text{,}

where

\widetilde{R}_j

is a rectangle obtained from

R_j

by extending by

\delta \epsilon

on all sides. See Figure 10.17.

A diagram of a gray rectangle R sub j inside a dashed rectangle tilde R sub j. A bold figure g of R sub j that is a deformed version of the gray rectangle is given and lies completely inside the dashed rectangle. The horizontal side of the gray rectangle is labeled as s sub 1 and the vertical side is labeled as s sub 2. The distance between the sides of the gray square and the dashed square is labeled as delta epsilon. A point inside all the rectangles is labeled as x sub j equals zero equals g of x sub j. A point y is labeled on the side of the gray rectangle and a nearby point within delta epsilon and on the side of the black rectangle is marked g of y. — Figure 10.17. Image of $R_j$ under $g$ lies inside $\widetilde{R}_j\text{.}$ A sample point $y \in R_j$ (on the boundary of $R_j$ in fact) is marked and $g(y)$ must lie within with a radius of $\delta\epsilon$ (also marked).

If the sides of

R_j

are

s_1,s_2,\ldots,s_n\text{,}

then

V(R_j) = s_1 s_2 \cdots s_n\text{.}

Recall

\delta \leq 2\sqrt{n} \, s_j\text{.}

Thus,

\begin{equation*} \begin{split} V(\widetilde{R}_j) & = (s_1+2\delta \epsilon ) (s_2+2\delta \epsilon ) \cdots (s_n+2\delta \epsilon ) \\ & \leq \bigl(s_1+4 \sqrt{n}\,s_1 \epsilon \bigr) \bigl(s_2+4 \sqrt{n}\,s_2 \epsilon \bigr) \cdots \bigl(s_n+4 \sqrt{n}\,s_n \epsilon \bigr) \\ & = s_1 \bigl(1+4 \sqrt{n}\, \epsilon \bigr) \, s_2 \bigl(1+4 \sqrt{n}\, \epsilon \bigr) \cdots s_n \bigl(1+4 \sqrt{n}\, \epsilon \bigr) = V(R_j) \, {\bigl(1+4\sqrt{n} \, \epsilon\bigr)}^n . \end{split} \end{equation*}

The claim is proved:

g(R_j) \subset \widetilde{R}_j

and

\begin{equation*} V\bigl(g(R_j)\bigr) \leq V(\widetilde{R}_j) \leq V(R_j) \, {\bigl(1+4\sqrt{n} \, \epsilon\bigr)}^n . \end{equation*}

Next, suppose

A \coloneqq g'(0)

is not necessarily the identity. Write

g = A \circ \widetilde{g}

where

\widetilde{g}'(0) = I\text{.}

We have that

\snorm{A^{-1}} \leq M

and so

\begin{equation*} \bnorm{\widetilde{g}'(x)-I} = \bnorm{A^{-1}g'(x)-A^{-1}A} \leq \bnorm{A^{-1}}\,\bnorm{g'(x)-A} < M \frac{\epsilon}{M} = \epsilon . \end{equation*}

In other words, the claim above applies to

\widetilde{g}

as what we needed in its proof was precisely that

\bnorm{\widetilde{g}'(x)-I} < \epsilon\text{.}

Therefore, we have that

\widetilde{g}(R_j)

is contained in a rectangle

\widetilde{R}_j

with

V(\widetilde{R}_j) \leq V(R_j) \, {\bigl(1+4\sqrt{n} \, \epsilon\bigr)}^n\text{.}

By Proposition 10.7.1,

V\bigl(A(\widetilde{R}_j)\bigr) = \babs{\det(A)} \, V(\widetilde{R}_j)\text{,}

and hence

\begin{equation*} \begin{split} V\bigl(g(R_j)\bigr) & = V\bigl(A\bigl(\widetilde{g}(R_j)\bigr)\bigr) \\ & \leq V\bigl(A(\widetilde{R}_j)\bigr) \\ & \leq \babs{\det(A)} \, V(R_j) \, {\bigl(1+4\sqrt{n} \, \epsilon\bigr)}^n \\ & = \babs{J_g(0)} \, V(R_j) \, {\bigl(1+4\sqrt{n} \, \epsilon\bigr)}^n . \end{split} \end{equation*}

Translation does not change volume, and therefore for every

R_j\text{,}

and

x_j \in R_j\text{,}

including when

x_j \neq 0

and

g(x_j) \neq 0\text{,}

we find

\begin{equation*} V\bigl(g(R_j)\bigr) \leq \babs{J_g(x_j)} \, V(R_j) \, {\bigl(1+4\sqrt{n} \, \epsilon\bigr)}^n . \end{equation*}

Write

f

f = f_+ - f_-

for two nonnegative Riemann integrable functions

f_+

and

f_-\text{:}

\begin{equation*} f_+(u) \coloneqq \max \bigl\{ f(u) , 0 \bigr\}, \qquad f_-(u) \coloneqq \max \bigl\{ -f(u) , 0 \bigr\} . \end{equation*}

So, if we prove the theorem for a nonnegative

f\text{,}

we obtain the theorem for arbitrary

f\text{.}

Therefore, suppose without loss of generality that

f(u) \geq 0

for all

u \in g(S)\text{.}

As the rectangles

R_1,R_2,\ldots,R_N

give a refinement of the partition

P\text{,}

\begin{equation*} \begin{split} \epsilon + \int_S f\bigl(g(x)\bigr) \, \babs{J_g(x)} \, dx & \geq \sum_{j=1}^N \biggl(\sup_{x \in R_j} f\bigl(g(x)\bigr) \, \babs{J_g(x)} \biggr) \, V(R_j) \\ & \geq \sum_{j=1}^N \biggl(\sup_{x \in R_j} f\bigl(g(x)\bigr) \biggr) \, \babs{J_g(x_j)} \, V(R_j) \\ & \geq \sum_{j=1}^N \biggl(\sup_{u \in g(R_j)} f(u) \biggr) \, V\bigl(g(R_j)\bigr) \frac{1}{{(1+4\sqrt{n} \, \epsilon)}^n} \\ & \geq \sum_{j=1}^N \left(\int_{g(R_j)}f(u) \,du \right) \frac{1}{{(1+4\sqrt{n} \, \epsilon)}^n} \\ & = \frac{1}{{(1+4\sqrt{n} \, \epsilon)}^n} \int_{g(S)} f(u) \,du . \end{split} \end{equation*}

The last equality follows because the overlaps of the rectangles are their boundaries, which are of measure zero, and hence the image of their boundaries is also measure zero. Let

\epsilon

go to zero to find

\begin{equation*} \int_S f\bigl(g(x)\bigr) \, \babs{J_g(x)} \, dx \geq \int_{g(S)} f(u) \,du . \end{equation*}

Recall that

g^{-1}

exists and

g^{-1}\bigl(g(S)\bigr) = S\text{.}

Also,

1 = J_{g\circ g^{-1}} = J_g\bigl(g^{-1}(u)\bigr) \,J_{g^{-1}}(u)

for

u \in g(S)\text{.}

\begin{equation*} \begin{split} \int_{g(S)} f(u) \, du & = \int_{g(S)} f\Bigl(g\bigl(g^{-1}(u)\bigr)\Bigr) \, \Babs{J_g\bigl(g^{-1}(u)\bigr)} \, \babs{J_{g^{-1}}(u)} \, du \\ & \geq \int_{g^{-1}(g(S))} f\bigl(g(x)\bigr) \, \babs{J_g(x)} \, dx = \int_{S} f\bigl(g(x)\bigr) \, \babs{J_g(x)} \, dx . \qedhere \end{split} \end{equation*}