1 ABSTRACT MEASURE THEORY Analysis Study Guide

1 Abstract Measure Theory

Def 12: 1 (a) A collection M of subsets of X is a σ-algebra if it has the following properties: i) ∅, X ∈ M , ii)If E ∈ M then Ec ∈ M , and iii) if Ei∞i=1 ⊆ M then ∪∞i=1Ei ∈ M . (b) If M is a σ-algebra on X, then thepair (X,M ) is called a measurable space. If M is understood, then X will be called a measurable space. Themembers of M are called measurable sets. (c) If (X,M ) and (Y,N ) are measurable spaces and f : X → Ysatisfies f−1(E) ∈ M for each E ∈ N , then f is called (M ,N )-measurable. If M and N are understood, wejust say f is measurable.

Example 13: (a) If X 6= ∅, then M = ∅, X is the trivial measurable space. (b) If X 6= ∅ then M = P(X) isthe discrete measurable space. (c) If X = R, ,then M = ∅, 0 ,R \ 0 ,R is a σ-algebra. (e) If X 6= ∅, thenM = E ⊆ X : E or Ec is countable is a σ-algebra.

Lma 14: (Disjoint Dissection) If Ej∞j=1 ⊆P(X) then Fk∞k=1 ⊆P(X) defined by F1 := E1 and Fk = Ek \ (∪k−11 Ej)

for k > 1 is a sequence of mutually disjoint sets such that ∪∞1 Fk = ∪∞1 Ej .Thm 15: 2 If E is a collection of subsets of X then there is a smallest σ-algebra (which is unique) M (E) containing E .Def 16: The unique smallest σ-algebra containing a subset E ⊆ P(X) is called the σ-algebra generated by E , and

we denote this σ-algebra by M (E).Def 17: Let (X, T ) be a topological space. The σ-algebra M (T ) is called the Borel σ-algebra on X. It is denoted by

B(X,T ) or BX . Elements of BX are called Borel sets.Def 18: Let (X, T ) be a topological space. Set Fσ := countable unions of closed sets in X,

Gδ := countable intersections of open sets in X, Fσδ := countable intersections of sets in Fσ, etc.Rmrk: All are subcollections of BX .Prop 19: BR is generated by any of the following : E1 = (a, b) : a < b, E2 = [a, b] : a < b, E3 = (a, b] : a < b, E4 =[a, b) : a < b, E5 = (a,∞) : a ∈ R, E6 = (−∞, a) : a ∈ R, E7 = [a,∞) : a ∈ R, or E8 = (−∞, a] : a ∈ R.

Rmrk: The Borel σ-algebra on R can be generated by E =

(a,∞] : a ∈ R

or E =

[−∞, a) : a ∈ R

. We also haveBR =

E ⊆ R : E ∩ R ∈ BR

.

Prop 20: Let (X,M ) and (Y,N ) be measure spaces. If N = M (E) for some E ⊆ P(X), then f : X → Y is(M ,N )-measurable if and only if f−1(E) ∈M for all E ∈ E .

Cor 21: If f : X → Y , with X and Y topological spaces, is continuous, then f is (BX ,BY )-measurable.Prop 22: If (X,M ) is a measurable space and f : X → R then, for all a ∈ R, TFAE: (a) f is measurable; (b)

f−1((a,∞)) ∈M ; (c) f−1([a,∞)) ∈M ; (d) f−1((−∞, a)) ∈M ; (e) f−1((−∞, a]) ∈M .Def 23: 3 Let (Yα,Nα)α∈A be a family of measurable spaces. If fα : Xα → Yα is a map for each α ∈ A, then the

σ-algebra or X generated by fαα∈A is the σ-algebra generated byf−1α (E) : α ∈ A and E ∈ Nα

.

Prop 24: Let (X,M ) be a measurable space, and let f, g : X → R be measurable. Then

(a) |f | and f2 are measurable;

(b) αf , α+ f are measurable for all α ∈ R;

(c) If f is never 0, then 1/f is measurable;

(d) the set x ∈ X : f(x) > g(x) ∈M ;

(e) f + g, f − g are measurable;

(f) f · g is measurable;

(g) If g is never 0 then f/g is measurable.

Cor 25: If f : X → C is measurable, then so is sgn(f), where sgn(z) = z/|z| if z 6= 0 and 0 if z = 0.Prop 26: If fj∞j=1 is a sequence of R-valued functions on (X,M ) then g1(x) = sup

j∈Nfj(x), g2(x) = inf

j∈Nfj(x), g3(x) =

lim supj→∞

fj(x), and g4(x) = lim infj→∞

fj(x) are measurable. If limj→∞ fj(x) = f(x) then at each x ∈ X, f(x) is

measurable.Cor 27: 4 If f, g : X → R are measurable, then x 7→ max f(x), g(x) and x 7→ min f(x), g(x) are measurable.Cor 28: If f : X → R is measurable, then so are f+, f−, and |f |.Def 29: A simple function on X is a measurable function with finite range, which is a subset of R (or C). In particular,

a simple function is bounded. If ϕ : X → R is simple, then Ran (ϕ) = a1, · · · , an, and for each j ∈ 1, · · · , n theset Ej = ϕ−1(aj) is measurable and ∪nj=1Ej = X. The standard representation for ϕ is ϕ =

∑nj=1 ajχEj .

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2 ABSTRACT INTEGRATION Analysis Study Guide

Thm 30: Let (X,M ) be a measurable space. If f : X → [0,∞] is measurable, then there is a sequence ϕj∞j=1 of simplefunctions such that (1) 0 ≤ ϕ1 ≤ ϕ2 ≤ · · · ; (2) limj→∞ ϕj(x) = f(x) for all x ∈ X; and (3), ϕj → f uniformly onany set where f is uniformly bounded. (An analogous result holds if f has complex range.)

Def 31: 5 Let (X,M ) be a measurable space.

(a) A (positive) measure on the σ-algebra M is a function µ : M → [0,∞] with µ(∅) = 0 and if Ej∞j=1 is asequence of mutually disjoint sets then µ(∪∞1 Ej) =

∑j µ(Ej).

(b) A measure space is a triple (X,M , µ) with µ a measure.

Def 32: Let (X,M , µ) be a measure space.

(a) µ is called semifinite if for each E ∈ M with µ(E) > 0 there is a subset F ∈ M such that F ⊆ E and0 < µ(F ) <∞.

(b) µ is called σ-finite if X = ∪∞j=1Ej for some Ej∞j=1 ⊆M and µ(Ej) <∞ for all j ∈ N.

(c) A set E ∈M is called σ-finite if E = ∪∞j=1Ej for some Ej∞j=1 ⊆M with µ(Ej) <∞ for j ∈ N.

(d) µ is called finite if µ(X) <∞.

(e) µ is called a probability measure if µ(X) = 1.

Example 33:

(a) Suppose X is nonempty. Define µ : P(X)→ [0,∞] by µ(E) = card(E) if E is finite, and µ(E) =∞ otherwise.Then µ is semifinite, µ is σ-finite iff X is countable, and E ⊆ X is σ-finite iff E is countable.

(b) Let X be nonempty. Define M = E ⊆ X : E or Ec is countable and µ : M → [0,∞) by µ(E) = 0 if E iscountable and µ(E) = 1 if E is uncountable.

(c) Let (X,M ) be a measure space. Let x0 ∈M be given. Define µ : M → [0,∞] by µ(E) = 1 if x0 ∈ E andµ(E) = 0 otherwise. This is called a point mass or Dirac measure at x0. We denote it by δx0 .

Def 34: If (X,M , µ) is a measure space, then a µ-null set, or null set, in X is a set such E such that µ(E) = 0.Def 35: If statement P is true for all x ∈ E with µ(X \ E) = 0 then we say that P holds almost everywhere in X.Def 36: Let (X,M , µ) be a measure space. The measure µ is called complete if whenever E ∈M is a null set we also

find F ∈M for each F ⊆ E.Thm 37: Suppose that (X,M ) is a measure space. Set N := N ∈M : µ(N) = 0, and

M := E ∪ F : E ∈M and F ⊆ N ∈ N . Then M is a σ-algebra and there exists a unique extension of µ, µ toM , where µ is complete.

Def 38: The measure space (X,M , µ) is called the completion of (X,M , µ).Thm 39: 6 Let (X,M , µ) be a measure space. Then µ has the following properties:

• (Monotonicity) If E,F ∈M and E ⊆ F then µ(E) ≤ µ(F ).

• (Countable subadditivity) If Ej∞j=1 ⊆M then µ(∪∞j=1Ej) ≤∑∞j=1 µ(Ej).

• (Continuity from below) Ej∞j=1 ⊆M and Ej ⊆ Ej+1 for all j then µ(∪∞j=1Ej) = limj→∞ µ(Ej).

• (Continuity from above) If Ej∞j=1 ⊆M and Ej ⊇ Ej+1 and µ(E1) <∞ then µ(∩∞j=1Ej) = limj→∞

µ(Ej).

2 Abstract Integration

Notation: Given a measurable space (X,M ), set L+ := f : X → [0,∞] : f is measurable.Def 40: 7 Let (X,M , µ) be a measure space.

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2 ABSTRACT INTEGRATION Analysis Study Guide

(a) Let ϕ ∈ L+ be a simple function with standard representation ϕ =∑nj=1 ajχEj . The (Lebesgue) integral

of ϕ with respect to µ is ∫X

ϕdµ :=n∑j=1

ajµ(Ej).

(b) Let f ∈ L+ be given. The (Lebesgue) integral of f with respect to µ is∫X

fdµ := sup∫

X

ϕdµ : ϕ ∈ L+ is simple and 0 ≤ ϕ ≤ f

(c) Let f ∈ L+ and A ∈M . The (Lebesgue) integral of f with respect to µ over A is∫A

fdµ =∫X

fχAdµ.

Prop 41: Let f ∈ L+ be given.

(a) If A ∈M and µ(A) = 0 then∫Afdµ = 0.

(b) If f(x) = 0 for a.e. x ∈ X then∫Xfdµ = 0.

Prop 42: Let ϕ ∈ L+ be a simple function. Define λ : M → [0,+∞] by λ(E) :=∫Eϕdµ. Then λ is a measure.

Prop 43: Let ϕ,ψ ∈ L+ and c ∈ [0,∞] be given, with ϕ,ψ simple. Then

(a)∫X

(cϕ)dµ = c∫Xϕdµ.

(b)∫X

(ϕ+ ψ)dµ =∫Xϕdµ+

∫Xψdµ.

(c) If ϕ(x) ≤ ψ(x) for a.e. x ∈ X then∫Xϕdµ ≤

∫Xψdµ.

Cor 44: 8 If f, g ∈ L+ and f ≤ g for a.e. x ∈ X then∫Xfdµ ≤

∫Xgdµ.

Thm 45: (Monotone Convergence Theorem) Let fj∞j=1 ⊆ L+ satisfying 0 ≤ f1 ≤ f2 ≤ · · · ≤ fj ≤ · · · . Definef(x) = lim

j→∞fj(x) for all x ∈ X. Then f ∈ L+ and

∫X

fdµ = limj→∞

∫X

fjdµ.

Cor 46: 9 Suppose that aj,k∞j,k=1 ⊆ [0,+∞] satisfies 0 ≤ aj,1 ≤ aj,2 ≤ · · · for each j ∈ N. Then

limk→∞

∞∑j=1

aj,k =∞∑j=1

limk→∞

aj,k.

Thm 47: If fj∞j=1 ⊆ L+ and f(x) =

∑∞j=1 fj(x) for each x ∈ X, then∫

X

fdµ =∞∑j=1

∫X

fjdµ.

Prop 48: If f ∈ L+ then∫Xfdµ = 0 if and only if f = 0 a.e.

Cor 49: If fj∞j=1 ⊆ L+ and f ∈ L+ satisfy fj(x) increases to f(x) for a.e. x ∈ X then∫

X

fdµ = limj→∞

∫X

fjdµ.

Lma 50: (Fatou’s Lemma) If fj∞j=1 ⊆ L+ then∫

X

(lim infj→∞

fj)dµ ≤ lim infj→∞

∫X

fjdµ.

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3 SOME FUNCTIONAL ANALYSIS Analysis Study Guide

Cor 51: 10 If fj∞j=1 ⊆ L+ and lim

j→∞fj(x) = f(x) for a.e. x ∈ X then

∫Xfdµ ≤ lim inf

j→∞

∫Xfjdµ.

Prop 52: If f ∈ L+ and∫Xfdµ <∞ then x ∈ X : f(x) = +∞ is a null set, and x ∈ X : f(x) > 0 is σ-finite.

Def 53: Define L1(X,M , µ) to be the collection of measurable functions f : X → R such that∫X|f |dµ <∞. Then we

say that f is integrable, or µ-integrable. If f is integrable, we define∫X

fdµ :=∫X

f+dµ−∫X

f−dµ.

Prop 54: If f ∈ L1, then∣∣∫Xfdµ

∣∣ ≤ ∫X|f |dµ.

Prop 55: Let f, g ∈ L1 be given. TFAE:

(a)∫Efdµ =

∫Egdµ for all E ∈M .

(b)∫X|f − g|dµ = 0

(c) f = g µ-a.e.

Def 56: We define L1(X,µ), or just L1(X), L1(µ), or L1, to be the collection of equivalence classes given in Proposition55.

Prop 57: Suppose that µ is a complete measure.

(a) If f is measurable and f = g µ-a.e. then g is measurable.

(b) If fj∞j=1 is a sequence of measurable functions and limj→∞ fj = f exists µ-a.e. x ∈ X then f is measurable.(cf. Proposition 2.6)

Prop 58: 11 Let (X,M , µ) be the completion of (X,M , µ). If f : X → R is M -measurable then there is g : X → Rthat is M -measurable such that g = f µ-a.e. By convention, we do not distinguish between L1(µ) and L1(µ).

Def 59: Define ρ1 : L1(µ)× L1(µ)→ [0,+∞] by ρ1(f, g) =∫X|f − g|dµ.

Prop 60: The function ρ1 is a metric on L1(µ).Thm 61: (Dominated Convergence Theorem) Let fj∞j=1 ⊆ L

1(µ) be a sequence satisfying

(a) limj→∞

fj(x) = f(x) µ-a.e. x ∈ X.

(b) There is a g ∈ L1(µ) such that |fj(x)| ≤ g(x) for all j ∈ N µ-a.e. x ∈ X.

Then f ∈ L1(µ) and ∫X

fdµ = limj→∞

∫X

fjdµ.

Def 62: If fj∞j=1 ⊆ L1(µ) and f ∈ L1(µ) satisfy limj→∞

ρ1(fj , f) = 0, then we write fj → f in L1 and we say that fj

converge strongly to f in L1.Cor 63: Under hypotheses in DCT, we actually have that fj → f in L1.

3 Some Functional Analysis

Def 64: 12 A seminorm on X is a function ||·|| : X → [0,∞) such that ||x+ y|| ≤ ||x||+ ||y|| for all x, y ∈ X and||λx|| = |λ| · ||x|| for all λ ∈ E and x ∈ X. If a seminorm satisfies ||x|| = 0 if and only if x = 0 then we say ||·|| is anorm on X.

Example 65: 13

(a) Rn is a v.s. The function ||·||p : Rn → [0,∞) defined by ||x||p :=(∑n

j=1 |xj |p)1/p

for p ∈ [1,∞) is a norm onRn. Also, ||·||∞ : Rn → [0,∞) defined by ||x||∞ = max |x1|, · · · , |xn| is also a norm on Rn.

(b) Define ||·||1 : L1(µ) → [0,∞) by ||f ||1 :=∫X|f |dµ. It is easily verified that (L1(µ), ||·||1) is a normed vector

space.

Mike Janssen

3 SOME FUNCTIONAL ANALYSIS Analysis Study Guide

Def 66: Two norms ||·|| and ||·||] on X are equivalent if there are constants C1, C2 > 0 finite such that C1 ||x|| ≤||x||] ≤ C2 ||x|| for all x ∈ X. Equivalent norms induce the same topologies.

Example 67: In Rn, each of the norms ||·||p for p ∈ [1,∞) are equivalent. In RN, the norms ||x||p =(∑∞

j=1 |xj |p)1/p

where x = xj∞j=1 ∈ RN are not equivalent as p varies.Def 68: If (X, ||·||) is a complete normed vector space with respect to the metric ρ||·|| then we call (X, ||·||), or just X,

a Banach space (or a B-space).

Def 69: Let (X, ||·||) be given. Let xj∞j=1 ⊆ X be given. The series∑∞j=1 xj converges to x ∈ X iff lim

N→∞

N∑j=1

xj = x

with respect to the norm; i.e., limN→∞

∣∣∣∣∣∣∑Nj=1 xj − x

∣∣∣∣∣∣ = 0. We say that xj∞j=1 is absolutely convergent if∞∑j=1

||xj || <∞.

Thm 70: (p. 152) A normed vector space is complete iff all absolutely convergent sequences converge in X.Thm 71: The normed vector space (L1(µ), ||·||) is a B-space.Prop 72: 14 The set of simple functions on L1(µ) is dense in L1(µ) with respect to the strong (norm) topology.Def 73: 15 For each p ∈ (0,∞), define Lp(X,µ) by Lp(X,µ) :=

f meas :

∫X|f |p dµ <∞

. Define L∞(X,µ) :=

f measurable : ess supx∈X |f(x)| <∞. Here, for f : X → R measurable, then

ess supx∈X f(x) := infa ∈ R : µ(x ∈ X : f(x) > a) = 0

.

In other words, f(x) ≤ ess supx∈X f(x) a.e. and ess supx∈X f(x) is the smallest number such that this inequalityholds.

Def 74: For each p ∈ (0,∞), define ||·||p : Lp(µ)→ [0,∞) by ||f ||p =(∫X|f |p dµ

)1/p. Define ||·||∞ : L∞(X,µ)→ [0,∞)by ||f ||∞ := ess supx∈X |f(x)|. The triangle inequality is only satisfied if p ≥ 1.

Thm 75: (Minkowski’s Inequality) Suppose that p ∈ [1,∞). Let f, g ∈ L+ be given. Then(∫X

(f + g)p dµ)1/p

≤(∫

X

fp dµ)1/p

+(∫

X

gp dµ)1/p

.

Thm 76: (Holder’s Inequality) Suppose that p, q ∈ (1,∞) satisfy 1/p+ 1/q = 1. Let f, g ∈ L+ be given.∫X

fg dµ ≤(∫

X

fp dµ)1/p(∫

X

gq dµ)1/q

.

Thm 77: (Young’s Inequality) Suppose p, q ∈ (1,∞) satisfy 1/p + 1/q = 1. Let a, b ∈ R be given. Then |ab| ≤(1/p)|a|p + (1/q)|b|q.

Rmrk: It follows from Minkowski that the functions ||·||p are norms in Lp for p ∈ [1,∞). The case for p = ∞ can beproved directly.

Thm 78: 16 The normed vector space (Lp(µ), ||·||p) is a B-space for all p ∈ [1,∞].Prop 79: For each p ∈ [1,∞], the simple functions are dense in Lp with respect to the norm topology.Prop 80: 17 If 1 ≤ p ≤ q ≤ r ≤ ∞ then Lp ∩ Lr ⊆ Lq and ||f ||q ≤ ||f ||

λp ||f ||

1−λr where 1/q = λ/p+ (1− λ)/r.

Prop 81: If µ is finite and 1 ≤ p ≤ q ≤ ∞, then Lq ⊆ Lp and ||f ||p ≤ ||f ||q µ(X)1/p−1/q.Modes of Convergence Let fj∞j=1 and f be measurable functions.

• We say fj → f uniformly in X if for all ε > 0 there exists an N(ε) ∈ N such that for all n ≥ N(ε) we have|fn(x)− f(x)| < ε.

• We say fj → f pointwise if for each x ∈ X and ε > 0 there is N(ε, x) ∈ N such for j ≥ N(ε, x) we have|fj(x)− f(x)| < ε.

• Convergence in Lp for p ∈ [1,∞) We say fj → f in Lp(X,µ) if for each ε > 0 there is N(ε) ∈ N such that||fj − f ||p < ε for all j ≥ N(ε).

• We say fj → f almost everywhere in X if there is some µ-null set E such that fj → f pointwise in X \E.

• We say fj → f in measure if for each ε > 0 we have limj→∞ µ (x ∈ X : |fj(x)− f(x)| ≥ ε) = 0.

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4 CONSTRUCTION OF MEASURES Analysis Study Guide

• Cauchy in measure18: We say that that fj∞j=1 is Cauchy in measure if for each ε > 0,limj,k→∞ µ (x ∈ X : |fj(x)− fk(x)| ≥ ε) = 0.

Thm 82: (Egoroff’s Theorem) Suppose that µ is finite and fj∞j=1 and f are measurable functions such that fj → f

a.e. in X. Then for each α > 0 there is a measurable set E ⊆ X such that µ(E) < α and fj → f uniformly on X \E.Thm 83: Suppose that fj∞j=1 is a sequence of measurable functions that are Cauchy in measure. Then there exists a

measurable f such that fj → f in measure.Thm 84: 19 If fj∞j=1 is a sequence of measurable functions such that fj → f in measure, with f measurable, then

there is a subsequence fjk∞k=1 ⊆ fj

∞j=1 such that fjk → f for µ-a.e. x ∈ X.

Thm 85: If fj∞j=1 is a sequence of measurable functions such that fj → f in measure and fj → g in measure formeasurable f, g, then f = g for µ-a.e. x ∈ X.

Thm 86: Let p ∈ [1,∞] be given. If fj∞j=1 ⊆ Lp(µ) such that fj → f in Lp for some f ∈ Lp then there is a subsequence

fjk∞k=1 ⊆ fj

∞j=1 such that fjk → f for µ-a.e. x ∈ X.

Example 87: Let X = N, M = P(N), and µ the counting measure. So for f : N → R we have∫X

f dµ =∞∑k=1

f(k)

(assuming the series converges absolutely). Suppose that fj(k) = k/j for all k, j ∈ N. Then fj(k)→ 0 pointwise in

N, fj 6→ 0 uniformly, given p ∈ [1,∞),∫

N|fj − 0|p dµ = 1/jp

∞∑k=1

kp diverges, and fj 6→ 0 in measure.

4 Construction of Measures

4.1 Caratheodory’s Approach

Thm (Ulam):20 Suppose that µ is a measure on R such that every subset E ⊆ R is measurable. If µ([n, n+ 1)) <∞for each n ∈ Z and µ(x) = 0 for each x ∈ R, then µ(E) = 0 for all E ⊆ R.

Def 88: A family of sets E ⊆P(X) is a semi-algebra (or elementary family) on X if it has the following properties:

(a) ∅, X ∈ E

(b) If E1, E2 ∈ E then E1 ∩ E2 ∈ E

(c) If E ∈ E then Ec = ∪ni=1Ei where Ei is a finite disjoint sequence of sets in E .

Example 89:

(a) I := open, half-closed, half-open, closed intervals in R.

(b) In := cartesian product of n elements of I.

Notation: I(a, b) will be used to denote any one of the intervals (a, b), (a, b], [a, b], or [a, b), for all a, b ∈ R which makesense. Define I =

I(a, b) : a ≤ b, a, b ∈ R

and In =

∏nj=1 I(aj , bj) : aj ≤ bj for j = 1, · · · , n

.

Def 90: 21 Given n ∈ N, define mn : In → [0,∞] by mn(∏n

j=1 I(aj , bj))

=∏nj=1(bj − aj). When n is understood, we

write m for mn.Def 91: Let E ⊆P(X) be a semi-algebra. A set function µ : E → [0,∞] is called

• monotone if µ(E) ≤ µ(F ) whenever E,F ∈ E and E ⊆ F .

• finitely additive if µ(∪nj=1Ej) =n∑j=1

µ(Ej) whenever Ejnj=1 ⊆ E is a sequence of mutually disjoint sets with

∪nj=1Ej ∈ E .

• countably additive if µ(∪∞j=1Ej) =∞∑j=1

µ(Ej) whenever Ej∞j=1 ⊆ E are mutually disjoint and ∪∞j=1Ej ∈ E .

• countably subadditive if µ(∪∞j=1Ej) ≤∞∑j=1

µ(Ej) whenever Ej∞j=1 ⊆ E and ∪∞j=1Ej ∈ E .

Mike Janssen

4 CONSTRUCTION OF MEASURES Analysis Study Guide 4.1 Caratheodory’s Approach

Example 92: It is possible to show that mn is monotone, finitely additive, and countably subadditive (and countablyadditive, but this is hard to show).

Def 93: A family of sets A ⊆P(X) is called an algebra if

(a) ∅, X ∈ A

(b) If E1, E2 ∈ A then E1 ∩ E2 ∈ A

(c) If E ∈ A then Ec ∈ A.

Prop 94: 22 Let C ⊆ P(X) be given. Then there exists a unique smallest algebra A(C) ⊆ P(X) such that C ⊆ A(C)and if F ⊆P(X) is an algebra containing C then A(C) ⊆ F .

Def 95: The algebra provided by Proposition 94 is called the algebra generated by C.Prop 96: Suppose that E ⊆P(X) is a semi-algebra. Then

A(E) =E ⊆ X : E = ∪nj=1Ej for some disjoint sequence Ejnj=1 ⊆ E

Example 97:A(I) = finite unions of intervals in R, and A(In) = finite unions of boxes in Rn.Thm 98: Suppose that µ is a finitely-additive, countably subadditive set function on a semi-algebra E such that µ(∅) = 0.

Then there exists a unique countably subadditive set function µ on A(E) extending µ.Thm 99: Let A be an algebra of sets and let µ : A → [0,∞] such that µ(∅) = 0 be given. Then µ is countably additive

if and only if µ is finitely additive and countably subadditive.Def 100: 23 Suppose that A ⊆P(X) is an algebra. A function µ : A → [0,∞] is a premeasure on A if

(a) µ(∅) = 0

(b) µ(∪∞j=1Ej

)=∞∑j=1

(Ej) whenever Ej∞j=1 is mutually disjoint and ∪∞j=1Ej ∈ A.

Notation: Denote the left-open and right-closed intervals in R by H, so

H := (a, b] : (a, b] ⊆ R ∪

(a,∞) : a ∈ R.

Now, H is a semi-algebra, and BR will be generated by H. Given a function F : R→ R, define F (∞) := limx→∞

F (x)

and F (−∞) := limx→−∞

F (x), provided the limits exist in R.

Prop 101: Let F : R→ R be a non-decreasing function. Define µF : H → [0,∞] by µF ((a, b]) := F (b)−F (a) if (a, b] ⊆ Rand µF ((a,∞)) := F (∞)− F (a) if a ∈ [−∞,∞). Then µF is well-defined. Moreover, if µF is right-continuous, thenµF is countably subadditive.

Prop 102: 24 Let F : R → R be non-decreasing and right-continuous. Define µF : H → [0,∞] as in Proposition 101.

Then µF : A(H) → [0,∞] defined by µF (∪nj=1Ij) =n∑j=1

µF (Ij) whenever Ijnj=1 ⊆ H is mutually disjoint is a

premeasure on A(H).Prop 103: Suppose that µ : H → [0,∞] is finitely additive and µ((a, b]) <∞ for each a, b ∈ R. Then there is a function

F : R → R that is nondecreasing such that µ((a, b]) = F (b) − F (a) for all a, b ∈ R. If µ is countably additive, thenF is right-continuous and µ = µF .

Def 104: 25 An outer measure on X is a monotone countably subadditive set function µ∗ : P(X)→ [0,∞] such thatµ∗(∅) = 0.

Prop 105: Let C ⊆P(X) and ρ : C→ [0,∞] such that ∅, X ∈ C and ρ(∅) = 0. For each E ⊆ X define

µ∗(E) := inf

∞∑j=1

ρ(Ej) : Ej ⊆ C, E ⊆ ∪∞j=1Ej

.Then µ∗ is an outer measure.

Def 106: We call µ∗ in Proposition 105 the outer measure induced by ρ.Def 107: If µ∗ is an outer measure on X then a set A ⊆ X is called µ∗-measurable if µ∗(E) = µ∗(E∩A) +µ∗(E∩Ac)

for all E ⊆ X.

Mike Janssen

4 CONSTRUCTION OF MEASURES Analysis Study Guide 4.1 Caratheodory’s Approach

Thm 108: (Caratheodory’s Theorem) If µ∗ is an outer measure on X, then the collection M of µ∗-measurable sets is aσ-algebra. Moreover, the restriction of µ∗ to M is a complete measure.

Prop 109: 26 If µ is a premeasure on an algebra A ⊆P(X) and µ∗ is the outer measure induced by µ then µ∗|A = µand every set of A is µ∗-measurable.

Thm 110: Let A ⊆P(X) be an algebra and let M (A) be the σ-algebra generated by A. Let µ be a premeasure on A.

(a) There is a measure µ on M (A) such that µ|A = µ. (In fact, µ = µ∗|M (A).

(b) If ν is another measure extending µ to M (A), then ν(E) ≤ µ(E) for all E ∈M (A), and µ(E) = ν(E) wheneverµ(E) <∞.

(c) If µ is σ-finite then µ is the unique extension of µ to a measure on M (A).

Rmrk: 27 Recall (Proposition 102) that if F : R→ R is a non-decreasing right-continuous function then µF : A(H)→[0,∞] is a premeasure on A(H). Since µF is σ-finite on R, there is a unique extension of µF to a measure µF onthe σ-algebra generated by A(H), which is BR. Actually, Caratheodory’s Theorem gives a complete measure on theσ-algebra of µ∗F -measurable sets. Usually we use µF for both the measure on BR and the measure on MµF (theµ∗F -measurable sets). The complete measure µF on MµF is called the Lebesgue-Stieltjes measure associatedwith F .

If E ∈MµF , then

µF (E) = inf

∞∑j=1

[F (bj)− F (aj)] : E ⊆ ∪∞j=1(aj , bj ]

.In fact,

µF (E) = inf

∞∑j=1

[F (bj)− F (aj)] : E ⊆ ∪∞j=1(aj , bj)

.Def 111: Let µF be a Lebesgue-Stieltjes measure on MµF . If g ∈ L1(µF ) then

∫Rg dµF is called the Lebesgue-Stieltjes

integral of g with respect to µF .Thm 112: If F : R→ R is non-decreasing and continuously differentiable, and g : R→ R is continuous, then∫

(a,b]

g dµF =∫ b

a

gF ′ dx.

Cor 113: If F (x) = x, then µF = m and ∫(a,b]

g dm =∫ b

a

g dx.

Def 114: Let µ be a Borel measure on X and let E ∈ BX be given. The measure µ is outer regular on E if

µ(E) = inf µ(U) : E ⊆ U and U is open.

The measure µ is inner regular on E if

µ(E) = sup µ(K) : K ⊆ E and K is compact.

µ is called regular if it is both inner regular and outer regular on all E ∈ BX .Def 115: A Borel measure µ on X is called a Radon measure if µ is

(a) finite on all compact sets

(b) outer regular on all E ∈ BX

(c) inner regular on all open sets.

Mike Janssen

4 CONSTRUCTION OF MEASURES Analysis Study Guide 4.2 Hausdorff Measures

Thm 116: Let µF be a Lebesgue-Stieltjes measure on R. Then µF is inner and outer regular on all E ∈ BR. In fact,µF is “inner and outer regular” on all E ∈MµF .

Cor 117: 28 If µF is a Lebesgue-Stieltjes measure then it is a Radon measure. In fact, it is a regular measure.Thm 118: Let µF be a Lebesgue-Stieltjes measure. If E ⊆ R, then TFAE:

(a) E ∈MµF .

(b) E = V \N1 for some Gδ set V and µF -null set N1.

(c) E = H ∪N2 for some Fσ set H and µF -null set N2.

Notation: 29 Given E ⊆ Rn, T an n × n matrix over R, and s ∈ Rn, put TE = Tx : x ∈ E and E + s =s+ x : x ∈ E.

Thm 119: If E ∈ L n then E + s ∈ L n and TE ∈ L n for all s ∈ Rn and T ∈ Rn×n. Moreover, m(E + s) = m(E) andm(TE) = |detT |m(E).

Cor 120: If E ∈ L then E + s ∈ L and rE ∈M . Moreover, m(E + s) = m(E) and m(E) = |r|m(E) (for s, r ∈ R).Here rE = rx : x ∈ E.

Rmrk: If µ is another Radon measure on Rn such that µ(E + s) = µ(E) for all s ∈ Rn and all E ∈ BR then there is ac ≥ 0 such that µ = cm.

Example:30 Examples of sets in L :

(a) Consider E = Q ∩ [0, 1] = ∪∞j=1 qj. Thus m(E) = 0 but m(E) = m([0, 1]) = 1. Thus E is topologically“large” but measure-theoretically “small”.

(b) Let ε ∈ (0, 1/2) be given. For each j ∈ N, set Uj = (qj − ε2−j , qj + ε2−j) ∩ [0, 1]. Put U = ∪∞j=1Uj . Then U is

open in the subspace topology. We have m(U) ≤∞∑j=1

m(Uj) =∞∑j=1

ε2−j+1 = 2ε. But m(U) = m([0, 1]) = 1. So

U is an open set with (arbitrarily) small measure.

(c) Consider [0, 1] \ U . This is a closed nowhere-dense set of measure 1− 2ε.

Example:31 Discussion of different strange measurable sets and their behaviors under strange functions. See notes.

4.2 Hausdorff Measures

We seek a way to measure low-dimensional subsets of Rn32. Note that the Lebesgue measure of a lower-dimensionalsubset of Rn is always 0. There is a continuum of Hausdorff measures in Rn that can be used to provide a refinedmeasure of Lebesgue null sets in Rn.

Let (X, ρ) be a metric space.Def 121: Let p ≥ 0 be given and δ > 0. Define Hp,δ, Hp : P(X)→ [0,∞] by

Hp,δ(E) := inf

∞∑j=1

(diam(Ej))p : E ⊆ ∪∞j=1Ej , diam(Ej) ≤ δ

and

Hp(E) := limδ→0+

Hp,δ(E).

The function Hp is called the p-dimensional Hausdorff (outer) measure.Rmrk: In the definition of Hp,δ, the sets Ej can be required to be open or closed.Rmrk:H0 is the counting measure.Rmrk: In Rn, the n-dimensional volume of an m-dimensional ball b is proportional to (diam(B))m. In fact, the volume

of an m-dimensional ball with the same diameter as an n-dimensional ball C is also proportional to (diam(C))m.Def 122: An outer measure µ∗ on (X, ρ) is called a metric outer measure if µ∗(A ∪ B) = µ∗(A) + µ∗(B) whenever

ρ(A,B) > 0. This is clearly more than disjointness. Recall ρ(A,B) = inf ρ(x, y) : x ∈ A, y ∈ B.Prop 123: If µ∗ is a metric outer measure, then every Borel set is µ∗-measurable.Prop 124: Let A be a Borel set. If Hp(A) < ∞, then Hq(A) = 0 for all q > p. If Hp(A) > 0 then Hq(A) = ∞ for

all q < p. It follows that inf p ≥ 0 : Hp(A) = 0 = sup p ≥ 0 : Hp(A) =∞. This value is called the Hausdorffdimension of A.

Mike Janssen

4 CONSTRUCTION OF MEASURES Analysis Study Guide 4.3 Product Spaces

4.3 Product Spaces

Def 125: 33 Let (Xα,Mαα∈A be measurable spaces. Then the product σ-algebra on X =∏α∈AXα is the

σ-algebra generated byπ−1α (Eα) : Eα ∈Mα, α ∈ A

where πα : X → Xα is the α-th coordinate map.

Notation: This product σ-algebra is denoted by⊗α∈A

Mα.

Prop 126: If A is countable, then⊗α∈A

Mα is the σ-algebra generated by ∏α∈A

Eα : Eα ∈Mα

.

Prop 127: Let X1, · · · , Xn be metric spaces. Let X =∏∞j=1Xj equipped with the product metric. Then

n⊗j=1

BXi ⊆ BX .

If each Xj is separable, thenn⊗j=1

BXj = BX .

Cor 128: BRn =n⊗j=1

BR.

Let (X,M , µ) and (Y,N , ν) be measure spaces.Def 129: If E ∈M , F ∈ N , then E × F is a measurable rectangle. We denote the set of measurable rectangles byR. Note that M ⊗N is the σ-algebra generated by R by Proposition 126. Observe that R is a semi-algebra.

Thm 130: Let π : R → [0,∞] be defined by π(E × F ) = µ(E)ν(V ). Then π is well-defined, countably additive, andπ(∅) = 0.

Thm 131: There exists a unique extension of π to a premeasure π on A(R).Thm 132: The premeasure π generates an outer measure π∗ on X × Y . The restriction of π∗ to M ⊗N is a measure

extending π. Moreover, if µ and ν are σ-finite then so is the measure π∗|M⊗N . This is the unique extension of π toa measure on M ⊗N .

Notation: We denote the measure π∗|M⊗N by µ× ν.Question: How does one measure E ∈M ⊗N ?Def 133: 34 If E ⊆ X × Y , then for each x ∈ X then the x-section of E is Ex := y ∈ Y : (x, y) ∈ E. For each

y ∈ Y , the y-section of E is Ey := x ∈ X : (x, y) ∈ E. If f : X × Y → R then the x-section fx and y-sectionfy of f are fx(y) = f(x, y) and fy(x) = f(x, y).

Prop 134:

(a) If E ∈M ⊗N , then Ex ∈ N for all x ∈ X and Ey ∈M for all y ∈ Y .

(b) If f is M ⊗ N -measurable then fx is N -measurable and fy is M -measurable for each x ∈ X, y ∈ Y ,respectively.

Def 135: A collection C ⊆P(X) is a monotone class if it has the following properties:

(a) If Ej∞j=1 ⊆ C and E1 ⊆ E2 ⊆ · · · , then ∪∞j=1Ej ∈ C.

(b) If Ej∞j=1 ⊆ C and E1 ⊇ E2 ⊇ · · · then ∩∞j=1Ej ∈ C.

Note: All σ-algebras are monotone classes. Given any E ⊆ P(X), there is a unique smallest monotone class, C(E),called the monotone class generated by E .

Thm 136: M ⊆P(X) is a σ-algebra if and only if M is a monotone class and an algebra.Fact: 35 If E ∈M ⊗N , we want

µ× ν(E) =∫X

ν(Ex) dµ(x)

=∫Y

µ(Ey) d ν(y).

Lma 137: (Monotone Class Lemma) If A ⊆P(X) is an algebra, then the monotone class C(A) and the σ-algebra M (A)are equal.

Thm 138: Suppose that (X,M , µ) and (Y,N , ν) are σ-finite spaces. Let E ∈M ⊗N be given. Then:

(i) x 7→ ν(Ex) and y 7→ µ(Ey) are measurable, and

Mike Janssen

5 DECOMPOSITION OF MEASURES Analysis Study Guide

(ii) We have

µ× ν(E) =∫X

ν(Ex) dµ(x)

=∫Y

µ(Ey) d ν(y).

Thm 139: (Fubini-Tonelli) Suppose (X,M , µ) and (Y,N , ν) are σ-finite.

(a) (Tonelli) If f ∈ L+(X × Y ) then g(x) =∫Y

fx d ν(y) and h(y) =∫X

fy dµ(x) are in L+(X) and L+(Y ),

respectively, and ∫X×Y

f(x, y) d(µ× ν) =∫X

∫Y

f d ν dµ =∫Y

∫X

f dµ d ν. (1)

(b) (Fubini) If f ∈ L1(µ × ν) then fx ∈ L1(ν) and fy ∈ L1(µ) for a.e. x ∈ X and a.e. y ∈ Y . Moreover, thefunctions g and h from (a) are in L1(µ) and L1(ν), respectively, and (1) holds.

Rmrk: To use (b), you need to show f ∈ L1. The common method for doing so is as follows: If f is measurable, then|f | ∈ L+, and so (a) implies ∫

X×Y|f |d(µ× ν) =

∫X

∫Y

|f |d ν dµ =∫Y

∫X

|f |dµ d ν,

and then show that one of the last two expressions is finite.

5 Decomposition of Measures

5.1 Intro and Definitions

Let (X,M ) be a measurable space.Def 140: 36 Let µ, ν be measures on M .

(a) µ and ν are mutually singular, and we write µ ⊥ ν, if there are disjoint sets Xµ and Xν in M such thatX = Xµ ∪Xν and µ(E) = µ(E ∩Xµ) and ν(E) = ν(E ∩Xν) for all E ∈M .

(b) We say that ν is absolutely continuous with respect to µ, and we write ν µ, if for every E ∈M , wehave ν(E) = 0 whenever µ(E) = 0.

(c) ν is diffuse with respect to µ if for each E ∈M we have ν(E) = 0 whenever µ(E) <∞.

Rmrk:

(a) If µ ⊥ ν, then µ(Xν) = 0 and ν(Xµ) = 0.

(b) If ν is diffuse with respect to µ, then ν µ.

(c) If f ∈ L+, then ν : M → [0,∞] defined by

ν(E) :=∫E

f dµ (2)

is a measure. By Proposition 41, we have that ν µ. It turns out that (2) characterizes all measures that areabsolutely continuous with respect to µ if µ is σ-finite, i.e., ν µ iff (2) holds for some f ∈ L+.

Mike Janssen

5 DECOMPOSITION OF MEASURES Analysis Study Guide 5.1 Intro and Definitions

Thm 141: Let µ, ν be measures on M with ν finite. Then ν µ iff for each ε > 0 there is a δ > 0 such that ν(E) < εwhenever E ∈M and µ(E) < δ.

Cor 142: 37 If f ∈ L1(µ), then for each ε > 0 there is a δ > 0 such that if µ(E) < δ then∣∣∣∣∫E

f dµ∣∣∣∣ < ε.

Notation: If ν(E) =∫Ef dµ, with f ∈ L+(µ), then we may write

dν

dµfor f , or we may write d ν for f dµ. Sometimes

we will refer to d ν as a measure.Thm 143: (Radon-Nikodym Theorem I) Let µ and ν be measures on M . Suppose that µ is σ-finite and ν µ. Then

there is an f ∈ L+ such that ν(E) =∫E

f dµ for each E ∈M . The function f is unique up to a µ-null set.

Lma 144: Let µ, ν be measures on M . Define νac : M → [0,+∞] by

νac(E) := sup∫

E

f dµ : f ∈ L+ and∫F

f dµ ≤ ν(f)∀F ⊆ E,F ∈M

.

Then νac is a measure, νac µ, and, for each E ∈M there is an admissible f such that νac(E) =∫E

f dµ. Moreover,

if νac is σ-finite, then f can be chosen independently of E.Lma 145: 38 Let µ, ν be finite measures on M . For each E ∈M , define

(ν − µ)+(E) := sup ν(F )− µ(F ) : F ⊆ E and F ∈M .

Then (ν − µ)+ is a measure, and for each E ∈M we have that

(ν − µ)+(E) := supν(F )− µ(F ) : F ⊆ E, F ∈M , and (µ− ν)+(F ) = 0

.

Def 146: 39 Let F = fαα∈A be a family of measurable functions fα : X → R. We call f : X → R an essentialsupremum of F if

(a) f is measurable

(b) f(x ≥ fα(x) for each α ∈ A and a.e. x ∈ X

(c) f(x) ≤ g(x) a.e. if g satisfies a and b.

Def 147: Let Eαα∈A ⊆ M be given. A set E ∈ M is called an essential union of Eαα∈A if χE is an essentialsupremum of χEαα∈A.

Rmrk: It can be shown that if χEαα∈A has an essential supremum then Eαα∈A has an essential union.Def 148: A measure µ on M is localizable if any family of measurable sets has an essential union.Rmrk: Any σ-finite measure is localizable.Thm 149: Let µ be a σ-finite measure on M . Let F = fαα∈A be a family of measurable functions, fα : X → R. Then

there is a countable set I ⊆ A such that the function x 7→ supα∈I

fα(x) is an essential supremum.

Thm 150: (Radon-Nikodym II) Let µ, ν be measures on M . Suppose that µ is localizable, ν µ and that

ν(E) = sup ν(E ∩ F ) : F ∈M and µ(F ) <∞

for all E ∈ M . Then for all E ∈ M there is an f ∈ L+(µ) such that ν(E) =∫Ef dµ. If µ is semifinite, then f is

unique up to a µ-null set.Thm 151: (Lebesgue Decomposition Theorem) Let µ, ν be a measure on M with µ and ν σ-finite. Then there exist

measures λ, ρ on M such that ρ µ, λ ⊥ µ and ν = ρ+λ. Moreover, this decomposition is unique, i.e., if ν = ρ+λwith ρ µ and λ ⊥ µ, then ρ = ρ and λ = λ.

Thm 152: 40 Let µ, ν be measures on M , with µ a σ-finite measure. Then

ν = νac + νs,

with νac µ given in Lemma 144 and

νs(E) := sup ν(F ) : F ∈M , F ⊆ E and µ(F ) = 0.

Mike Janssen

5 DECOMPOSITION OF MEASURES Analysis Study Guide 5.2 Signed Measures

5.2 Signed Measures

Def 153: A signed measure on (X,M ) is a function ν : M → [−∞,∞] such that

(a) ν(∅) = 0

(b) ν takes at most one of the values +∞ or −∞; i.e., ν : M → (−∞,+∞] or ν : M → [−∞,+∞)

(c) If En∞n=1 ⊆ M are mutually disjoint, then ν(∪∞n=1En) =∞∑n=1

ν(En), and the series converges absolutely if

|ν(∪n∈NEn)| <∞.

Rmrk: Positive measures are signed measures.Rmrk: Condition (c) can be simplified to just countable additivity, using a rearrangement of En∞n=1.Example 154:

(a) If µ1, µ2 are positive measures with at least one finite, then ν = µ1 − µ2 is a signed measure.

(b) If f is a measurable function, either∫X

f+ dµ or∫X

f− dµ is finite, then

ν(E) =∫E

f dµ =∫E

f+ dµ−∫E

f− dµ

is a signed measure. Such f are called extended µ-integrable.

Prop 155: Let ν be a signed measure on M . (Continuity from below) If En∞n=1 ⊆ M and E1 ⊆ E2 ⊆ · · · thenlimn→∞

ν(En) = ν(∪n∈NEn). (Continuity from above) If En∞n=1 ⊆ M with E1 ⊇ E2 ⊇ · · · and |ν(E1)| < ∞ then

limn→∞

ν(En) = ν(∩∞n=1En).Def 156: Let ν be a signed measure. A set E ∈M is called

(a) positive if ν(F ) ≥ 0 for all F ∈M with F ⊆ E;

(b) negative if ν(F ) ≤ 0 for all F ∈M such that F ⊆ E;

(c) null if ν(F ) = 0 for all F ∈M such that F ⊆ E.

Example 157: Let f : R → R be an odd function in L1(m). Further suppose f(x) > 0 for x > 0. Define ν by

ν(E) :=∫E

f dm for all E ∈ BR. Then E ∈ BR is a positive set if E ⊆ [0,∞), except possibly some m-null set;

E ∈ BR is a negative set if E ⊆ (−∞, 0], except possibly some m-null set; and E is a null set if m(E) = 0.Prop 158: Let ν be a signed measure on M . Let E ∈M be such that 0 < ν(E) <∞. Then there is a positive subset

F of E such that ν(F ) > 0.Thm 159: (Hahn Decomposition Theorem) Let ν be a signed measure on M . Then there are sets P,N ∈M such that

P ∩N = ∅, P ∪N = X, P is positive, and N is negative.Def 160: A pair of sets P and N satisfying the conclusion of Theorem 159 is called a Hahn decomposition for ν.Thm 161: 41 (Jordan Decomposition Theorem) Let ν be a signed measure on M . There are unique measures ν+ and

ν− on M such that ν = ν+ − ν− and ν+ ⊥ ν−.Def 162: Let ν be a signed measure on M . The decomposition in Theorem 161 is called the Jordan decomposition

of ν. The measure ν+ is called the positive (upper) variation and the measure ν− is called the negative (lower)variation. The positive measure |ν| = ν+ + ν− is called the total variation.

Def 163: Let ν, ω be signed measures on M .

(a) E ∈M is σ-finite (with respect to ν) if it is σ-finite for |ν|.

(b) ν is σ-finite if |ν| is.

(c) ν and ω are mutually singular if |ν| ⊥ |ω|.

(d) ν is absolutely continuous with respect to ω if |ν| |ω|.

Mike Janssen

6 DIFFERENTIATION Analysis Study Guide

Def 164: Suppose that ν is a signed measure on M . Set L1(ν) = L1(ν+) ∩ L1(ν−), and define∫X

f d ν =∫X

f d ν+ −∫X

f d ν−

for all f ∈ L1(ν).Thm 165: (Lebesgue-Radon-Nikodym Theorem) Let ν be a σ-finite signed measure on M and let µ be a positive σ-finite

measure on M . Then there are unique σ-finite signed measures λ, ρ on M such that

λ ⊥ µ, ρ µ, and ν = ρ+ λ

Moreover, there is an extended µ-integrable function f such that d ρ = f dµ. The function f is unique up to a µ-nullset. The decomposition of ν into λ and ρ is unique.

Example 166: 42

(a) Let ν be a σ-finite signed measure. Let P,N be a Hahn decomposition for ν. Then ν(E) = ν+(E)− ν−(E) =∫E

χP d ν+ −∫E

χN d ν− =∫E

[χP − χN ] d |ν|. Thus ν |ν| andd ν

d |ν|= χP − χN .

(b) Let F : R→ R be given by

F (x) =

0, x < 03− e−x, 0 ≤ x < 14− e−x, 1 ≤ x <∞.

Then F is nondecreasing and right-continuous. The Lebesgue-Stieltjes measure µF satisfies is µF (E) =∫E

e−x dµ+ 2δ0 + δ1.

Example 167: 43

(a) If ψ is the Cantor-Vitali function then set

F (x) =

0 x < 0ψ(x) 0 ≤ x < 11 x > 1

which is a continuous, nondecreasing function. If µF is the associated Lebesgue-Stieltjes measure, then it canbe shown that µF ⊥ m.

(b) The measure µ : P(R) → [0,∞] defined by µ(E) := card(E ∩ Z) is singular with respect to m. The measureµ is discrete.

6 Differentiation

We will look at using the LRN Theorem in particular contexts. We first consider the setting where (X,M ) =

(Rn,BRn). In this setting, there are ways of identifyingd νdm

more explicitly.

Def 168: A family Err>0 ⊆ BRn is said to shrink nicely to x ∈ Rn if

(a) Er ⊆ B(r, x) for each r > 0

(b) There exists α > 0 such that m(Er) ≥ αm(B(r, x)) for all r > 0.

Def 169: 44 A measurable function f : Rn → R is locally integrable with respect to m if∫K

|f |dm < ∞ for every

compact set K ∈ BRn . The space of locally integrable functions on Rn is called L1loc(R), or just L1

loc.

Mike Janssen

6 DIFFERENTIATION Analysis Study Guide

Def 170: Let f ∈ L1loc be given. For each bounded set E ∈ BRn with m(E) > 0 we define the mean value (or average

value) of f over E by1

m(E)

∫E

f dm.

Notation:−∫f dm =

1m(E)

∫E

f dm (assuming 0 < m(E) <∞)

Thm 171: (Lebesgue Differentiation Theorem) Suppose that f ∈ L1loc. For each Lebesgue point x ∈ Rn of f , we have

limr→0+

−∫Er

|f(y)− f(x)|dm(y) = 0

andlimr→0+

−∫Er

f(y) dm(y) = f(x)

for each Err>0 ⊆ BRn that shrink nicely to x.Recall that the elements of L1 (and L1

loc) are actually equivalence classes of functions.

Thm 172: Let f ∈ L1loc be given. Then for a.e. x ∈ Rn, lim

r→0+−∫B(r,x)

f dm = f(x). THis means that, given a representative

g for the equivalence class of f , we will find that limr→0+

−∫B(r,x)

g dm = f(x) for a.e. x ∈ Rn. This limit is independent

of the choice of the equivalence class representative. The function

f∗(x) :=

limr→0+

−∫B(r,x)

f dm if the limit exists

0 otherwise

is called the precise representative for the equivalence class f ∈ L1loc.

Def 173: For each f ∈ L1loc (f is a representative of an equivalence class), the set

Lf :=

x ∈ Rn : lim

r→0+−∫B(r,x)

|f(y)− f(x)|dm(y) = 0

is called the Lebesgue set of f , and its points are called Lebesgue points of f .Thm 174: If f ∈ L1

loc, then m(Rn \ Lf ) = 0.Rmrk: Theorem 174 states that f ∈ L1

loc can be identified with the precise representative for a.e. x ∈ Rn. Theorem 171(LD) says that we can replace B(r, x)r>0 with Err>0 that shrink nicely to x.

Thm 175: 45 If f ∈ L1(m), then for each ε > 0 there exists a continuous g : Rn → R such that ||f − g||L1 < ε.Def 176: 46 Let f ∈ L1

loc be given. The Hardy-Littlewood Maximal function Hf : Rn → [0,∞] is given by

Hf(x) = supr>0−∫B(r,x)

|f |dm.

Prop 177: Let f, g ∈ L1loc be given. Then

(i) 0 ≤ Hf(x) ≤ +∞ for all x ∈ R.

(ii) H(f + g) ≤ Hf +Hg.

(iii) H(cf) = |c|Hf for all c ∈ R.

(iv) Hf is lower semi-continuous.

(v) Hf is BRn -measurable.

Rmrk: Recall that a function h : Rn → R is lower semi-continuous if the set x ∈ Rn : h(x) > a is open for eacha ∈ R. Equivalently, h is lower semi-continuous if lim inf

y→xh(y) ≥ h(x).

Mike Janssen

6 DIFFERENTIATION Analysis Study Guide 6.1 Functions of Bounded Variation

Example 178: 47 Consider χ[0,1] ∈ L1loc. (In fact, χ[0,1] ∈ L1.) Then

Hχ[0,1] = supr>0−∫

(x−r,x+r)

χ[0,1] dm

= supr>0

12rm([0, 1] ∩ (x− r, x+ r))

=

1

2(1−x) if x ≤ 0

1 if x ∈ (0, 1)1

2x if x ≥ 1.

Observe that χ[0,1] ∈ L1 but Hχ[0,1] /∈ L1. In general, Hf /∈ L1(R) unless f = 0 for a.e. x ∈ Rn.Thm 179: (Chebyshev’s Inequality): Let (X,M , µ) be a measure space. If f ∈ Lp(µ) for some p ∈ [1,∞), then for all

α > 0, µ (x ∈ X : |f(x)| > α) ≤ 1αp||f ||pL1 .

Def 180: Let (X,M , µ) be a measure space. For a measurable function f : X → R, define

[f ]p := supα>0

(αpµ(x ∈ X : |f(x)| > α))

for p ∈ [1,∞). We say that f is in weak-Lp if [f ]p <∞.Rmrk: [·]p is not a norm, as it does not satisfy the triangle inequality.Rmrk: Lp ( weak-Lp and [f ]p ≤ ||f ||p. Although Hf /∈ L1 in general, it will belong to weak-L1 if f ∈ L1.Thm 181: (Maximal Theorem) There is a constant c > 0 such that for each f ∈ L1 and all α > 0 such that for each

f ∈ L1 and all α > 0, m(x ∈ X : Hf(x) > α) ≤ c

α||f ||L1 .

Lma 182: (Simple Vitali Covering Lemma) Let C be a collection of open balls in Rn and set U :=⋃B∈C

B. If τ < m(U)

then there exists disjoint balls Bjkj=1 ⊆ C such thatk∑j=1

m(Bj) >τ

3n.

Def 183: 48 A signed measure ν on BRn is regular if

(a) |ν|(K) <∞ for each compact K ⊆ Rn.

(b) |ν|(E) = inf |ν|(U) : U is open and E ⊆ U for all E ∈ BRn .

Rmrk: If ν is positive, then Definition 183 is equivalent to Definition 114.Rmrk: If ν is regular, then |ν| is a Radon measure.Rmrk: If ν is regular, then ν is σ-finite.Rmrk: If d ν = f dm then ν is regular iff |f | ∈ L1

loc.Thm 184: Let ν be a regular signed measure on BRn . Let d ν = dλ+ f dm be its Lebesgue-Radon-Nikodym decompo-

sition with respect to m. Then for m-a.e. x ∈ Rn,

limr→0+

ν(Er)m(Er)

= f(x)

for every family Err>0 ⊆ BRn that shrinks nicely to x.Note: ν regular implies ν σ-finite implies the Lebesgue-Radon-Nikodym Theorem can be used.Thm 185: 49 (Lusin’s Theorem) Let f : Rn → R be a Borel-measurable function. Suppose there is an A ∈ BRn such

that m(A) < ∞ and f(x) = 0 for all x ∈ Ac. For each ε > 0 there exists continuous g : Rn → R such thatsupx∈Rn

|g(x)| ≤ supx∈Rn

|f(x)| and m (x ∈ Rn : g(x) 6= f(x)) < ε.

6.1 Functions of Bounded Variation

Thm 186: 50 (Differentiability of Monotone Functions) Let F : R → R be a nondecreasing function, and defineG(x) = F (x+).

(a) The points of discontinuity for F constitute an m-null set. In fact, the set is countable.

Mike Janssen

6 DIFFERENTIATION Analysis Study Guide 6.1 Functions of Bounded Variation

(b) F is differentiable at a.e. x ∈ R and F ′ = G′ a.e. in R.

Def 187: 51 Let F : R→ R be given. Define TF : R→ [0,∞] by

TF (x) = sup

n∑j=1

|F (xj)− F (xj−1)| : n ∈ N and −∞ < x0 < x1 < · · · < xn = x

.We call TF the total variation function of F . If [a, b] ⊆ R, then TF (b)− TF (a) is the total variation of F over[a, b].

Def 188: Let F : R→ R be given. If TF (∞) <∞ then we say that F is of bounded variation on R. We set

BV(R) := F : R→ R : TF (∞) <∞.

Def 189: Let F : [a, b]→ R be given. If TF (b)− TF (a) <∞ then we say that F is of bounded variation over [a, b].We set

BV([a, b]) := F : R→ R : TF (b)− TF (a) <∞.

Rmrk: TF is increasing.

Rmrk: If F ∈ BV(R), then TF (b)− TF (a) = sup

n∑j=1

|F (xj)− F (xj−1) : n ∈ N and a = x0 < · · · < xn = b

.

Rmrk: BV(R) and BV([a, b]) are vector spaces.Rmrk: If F ∈ BV([a, b]) then

F (x) :=

F (a) x ≤ aF (x) x ∈ [a, b]F (b) x ≥ b

∈ BV(R).

Example 190:

(a) All bounded monotone functions are BV(X) for X ∈ R, [a, b].

(b) The function sin(x) is not in BV(R), but sin(x) · χK ∈ BV(R) for any compact K ⊆ R.

(c) The function

F (x) =

xk sin(1/x) x > 00 x ≤ 0

is not in BV if k = 0, 1.

Lma 191: If F ∈ BV(R), then x 7→ TF (x) + F (x) and x 7→ TF (x)− F (x) are both nondecreasing.

Thm 192: If F ∈ BV(R), then F =12

(TF + F )− 12

(TF − F ), and12

(TF + F ) and12

(TF − F ) are nondecreasing. Also,

if F is the difference of 2 nondecreasing bounded functions, then F ∈ BV(R).Thm 193:

(a) If F ∈ BV(R), then F (x+) and F (x−) exist for all x ∈ R, and F (−∞) and F (+∞) exist as well.

(b) If F ∈ BV(R), then F is discontinuous on a set of at most countably many points.

(c) If F ∈ BV(R) and G(x) := F (x+) then F ′ exists a.e. and F ′ = G′ a.e. in R.

Def 194: If F ∈ BV(R), then the decomposition F =12

(TF + F ) − 12

(TF − F ) is called the Jordan decomposition

of F . Moreover,12

(TF +F ) is called the positive variation of F and12

(TF −F ) is called the negative variationof F .

Def 195: 52 SetNBV(R) := F ∈ BV(R) : F (−∞) = 0 and F is right-continuous.

Lma 196: If F ∈ BV(R) then TF (−∞) = 0. If F ∈ BV(R) is right-continuous, then TF ∈ NBV(R).

Mike Janssen

7 CHANGE OF VARIABLES Analysis Study Guide

Thm 197: If µ is a finite signed Borel measure and F : R → R is defined by F (x) = µ((−∞, x]) for all x ∈ R thenF ∈ NBV. IF F ∈ NBV then there is a unique signed Borel measure µF such that µF ((−∞, x]) = F (x) for all x ∈ R.Moreover, |µF | = µTF ; i.e., |µF |((−∞, x]) = TF (x) for all x ∈ R.

Prop 198: 53 Let F ∈ BV be right-continuous. Then F ′ ∈ L1. Moreover, if F ∈ NBV then

(a) µF ⊥ m⇔ F ′ = 0 for m-a.e. x ∈ R.

(b) µF m⇔ F (x) =∫

(−∞,x]

F ′ dm for all x ∈ R.

Def 199: We say that F : R → R is absolutely continuous if for each ε > 0 there is a δ > 0 such that whenever(aj , bj)Nj=1 are disjoint intervals with

∑Nj=1 bj − aj < δ we also have

∑Nj=1 |F (bj) − F (aj)| < ε. We say F is

absolutely continuous on [a, b] if the above property holds with intervals taken as subsets of [a, b].Prop 200: If F ∈ NBV, then F is absolutely continuous if and only if µF m.

Cor 201: 54 If f ∈ L1(m), then F (x) :=∫

(−∞,x]

f dm ∈ NBV is absolutely continuous and F ′ = f a.e. Also, if

F ∈ NBV is absolutely continuous then F ′ ∈ L1 and F (x) =∫

(−∞,x]

F ′ dm.

Thm 202: (FTC for Lebesgue Integrals) Let [a, b] ⊆ R and F : [a, b]→ R be given. TFAE:

(a) F is absolutely continuous on [a, b]

(b) F (x) = F (a) +∫

(a,x]

f dm for some f ∈ L1([a, b],m)

(c) F ′ exists a.e. in [a, b], F ′ ∈ L1([a, b],m) and F (x) = F (a) +∫

(a,x]

F ′ dm.

Def 203: If F ∈ NBV, then the integral of a Borel-measurable function with respect to µF is called a Lebesgue-

Stieltjes integral (if it makes sense; see Definition 164). We may denote the integral by∫

Rg dF or

∫Rg(x) dF (x).

If F is absolutely continuous, then∫

Rg dF =

∫RgF ′ dm.

Thm 204: Suppose F,G ∈ NBV, and at least one of F or G is continuous. Then for any (a, b] ⊆ R, we have∫(a,b]

F dG = [F (b)G(b)− F (a)G(a)]−∫

(a,b]

GdF.

7 Change of Variables

Prop 205: 55 Let (X,M , µ) be a measure space and (Y,N ) a measurable space. Let T : X → Y be measurable,and define µ T−1 : N → [0,∞] by (µ T−1)(E) = µ(T−1(E)). Then µ T−1 is a measure on N .

Def 206: The measure µ T−1 is called the push forward of µ through T , or the measure induced by µ and T .Example 207: Let T : [0, 2π)→ R2 be given by T (t) = (cos (t) , sin (t)). Then T is a bijection between [0, 2π) and S1.

Let BS1 be the σ-algebra generated by T−1 : S1 → [0, 2π) (cf. Definition 23). Then T is measurable. The surfaceLebesgue measure on S1 is m T−1 (the push forward of m through T ). If E ∈ BS1 , then (m T−1)(E) gives the“total angle” subtended by E.

Thm 208: (General Change of Variables) Let (X,M , µ) be a measure space and (Y,N ) be a measurable space. Supposethat T : X → Y is a measurable function. Then for all N -measurable functions f : Y → R, we have∫

X

f(T (x)) dµ =∫Y

f(y) d(µ T−1)

provided one of the integrals makes sense.Cor 209: Let (X,M , µ) and (Y,N , ν) be measure spaces, and let T : X → Y be measurable. If ν µ T−1, then

there is a measurable ϕ : Y → R such that for each f ∈ L1(ν) satisfying f T ∈ L1(µ), we have∫E

f(y) d ν(y) =∫T−1(E)

f(T (x))ϕ(T (x)) dµ(x)

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8 DUAL SPACES Analysis Study Guide

for all E ∈ N .Thm 210: 56 Suppose that T : [a, b] → [c, d] is an increasing, absolutely continuous bijection. Let f ∈ L1([c, d],m) be

given such that f T ∈ L1([a, b]). Then∫[a,b]

f(T (x))T ′(x) dm(x) =∫

[c,d]

f(y) dm(y).

Thm 211: Let T : Rn → Rn be a bijective linear transformation. If f ∈ L1(m), then f T ∈ L1(m) and∫Rnf dm = |detT |

∫Rnf T dm.

Notation: 57 Let Ω ⊆ Rn be an open set. If G : Ω → Rn has continuously differentiable components, i.e., G =

(G1, · · · , Gn) with G1, · · · , Gn ∈ C1(Ω). Then DxG : Ω → Rn×n is given by [DxG]ij :=∂Gi∂xj

. We say G is a

C1-diffeomorphism if G is bijective and G and G−1 are both continuously differentiable.Thm 212: (Change of Variables) Suppose that Ω ⊆ Rn is open and that G : Ω→ Rn is a C1 diffeomorphism.

(a) If f ∈ L1(G(Ω),m), then ∫G(Ω)

f(y) dm(y) =∫

Ω

f(G(x))|detDxG(x)|dm(x).

(b) If E ⊆ Ω and E ∈ L n then G(E) ∈ L n and m(G(E)) =∫E

|detDxG(x)|dm(x).

Integration in Polar Coordinates

Set Sn−1 := x ∈ Rn : ||x|| = 1. For x ∈ Rn \ 0, define r(x) = ||x|| and θ(x) = x/ ||x|| ∈ §n−1. It can be verifiedthat the map G : (0,∞)× Sn−1 → Rn \ 0 given by G(r, θ) = rθ is a C1-diffeomorphism. Since G is a bijection andcontinuous we may define a measure m∗ on B(0,∞)×Sn−1 by m∗(E) = m(G(E)). Define the measure ρn on B(0,∞)

by ρn(E) =∫E

rn−1 dm(r).

Thm 213: There is a unique Borel measure σn−1 on Sn−1 such that m∗ = ρn × σn−1.

8 Dual Spaces

Def 214: 58 If (X , ||·||X ) is a normed vector space, the space of continuous linear functionals (L(X ,R), |||·|||) iscalled the (continuous) dual space of X . It is denoted by X ∗ or (X ∗, ||·||X ∗).

Rmrk: X ∗ is always a B-space.Example 215: Suppose that X = Rn and ||·|| =

√x · x. For each ϕ ∈ X ∗ there is a unique y ∈ Rn such that

ϕ(x) = y · x for all x ∈ Rn. Clearly the functional x 7→ y · x is in X ∗. Thus, X ∗ can be identified with X itself. Itcan be shown that X and X ∗ are isometric.

Thm 216: (Riesz Representation Theorem for Lp spaces) Suppose that p, q ∈ (1,∞) satisfy 1/p + 1/q = 1. Then for

each ϕ ∈ (Lp)∗ there is a g ∈ Lq such that ϕ(f) =∫X

fg dµ for all f ∈ Lp. If µ is σ-finite, then the result holds

when p = 1. (In which case q =∞).Rmrk: It will be shown (later) that (Lp)∗ ∼= Lq.Thm 217: 59 Suppose p, q ∈ [1,∞] satisfy 1/p+ 1/q = 1, and that g : X → R satisfies

(i) fg ∈ L1 for all f ∈ Σ :=h ∈ L1 : h is simple

.

(ii) Mq(g) := sup∣∣∣∣∫

X

fg dµ∣∣∣∣ : f ∈ Σ ∩ Lp and ||f ||Lp ≤ 1

<∞.

(iii) Either Sg := x ∈ X : |g(x)| 6= 0 is σ-finite or µ is semi-finite.

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8 DUAL SPACES Analysis Study Guide

Then g ∈ Lq. Moreover, ||g||Lq = Mq(g).Cor 218: 60 If p ∈ (1,∞), then Lp is reflexive; i.e., (Lp)∗∗ ∼= Lp.Def 219: A directed set A is a non-empty set with a relation . such that α . α for all α ∈ A; if α . β and β . γ

then α . γ; if α, β ∈ A then there is γ ∈ A such that α . γ and β . γ. An element of a directed set is called anindex.

Example 220:

(a) Any nonempty subset of R with the usual ordering is a directed set.

(b) Let B be a neighborhood basis for a topology on X; if B ⊆ T and for each x ∈ X there is N ⊆ B suchthat x ∈ V for all V ∈ N and, if U ∈ T and x ∈ U there is V ∈ N such that V ⊆ U . For each x ∈ X, setNx := U ∈ B : x ∈ U. Fix x ∈ X. If we say that U . V whenever U, V ∈ Nx and U ⊇ V then Nx is adirected set.

Def 221: A net in X is a function x : A→ X with A a directed set. We denote the function by 〈xα〉α∈A. The set A iscalled the index set.

Def 222: Let (X, T ) be a topological space and let E ⊆ X be given. Let 〈xα〉α∈A be a net. We say

(a) 〈xα〉 is eventually in E if there is an α0 ∈ A such that xα ∈ E whenever α0 . α.

(b) 〈xα〉 is frequently in E if for all α ∈ A there is a β ∈ A such that α . β and xβ ∈ E.

(c) 〈xα〉 converges to a point x ∈ X if for every U ∈ T with x ∈ U , 〈xα〉 is eventually in U . For brevity, wewrite xα → x.

Prop 223: If (X, T ) and (Y,S) are topological spaces and f : X → Y then f is continuous at x ∈ X iff for each net〈xα〉 converges to x we have 〈f(xα)〉 converges to f(x).

Def 224: LetX be a set and (Yα, Tα)α∈A a family of topological spaces. Given a family of mappings fα : X → Yαα∈Athere is a unique weakest topology on X that makes each fα continuous. This topology is called the weak topologygenerated by fα.

Def 225: Let (X , ||·||) be a normed vector space. The weak topology on X is the weak topology generated by X ∗.Convergence in this topology is called weak convergence.

Rmrk: If 〈xα〉 ⊆ X is a net, we say xα → x strongly iff for each ε > 0, eventually ||xα − x|| < ε. On the other hand,xα → x weakly (written xα x) iff f(xα) → f(x) for all f ∈ X ∗. Recall that X can be identified as a subset ofX ∗∗ so each x ∈X can be identified with a linear functional on X ∗.

Def 226: Let (X , ||·||) be a normed vector space. The weak∗ topology on X ∗ is the weak topology generated by X(identified as a subset of X∗∗). Convergence in the weak∗ topology is called weak∗ convergence.

Rmrk: If 〈fα〉 ⊆X ∗ is a net, then fα → f weak∗ (written fα ∗ f) iff fα(x)→ f(x) for all x ∈X .

Def 227: 61 A subnet of a net 〈xα〉α∈A is a net 〈yβ〉β∈B together with a map β 7→ αβ from B to A such that for eachα0 ∈ A there is a β0 ∈ B such that α0 . αβ for all β ∈ B with β0 . β. Moreover, we require that yβ = xαβ .

Rmrk: Bounded nets in X ∗ must have a weak∗ convergent subnet.Cor 228: (to Riesz’s Theorem) If µ is σ-finite and 〈fα〉 is a bounded net in L∞. Then there is a subnet 〈gα〉 of 〈fα〉

and a function g in L∞ such that gβ ∗ g in L∞; i.e.,∫X

gβf dµ→∫X

gf dµ ∀f ∈ L1.

Cor 229: (to Riesz’s Theorem) If p ∈ (1,∞), q =p

p− 1, and 〈fα〉 is a bounded net in Lp then there exists a subnet

〈gβ〉 of 〈fα〉 and a g ∈ Lp such that gβ g in Lp; i.e.,∫X

gβf dµ→∫X

gf dµ ∀f ∈ Lq.

Cor 230: If p ∈ (1,∞) and fn∞n=1 is bounded in Lp then there exists a subsequence fnk∞k=1 such that fnk → f in

Lp for some f ∈ Lp.Using the Uniform Boundedness Principle (5.13 in Folland), we also haveCor 231: If p ∈ (1,∞) and fn∞n=1 in Lp such that fn f in Lp, then sup

n∈N||fn||Lp <∞.

See notes for a good counterexample of when the previous Corollary fails when p = 1.

Mike Janssen

8 DUAL SPACES Analysis Study Guide

(a)

(b)

(c)

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NOTES Analysis Study Guide NOTES

Notes1Math 921 Class notes, August 29, 20082Math 921 Class notes, September 3, 20083Math 921 Class notes, September 5, 20084Math 921 Class notes, September 8, 20085Math 921 Class notes, September 10, 20086Math 921 Class notes, September 12, 20087Math 921 Class notes, September 15, 20088Math 921 Class notes, September 19, 20089Math 921 Class notes, September 22, 2008

10Math 921 Class notes, September 24, 200811Math 921 Class notes, September 26, 200812Math 921 Class notes, September 29, 200813Math 921 Class notes, October 1, 200814Math 921 Class notes, October 3, 200815Math 921 Class notes, October 6, 200816Math 921 Class notes, October 8, 200817Math 921 Class notes, October 10, 200818Math 921 Class notes, October 13, 200819Math 921 Class notes, October 15, 200820Math 921 Class notes, October 24, 200821Math 921 Class notes, October 27, 200822Math 921 Class notes, October 29, 200823Math 921 Class notes, October 31, 200824Math 921 Class notes, November 5, 200825Math 921 Class notes, November 7, 200826Math 921 Class notes, November 10, 200827Math 921 Class notes, November 14, 200828Math 921 Class notes, November 17, 200829Math 921 Class notes, November 19, 2008

30Math 921 Class notes, November 21, 200831Math 921 Class notes, November 24, 200832Math 921 Class notes, December 1, 200833Math 921 Class notes, December 3, 200834Math 921 Class notes, December 5, 200835Math 921 Class notes, December 8, 200836Math 922 Class notes, January 12, 200837Math 922 Class notes, January 14, 200938Math 922 Class notes, January 21, 200939Math 922 Class notes, January 28, 200940Math 922 Class notes, January 30, 200941Math 922 Class notes, February 4, 200942Math 922 Class notes, February 6, 200943Math 922 Class notes, February 9, 200944Math 922 Class notes, February 11, 200945Math 922 Class notes, February 13, 200946Math 922 Class notes, February 16, 200947Math 922 Class notes, February 18, 200948Math 922 Class notes, February 23, 200949Math 922 Class notes, February 25, 200950Math 922 Class notes, February 27, 200951Math 922 Class notes, March 2, 200952Math 922 Class notes, March 6, 200953Math 902 Class notes, March 9, 200954Math 922 Class notes, March 11, 200955Math 902 Class notes, March 23, 200956Math 922 Class notes, March 25, 200957Math 922 Class notes, March 27, 200958Math 922 Class notes, April 1, 200959Math 922 Class notes, April 3, 200960Math 922 Class notes, April 10, 200961Math 922 Class notes, April 13, 2009

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9 NOTES Analysis Study Guide

9 Notes

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9 NOTES Analysis Study Guide

Mike Janssen

9 NOTES Analysis Study Guide

Mike Janssen

9 NOTES Analysis Study Guide

Mike Janssen

9 NOTES Analysis Study Guide

Mike Janssen