Hausdorff’s Description of Lebesgue’s Measure

Chapter 4 Products of Measures

4.2 Steiner Symmetrization

4.2.2. Hausdorff’s Description of Lebesgue’s Measure

onR^N. As distinguished from the one given in§2.2.2, this one is completely coordinate free.

ForC ⊆ P(R^N), set kCk= sup{diam(C) : C∈ C}, and, given δ >0, define H^N(Γ) = lim

δ&0H^N,δ(Γ), where (4.2.5)

H^N,δ(Γ)≡inf (

X

C∈C

Ω_Nrad(C)^N : Ca countable cover of Γ with kCk ≤δ )

.

I emphasize that I have placed no restriction on the sets C making up the covers C. On the other hand, it should be clear (cf. Exercise 4.2.9) that H^N(Γ) would have been unchanged if I had restricted myself to covers either by closed sets or by open sets. Also, it is obvious that H^N,δ(Γ) is a non- increasing function of δ >0 and therefore that there is no question that the indicated limit exists.

Directly from its definition, one sees thatH^N,δ is monotone and countably subadditive in the sense that

H^N,δ1(Γ1)≤H^N,δ2(Γ2) wheneverδ2≤δ1 and Γ1⊆Γ2

and

H^N,δ

∞

[

n=1

Γn

!

≤

∞

X

n=1

H^N,δ(Γn) for all{Γn: n≥1} ⊆ P R^N . Indeed, the first of these is completely trivial, and the second follows by choosing, for a given >0,{Cn : n≥1}for which

kC_nk ≤δ, Γ_n⊆[

C_m, and X

C∈Cn

Ω_N rad(C)^N

≤H^N,δ(Γ_n) + 2⁻ⁿ

and noting that H^N,δ

∞

[

n=1

Γn

!

≤

∞

X

n=1

X

C∈Cn

ΩN(rad(C)N

≤

∞

X

n=1

H^N,δ Γn

+.

Moreover, because λ_RN(Γ)≤X

C∈C

λ_RN(C) for any countable coverC ⊆ B_RN of Γ,

the inequality λ_RN(Γ) ≤ H^N,δ(Γ) ≤ H^N(Γ) is an essentially trivial conse- quence of (4.2.1). In order to prove the opposite inequality, I will use the following lemma.

Lemma 4.2.6. For anyδ >0and open setGinR^N withλ_RN(G)<∞,there exists a sequence {B_n : n≥1} of mutually disjoint closed balls contained in Gwith the properties thatrad(Bn)< ^δ₂ and

λ_RN G\

∞

[

n=1

Bn

!

= 0.

Proof: If G = ∅, there is nothing to do. Thus, assume that G 6= ∅, and set G₀ =G. Using Lemma 2.2.12, choose a countable, exact cover C0 ofG₀

4 Products of Measures

by non-overlapping squaresQwith diameter strictly less thanδ. Next, given Q∈ C0, choosea= (a1, . . . , aN)∈R^N andρ∈[0, δ) for which

Q=

N

Y

i=1

a_i−ρ, a_i+ρ ,

and set BQ =B a,^ρ₂

. Clearly, theBQ’s are mutually disjoint closed balls.

At the same time, there is a dimensional constant αN ∈ (0,1) for which λ_RN(BQ)≥αNvol(Q). Therefore there exists a finite subset

B0,1, . . . , B0,n₀

⊆

B_Q: Q∈ C₀ for which

λ_RN G₀\

n0

[

m=1

B_0,m

!

≤β_Nλ_RN(G₀), whereβ_N ≡1−^α₂^N ∈(0,1).

Now setG1=G0\Sⁿ0

1 B0,m. Noting thatG1is again non-empty and open, we can repeat the preceding argument to find a finite collection of mutually disjoint closed balls B1,m⊂G1,1≤m≤n1,in such a way that

λ_RN G1\

n₁

[

m=1

B1,m

!

≤βNλ_RN(G1).

More generally, we can use induction on`∈Z<sup>+</sup> to construct open setsG`⊆ G_{`−1</sub> and finite collections B<sub>`,1}, . . . , B_`,n` of mutually disjoint closed balls B ⊂G_` so that

λ_RN G_`+1

≤β_Nλ_RN G_`

whereG_{`+1</sub>=G<sub>`}\

n_`

[

m=1

B_`,m.

Clearly the collection

B`,m: `∈Nand 1≤m≤n`

has the required properties.

Theorem 4.2.7. For every Γ∈ Bandδ >0,H^N(Γ) =λ_RN(Γ) =H^N,δ(Γ).

Proof: As has been already pointed out, the inequalityλ_RN(Γ)≤H^N,δ(Γ)

≤H^N(Γ) is an immediate consequence of the definition combined with (4.2.1).

To get the opposite inequality, it suffices to show that λ_RN(Γ)≥H^N,δ(Γ) for every δ > 0 and Γ ∈ B_RN. To this end, first observe that H^N,δ(Γ) = 0 if λ_RN(Γ) = 0. Indeed, ifλ_RN(Γ) = 0, then, for each >0 we can first find an openG⊇Γ withλ_RN(G)< and then, by Lemma 2.2.12, a countable, exact

cover C of G by non-overlapping squares Q of diameter less than δ, which means that

H^N,δ(Γ)≤H^N,δ(G)≤ X

Q∈C

Ω_Nrad(Q)^N

≤N^N2Ω_N 2^N

X

Q∈C

λ_RN(Q)< N^N2Ω_N 2^N .

Next, becauseH^N,δis countably subadditive andλ_RN is countably additive, it suffices to check H^N,δ(Γ) ≤ λ_RN(Γ) for bounded sets Γ ∈ B_RN. Thus, suppose that Γ∈ B_RN is bounded, and letGbe any open superset of Γ with λ_RN(G) < ∞. By Lemma 4.2.6, there exists a sequence {Bn : n ≥ 1} of mutually disjoint closed balls contained in Gfor which diam(Bn)≤δand

λ_RN(G\A) = 0 whereA=

∞

[

n=1

Bn. Hence, because

H^N,δ(Γ)≤H^N,δ(G)≤H^N,δ(A) +H^N,δ(G\A) =H^N,δ(A), we see that

H^N,δ(Γ)≤

∞

X

n=1

Ω_Nrad(B_n)^N =

∞

X

n=1

λ_RN(B_n) =λ_RN(A) =λ_RN(G) for every openG⊇Γ, and, after taking the infimum over suchG’s, we arrive at the desired conclusion.

Knowing Theorem 4.2.7, the reader may be wondering why I did not simply define

H^N(Γ) = inf (

X

C∈C

ΩNrad(C)^N : C a countable cover of Γ )

.

Indeed, as the theorem shows, this definition is the same as the more com- plicated one that I gave and would have obviated the need for introducing H^N,δ and taking the limit as δ & 0. The reason for introducing the more complicated definition will not become clear until §8.3.3, where H^N will be seen as a member of a continuously parametrized family of measures onR^N, and it is the only member for which the more complicated definition is not needed.

4 Products of Measures Exercises for §4.2

Exercises 4.2.8. Assume thatN ≥2. Using the definition ofH^N in (4.2.5), give a direct (i.e., one that does not use Theorem 4.2.7) proof thatH^N(P) = 0 for every hyperplaneP (i.e., the translate of an (N−1)-dimensional subspace) in R^N.

Exercise 4.2.9. Show that, for each δ > 0, H^N,δ(Γ) is unchanged when one restricts to covers by closed sets, and from this conclude that H^N(Γ) is unchanged if one restricts to covers by open sets or by closed sets. Less obvious than the preceding is the fact that, at least for Γ ∈ B_RN, one can restrict to covers by closed balls or by open balls. To be precise, letH^N,δ_b (Γ) be the quantity in (4.2.5) obtained by restricting to covers by closed balls.

Given Γ∈ B_RN, use Lemma 4.2.6 and argue as in the proof of Theorem 4.2.7 to show that, for eachδ >0,H^N,δ_b (Γ) =H^N,δ(Γ). Finally, conclude from this that the same equality holds if one restricts to covers by open balls.

chapter

Changes of Variable

I have now developed the basic theory of Lebesgue integration, but I have provided nearly no tools with which to compute the integrals that have been shown to exist. The purpose of the present chapter is to introduce a technique that often makes evaluation, or at least estimation, possible. The technique is that of changing variables.

§5.1 Riemann vs. Lebesgue, Distributions, and Polar Coordinates The content of this section is applications of the following general principle.

Let (E,B, µ) be a measure space, (E⁰,B⁰) a measurable space, Φ : E −→

E⁰ a measurable map, and recall (cf. part (iii) of Exercise 2.1.19) that the pushforward ofµunder Φ is the measure Φ_∗µon (E,B⁰) given by Φ_∗µ(Γ⁰) = µ Φ⁻¹(Γ⁰)

for Γ⁰ ∈ B⁰. The following lemma is an essentially immediate consequence of this definition.

Lemma 5.1.1. Refer to the preceding. Then, for every non-negative mea- surable functionϕon(E⁰,B⁰),

(5.1.2)

Z

E⁰

ϕ d Φ_∗µ

= Z

E

ϕ◦Φdµ.

Moreover, ϕ∈L¹(Φ_∗µ;R) if and only ifϕ◦Φ∈L¹(µ;R), and(5.1.2) holds for allϕ∈L¹(Φ_∗µ;R).

Proof: Clearly it suffices to prove the first assertion. To this end, note that, by definition, (5.1.2) holds when ϕ is the indicator of a set Γ⁰ ∈ B⁰. Hence, it also holds when ϕis a non-negative measurable simple function on (E⁰,B⁰), and so, by the Monotone Convergence Theorem, it must hold for all non-negative measurable functions on (E⁰,B⁰).

§5.1.1. Riemann vs. Lebesgue: My first significant example of a change of variables will relate integrals over an arbitrary measure space to integrals over the real line. Namely, given a measurable R-valued function f on a measure space (E,B, µ), define the distribution of f under µ to be the measure µf ≡ f_∗µ on (R,B

R).¹ We then have that, for any non-negative

1In probability theory, distributions take on particular significance. In fact, from the point of view of a probabilistic purist, it is the distribution of a function, as opposed to the function itself, that is its distinguishing feature.

DOI 10.1007/978-1-4614-1135-2_5, © Springer Science+Business Media, LLC 2011,

D.W. Stroock Essentials of Integration Theory f Analysis, Graduate Texts in Mathematicsor 262, 113

5 Changes of Variable measurable ϕon (R,B

R), (5.1.3)

Z

E

ϕ◦f(x)µ(dx) = Z

R

ϕ(t)µ_f(dt).

The reason why it is sometimes useful to make this change of variables is that the integral on the right-hand side of (5.1.3) can often be evaluated as the limit of Riemann integrals, to which all the powerful tools of calculus apply. In order to see how the right-hand side of (5.1.3) leads to Riemann integrals, I will prove a general fact about the relationship between Lebesgue and Riemann integrals on the line. From a theoretical standpoint, the most interesting feature of this result is that it shows that a complete description of the class of Riemann integrable functions in terms of continuity properties defies a totally Riemannian solution and requires the Lebesgue notion ofalmost everywhere.

Theorem 5.1.4. Letν be a finite measure on (a, b],B(a,b]

, where −∞<

a < b <∞, and set ψ(t) =ν((a, t])fort∈[a, b] (ψ(a) =ν(∅) = 0). Then,ψ is right-continuous on[a, b), non-decreasing on [a, b], and, for each t ∈(a, b], ψ(t)−ψ(t−) = ν({t}). Furthermore, if ϕ is a bounded function on [a, b], then ϕ is Riemann integrable on [a, b] with respect to ψ if and only if ϕ is continuous(a.e.,ν)on(a, b], in which case,ϕis measurable on (a, b],B(a,b]

ν and

(5.1.5)

Z

(a,b]

ϕ d ν= (R) Z

[a,b]

ϕ(t)dψ(t).

Proof: It will be convenient to think ofν as being defined on [a, b],B_[a,b]

byν(Γ)≡ν(Γ∩(a, b]) for Γ∈ B_[a,b]. Thus, we will do so.

Obviously ψis right-continuous and non-increasing on [a, b], and therefore (cf. Exercise 2.1.17) it is also B_[a,b]-measurable there. Moreover, for each t∈(a, b],ψ(t)−ψ(t−) = lim_s%tν (s, t]

=ν({t}).

Now assume that ϕ is Riemann integrable on [a, b] with respect to ψ. To see thatϕis continuous (a.e.,ν) on (a, b], choose (cf. Lemma 1.2.14), for each n≥1, a finite, non-overlapping, exact coverCn of [a, b] by closed intervals I such thatkCnk<_n¹ and, for eachI∈ Cn,ψis continuous at the left endpoint I⁻. If ∆ =S∞

n=1{I⁻:I∈ Cn}, thenν(∆) = 0. Givenm≥1, letCm,n be the set of those I ∈ C_n for which sup_Iϕ−inf_Iϕ≥ _m¹. It is then easy to check that

t∈(a, b]\∆ :ϕis not continuous att ⊆

∞

[

m=1

∞

\

n=1

[Cm,n.

But, by Exercises 2.1.22 and 1.2.19, for each m≥1, ν

^∞

\

n=1

[Cm,n

≤ lim

n→∞

X

I∈C_m,n

∆Iψ= 0,

and thereforeν S∞ m=1

T∞ n=1

SCm,n

= 0. Hence, we have now shown thatϕ is continuous (a.e.,ν) on (a, b].

Conversely, suppose thatϕis continuous (a.e.,ν) on (a, b]. Given a sequence {Cn : n ≥ 1} of finite, non-overlapping, exact covers of [a, b] by compact intervals I with kCnk −→ 0, for each n ≥ 1, define ϕ_n(t) = sup_Iϕ and ϕn(t) = inf_Iϕ for t ∈ I\ {I⁻} and I ∈ C_n. Clearly, both ϕ_n and ϕ

n are measurable on (a, b],B_(a,b]

. Moreover, inf

(a,b]ϕ≤ϕ

n ≤ϕ≤ϕ_n≤sup

(a,b]

ϕ for alln≥1. Finally, sinceϕis continuous (a.e.,ν),

ϕ= lim

n→∞ϕ

n= lim

n→∞ϕ_n (a.e.,ν);

and so, not only is ϕ equal to lim_n→∞ϕ

n (a.e.,ν) and therefore measurable on ((a, b],B_(a,b]^ν), but also

n→∞lim Z

(a,b]

ϕndν= Z

(a,b]

ϕ d ν= lim

n→∞

Z

(a,b]

ϕ_ndν.

In particular, we conclude that X

I∈Cn

sup

I

ϕ−inf

I ϕ

∆Iψ= Z

(a,b]

ϕ_n−ϕ_n

dν −→0

as n → ∞. From this it is clear both that ϕis Riemann integrable on [a, b]

with respect toψand that (5.1.3) holds.

I am now ready to prove the main result to which this subsection is devoted.

Theorem 5.1.6. Let (E,B, µ) be a measure space and f : E −→ [0,∞]

a B-measurable function. Then t ∈ (0,∞) 7−→ µ(f > t) ∈ [0,∞] is a right-continuous, non-increasing function. In particular, it is measurable on

(0,∞),B_(0,∞)

and has at most a countable number of discontinuities. Next, assume that ϕ ∈ C [0,∞);R

∩C¹ (0,∞);R

is a non-decreasing function satisfyingϕ(0) = 0< ϕ(t), t >0, and setϕ(∞) = limt→∞ϕ(t). Then (5.1.7)

Z

E

ϕ◦f(x)µ(dx) = Z

(0,∞)

ϕ⁰(t)µ(f > t)λ_R(dt).

Hence, either µ(f > δ) = ∞ for some δ ∈(0,∞), in which case both sides of (5.1.7) are infinite, or, for each 0 < δ < r < ∞, the map t ∈ [δ, r] 7−→

ϕ⁰(t)µ(f > t)is Riemann integrable and Z

E

ϕ◦f(x)µ(dx) = lim

δ&0 r%∞

(R) Z

[δ,r]

ϕ⁰(t)µ(f > t)dt.

5 Changes of Variable

Proof: It is clear that t ∈ (0,∞) 7−→ µ(f > t) is right-continuous, non- increasing, and thereforeB_(0,∞)-measurable.

We turn next to the proof of (5.1.7). Since, by Theorems 5.1.4 and 1.2.3, lim

α&0

Z

(α,δ]

ϕ⁰(t)λ_R(dt) = lim

α&0(R) Z δ

α

ϕ⁰(t)dt=ϕ(δ) and therefore

Z

E

ϕ◦f dµ

∧ Z

(0,∞)

ϕ⁰(t)µ(f > t)λ_R(dt)

!

≥ϕ(δ)µ(f > δ),

it is obvious that both sides of (5.1.7) are infinite when µ(f > δ) = ∞ for some δ >0. Thus we will assume thatµ(f > δ)<∞for everyδ >0. Then the restriction ofµf toB_[δ,∞)is a finite measure for everyδ >0. Givenδ >0, setψδ(t) =µf((δ, t]) fort∈[δ,∞) and apply Theorems 5.1.4 and 1.2.3 to see that

Z

{δ<f≤r}

ϕ◦f dµ= Z

(δ,r]

ϕ dµf = (R) Z

[δ,r]

ϕ(t)dψδ(t)

=ϕ(r)ψδ(r)−(R) Z

[δ,r]

ψδ(t)ϕ⁰(t)dt

=ϕ(δ)ψδ(r) + (R) Z

[δ,r]

ψδ(r)−ψδ(t) ϕ⁰(t)dt

=ϕ(δ)µ(δ < f ≤r) + (R) Z

[δ,r]

µ(t < f≤r)ϕ⁰(t)dt

=ϕ(δ)µ(δ < f ≤r) + Z

(δ,r]

µ(t < f≤r)ϕ⁰(t)λ_R(dt) for each r ∈ (δ,∞). Hence, after simple arithmetic manipulation and an application of the Monotone Convergence Theorem, we get

Z

{δ<f <∞}

ϕ◦f−ϕ(δ) dµ=

Z

(δ,∞)

ϕ⁰(t)µ(t < f <∞)λ_R(dt) whenr% ∞. At the same time, it is clear that

Z

{f=∞}

ϕ◦f−ϕ(δ) dµ=

ϕ(∞)−ϕ(δ)

µ(f =∞)

= Z

(δ,∞)

µ(f =∞)ϕ⁰(t)λ_R(dt);

and, after combining these, we arrive at Z

{f >δ}

ϕ◦f(x)−ϕ(δ)

µ(dx) = Z

(δ,∞)

µ(f > t)ϕ⁰(t)λ_R(dt).

Thus, (5.1.7) will be proved once we show that lim

δ&0

Z

{f >δ}

ϕ◦f(x)−ϕ(δ)

µ(dx) = Z

E

ϕ◦f(x)µ(dx).

But if 0< δ1< δ2<∞, then 0≤ ϕ◦f−ϕ(δ2)

1_{{f >δ2}}≤ ϕ◦f−ϕ(δ1)

1_{{f >δ1}},

and so the required convergence follows by the Monotone Convergence Theo- rem.

Finally, to prove the last part of the theorem whenµ(f > δ)<∞for every δ >0, simply note that

Z

(0,∞)

ϕ⁰(t)µ(f > t)λ_R(dt) = lim

δ&0 r%∞

Z

(δ,r]

ϕ⁰(t)µ(f > t)λ_R(dt),

and apply Theorem 5.1.4.

§5.1.2. Polar Coordinates: In this subsection I will examine the change of variables that is referred to casually as “switching to polar coordinates.”

Warning: From now on, at least whenever the meaning is clear from the con- text, I will use the notation “dx” instead of the more cumbersome “λ_RN(dx)”

when doing Lebesgue integration with respect to Lebesgue measure on R^N. Let S^N−1 denote the unit (N−1)-sphere{x∈ R^N : |x| = 1} in R^N, and define Φ :R^N \ {0} −→S^N−1 by Φ(x) = _|x|^x. Clearly Φ is continuous. Next, define the surface measure λ_SN−1 on S^N−1 to be the image under Φ of N λ_RN restricted to B_B

RN(0,1)\{0}. Because Φ(rx) = Φ(x) for allr > 0 and x∈R^N \ {0}, one can use Theorem 2.2.15 to check that

Z

BRN(0,r)\{0}

f◦Φ(x)dx=r^N Z

BRN(0,1)\{0}

f ◦Φ(x)dx

and therefore that (∗)

Z

BRN(0,r)\{0}

f ◦Φ(x)dx= r^N N

Z

S^N−1

f(ω)λ_SN−1(dω)

for every non-negative measurable f on (S^N−1,B_S^N−1). In particular, if (cf.

(iii) in Exercise 5.1.13) ω_N−1≡λ_SN−1 S^N−1

is the surface area ofS^N−1,

5 Changes of Variable

we have that ΩN = ^ωN−1_N , where ΩN is the volume (i.e., Lebesgue measure) of the unit ball inR^N.

Next, define Ψ : (0,∞)×S^N−1−→R^N\ {0}by Ψ(r, ω) =rω. Note that Ψ is a homeomorphism from (0,∞)×S^N−1 ontoR^N \ {0} and that its inverse is given by Ψ⁻¹(x) = (|x|,Φ(x)). That is, the components of Ψ⁻¹(x) are the polar coordinates of the point x∈R^N \ {0}. Finally, define the measure RN on (0,∞),B_(0,∞)

byRN(Γ) =R

Γr^N−1dr.

The importance here of these considerations is contained in the following result.

Theorem 5.1.8. Referring to the preceding, one has that λ_RN = Ψ∗ RN ×λ_S^N−1

onB_RN\{0}.

In particular, if f is a non-negative, measurable function on(R^N,B_RN), then

(5.1.9)

Z

R^N

f(x)dx= Z

(0,∞)

r^N−1 Z

S^N−1

f(rω)λ_S^N−1(dω)

dr

= Z

S^N−1

Z

(0,∞)

f(rω)r^N−1dr

!

λ_SN−1(dω).

Proof: By Corollary 3.2.15 and Theorem 4.1.6, all that we have to do is check that the first equality in (5.1.9) holds for every

f ∈Cc R^N;R

≡

f ∈C R^N;R

:f ≡0 off of some compact set . To this end, let f ∈C_c(R^N;R) be given and setF(r) =R

BRN(0,r)f(x)dx for r >0. Then, by (∗), for allr, h >0,F(r+h)−F(r) equals

Z

BRN(0,r+h)\B

RN(0,r)

f(x)dx

=

Z

BRN(0,r+h)\B

RN(0,r)

f◦Ψ(r,Φ(x))dx

+

Z

BRN(0,r+h)\B

RN(0,r)

f(x)−f◦Ψ(r,Φ(x)) dx

= (r+h)^N−r^N N

Z

S^N−1

f◦Ψ(r, ω)λ_SN−1(dω) +o(h),

where “o(h)” denotes a function that tends to 0 faster than h. Hence, F is continuously differentiable on (0,∞) and its derivative at r ∈ (0,∞) is given by r^N−1R

S^N−1f ◦Ψ(r, ω)λ_S^N−1(dω). SinceF(r) −→0 as r & 0, the desired result now follows from Theorem 5.1.4 and the Fundamental Theorem of Calculus.

Exercises for §5.1

Exercise 5.1.10. Suppose thatF :R−→Ris a bounded, continuous, non- decreasing function that tends to 0 at−∞, and letµF be the Borel measure onRfor whichFis the distribution function. Show that for−∞< a < b <∞

(R) Z

[a,b]

ϕ◦F(s)dF(s) = (R) Z

[F(a),F(b)]

ϕ(t)dt ifϕ∈C [F(a), F(b)];R , which is the classicalchange of variables formula. Next, apply Theorem 5.1.4 to conclude from this that

Z

R

ϕ◦F dµ_F = Z

[0,F(∞)]

ϕ(t)dt for allB_R-measurableϕ:R−→[0,∞).

Exercise 5.1.11. A particularly important case of Theorem 5.1.6 is the one in which ϕ(t) =t^p for somep∈(0,∞), in which case (5.1.7) yields

(5.1.12)

Z

E

|f(x)|^pµ(dx) =p Z

(0,∞)

t^p−1µ |f|> t dt.

Use (5.1.12) to show that |f|^p isµ-integrable if and only if

∞

X

n=1

1

n^p+1µ |f|> ¹_n +

∞

X

n=1

n^p−1µ |f|> n

<∞.

Compare this result to the one obtained in the last part of Exercise 3.2.19.

Exercise 5.1.13. Perform the calculations outlined in the following.

(i) Justify Gauss’s trick (cf. Exercise 5.2.18) Z

R

e^−|x|

2 2 dx

²

= Z

R²

e^−|x|

2

2 dx= 2π Z

(0,∞)

re^−r

2

2 dr= 2π

and conclude that for any N ∈Z⁺ and symmetricN×N matrix Athat is strictly positive definite (i.e., all the eigenvalues of Aare strictly positive),

Z

R^N

exp

−1

2(x, A⁻¹x)_RN

dx= (2π)^N2 det(A)¹₂ . (ii) Define Γ(γ) = R

(0,∞)t^γ−1e^−tdt for γ ∈ (0,∞). Show that, for any γ ∈ (0,∞), Γ(γ+ 1) = γΓ(γ). Also, show that Γ ¹₂

= π¹². The function Γ(·) is called Euler’sgamma function. Notice that it provides an extension of the factorial function in the sense that Γ(n+ 1) =n! for integersn≥0.

5 Changes of Variable (iii) Show that

ωN−1≡λ_S^N−1(S^N−1) = 2π^N2 Γ ^N₂,

and conclude that the volume ΩN of theN-dimensional unit ball is given by ΩN = π^N2

Γ ^N₂ + 1. Hint: Use polar coordinates to compute R

Re^−x22dx^N

=R

R^Ne^−|x|

2 2 dx.

(iv) Givenα, β∈(0,∞), show that Z

(0,∞)

t⁻¹²exp

−α²t−β² t

dt= π¹²e^−2αβ

α .

Finally, use the preceding to show that Z

(0,∞)

t⁻³²exp

−α²t−β² t

dt= π¹²e^−2αβ

β .

Hint: Define F(s) for s ∈ R to be the unique t ∈ (0,∞) satisfying s = αt¹² −βt⁻¹², and use Exercise 5.1.10 to show that

Z

(0,∞)

t⁻¹²exp

−α²t−β² t

dt=e^−2αβ α

Z

R

e^−s2ds.

Exercise 5.1.14. This exercise deals with a few of the elementary properties ofλ_SN−1.

(i) Show that if Γ is a non-empty, open subset ofS^N−1, thenλ_SN−1(Γ)>0.

Next, show that λ_SN−1 isorthogonally invariant. That is, show that ifO is an N×N orthogonal matrix and T_O is the associated transformation on R^N (cf. the paragraph preceding Theorem 2.2.15), then T_O

∗λ_SN−1 =λ_SN−1. Finally, use this fact to show that

Z

S^N−1

(ξ, ω)_RNλ_SN−1(dω) = 0

and Z

S^N−1

ξ, ω

R^N η, ω

R^Nλ_SN−1(dω) = ΩN(ξ, η)_RN

for any ξ, η∈R^N.

Hint: In proving these, for given ξ ∈ R^N \ {0}, consider the orthogonal transformation that sendsξto−ξbut acts as the identity on the orthogonal complement of ξ.

(ii) Define Ψ : [0,2π] −→ S¹ by Ψ(θ) = _cosθ

sinθ

, and set µ = Φ_∗λ_[0,2π]. Given any rotation invariant finite measure ν on (S¹,B_S1), show that ν =

ν(S¹)

2π µ. In particular, conclude thatλ_S1=µ. (Cf. Exercise 5.2.18 below.) Hint: Define

Oθ=

cosθ sinθ

−sinθ cosθ

for θ∈[0,2π], and note that

Z

S¹

f dν= 1 2π

Z

[0,2π]

Z

S¹

f◦T_Oθ(ω)ν(dω)

dθ.

§5.2 Jacobi’s Transformation and Surface Measure

I begin this section with a derivation of Jacobi’s famous generalization to non-linear maps of the result in Theorem 2.2.15. I will then apply Jacobi’s result to obtain surface measure bydifferentiating Lebesgue measure across a smooth hypersurface. Throughout, I will be assuming thatN ≥2.

§5.2.1. Jacobi’s Transformation Formula: Given an open setG⊆R^N and a continuously differentiable map

x∈G7−→Φ(x) =





 Φ₁(x)

... ΦN(x)





∈R^N,

define theJacobian matrix ^∂Φ_∂x(x)ofΦatxto be theN×N matrix whose jth column is the vector

∂_xjΦ(x)≡









∂Φ1

∂x_j

...

∂ΦN

∂x_j







 .

In addition, call JΦ(x)≡

det ^∂Φ_∂x(x)

theJacobian of Φ atx.

Lemma 5.2.1. If Gis an open set in R^N and Φ an element of C¹(G;R^N) whose Jacobian never vanishes onG, then Φmaps open (orBG-measurable) subsets ofGinto open (orB_RN-measurable) sets inR^N. In addition, if Γ∈ BG

with λ_RN(Γ) = 0, then λ_RN Φ(Γ)

= 0; and ifΓ ∈ BΦ(G) with λ_RN(Γ) = 0, thenλ_RN Φ⁻¹(Γ)

= 0. In particular, Γ∈ BG

λRN

if and only if Φ(Γ)∈ B_Φ(G)^λRN.

Proof: By the Inverse Function Theorem,¹ for eachx∈Gthere is an open neighborhood U ⊆G of x such that Φ U is invertible and its inverse has

1See, for example, W. Rudin’sPrinciples of Mathematical Analysis, McGraw Hill (1976).

5 Changes of Variable

first derivatives that are bounded and continuous. Hence, Gcan be written as the union of a countable number of open sets on each of which Φ admits an inverse having bounded continuous first-order derivatives; and so, without loss of generality, we may and will assume that Φ∈C_b¹(G;R^N) and that it admits an inverse Φ⁻¹ ∈ C_b¹ Φ(G);R^N

. But, in that case, both Φ and Φ⁻¹ take open sets to open sets and thereforeB_RN-measurable sets toB_RN-measurable sets. Finally, to see that Φ takes sets ofλ_RN-measure 0 to sets ofλ_RN-measure 0, note that Φ is uniformly Lipschitz continuous and apply Exercise 2.2.30.

The argument for Φ⁻¹ is precisely the same.

A continuously differentiable map Φ on an open set U ⊆R^N into R^N is called a diffeomorphism if it is injective (i.e., one-to-one) and JΦ never vanishes. If Φ is a diffeomorphism on the open setU and ifW = Φ(U), then I will say that Φ isdiffeomorphic fromU ontoW.

In what follows, and elsewhere, for any given set Γ⊆R^N and δ >0, I will use

Γ^(δ)≡

x∈R^N : |x−Γ|< δ}

to denote theopen δ-hull of Γ.

Theorem 5.2.2 (Jacobi’s Transformation Formula). LetGbe an open set in R^N and Φ an element of C²(G;R^N).² If the JacobianJΦof Φ never vanishes, then, for every measurable functionf on Φ(G),B_Φ(G)^λRN

,f◦Φis measurable on G,BG

λRN and (5.2.3)

Z

Φ(G)

f(y)dy≤ Z

G

f◦Φ(x)JΦ(x)dx

whenever f is non-negative. Moreover, if Φis a diffeomorphism on G, then (5.2.3)can be replaced by

(5.2.4)

Z

Φ(G)

f(y)dy= Z

G

f ◦Φ(x)JΦ(x)dx.

Proof: First note that (5.2.4) is a consequence of (5.2.3) when Φ is one-to- one. Indeed, if Φ is one-to-one, then the Inverse Function Theorem guarantees that Φ⁻¹∈C²(Φ(G);R^N). In addition, by the chain rule,

∂Φ⁻¹

∂y (y) = ∂Φ

∂x Φ⁻¹(y) −1

for y∈Φ(G).

Hence one can apply (5.2.3) to Φ⁻¹ and thereby obtain Z

G

f◦Φ(x)JΦ(x)dx≤ Z

Φ(G)

f(y) (JΦ)◦Φ⁻¹(y)JΦ⁻¹(y)dy= Z

Φ(G)

f(y)dy,

2By being a little more careful, one can get the same conclusion for Φ∈C¹(G,R^N).

which, in conjunction with (5.2.3), yields (5.2.4).

Next observe that it suffices to prove (5.2.3) under the assumptions thatG is bounded, Φ onGhas bounded first and second order derivatives, andJΦ is uniformly positive on G. In fact, if this is not already the case, then one can choose a non-decreasing sequence of bounded open sets G_n such that ΦG_n has these properties for eachn≥1 andG_n%G. Clearly, the result for Φ on Gfollows from the result for ΦG_n onG_n for everyn≥1. Thus, from now on, we will assume that G is bounded, the first and second derivatives of Φ are bounded, and JΦ is uniformly positive onG.

Set Q = Q(c;r) = QN

1[c_i−r, c_i+r] ⊆ G, the square of side length 2r centered at c= (c1, . . . , cN)∈R^N. Then, by Taylor’s Theorem, there is an L∈[0,∞) (depending only on the bound on the second derivatives of Φ) such that³

Φ Q(c;r)

⊆T_Φ(c) T∂Φ

∂x(c) Q(0;r)(Lr²) .

(Cf. §2.2.2 for the notation here.) At the same time, there is an M < ∞ (depending only onL, the lower bound onJΦ, and the upper bounds on the first derivatives of Φ) such that

T∂Φ

∂x(c)Q(0;r)(Lr²)

⊆T∂Φ

∂x(c)Q 0, r+M r² . Hence, by Theorem 2.2.15,

λ_RN Φ(Q)

≤JΦ(c)λ_RN Q(0, r+M r²)

= (1 +M r)^NJΦ(c)λ_RN(Q).

Now define µ(Γ) =R

ΓJΦ(x)dx for Γ∈ B_G^λRN, set ν = Φ_∗µ, note that ν is finite, and apply Theorem 2.1.15 to see that it is regular. Given an open set H ⊆ Φ(G), use Lemma 2.2.12 to choose, for each m∈ Z⁺, a countable, non-overlapping, exact coverCmof Φ⁻¹(H) by squaresQwith diam(Q)<_m¹. Then, by the preceding paragraph,

λ_RN(H)≤ X

Q∈C_m

λ_RN Φ(Q)

≤

1 + M m

^N X

Q∈C_m

JΦ(cQ)λ_RN(Q), where cQ denotes the center of the square Q. After lettingm → ∞ in the preceding, we conclude that λ_RN(H) ≤ ν(H) for open H ⊆ Φ(G); and so, because λ_RN and ν are regular, it follows that λ_RN(Γ) ≤ ν(Γ) for all Γ ∈ BΦ(G)

λRN

.

Starting from the preceding, working first with simple functions, and then passing to monotone limits, one concludes that (5.2.3) holds for all non- negative, measurable functionsf on Φ(G),BΦ(G)

λRN

.

As an essentially immediate consequence of Theorem 5.2.2, we have the following.

3It is at this point that the argument has to be modified when Φ is only once continuously differentiable. Namely, the estimate that follows must be replaced by one involving the modulus of continuity of Φ’s first derivatives.