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Faye Jackson MATH 597 - TOC
Faye Jackson
- MATH Notes on - April 18, (Intro to Graduate Real Analysis)
- I. Introduction & Syllabus C o n t e n t s
- II. Abstract Measure Theory
- II.1. Motivation
- II.2. σ-algebras
- II.3. Measures
- II.4. Building Measures
- II.5. Borel measures on R
- II.6. Lebesgue-Stieltjes measures on R
- II.7. Regularity properties of Lebesgue-Stieltjes measures
- III. Integration
- III.1. Measurable functions
- III.2. Integration of nonnegative functions
- III.3. Integration of complex functions
- III.4. L^1 Spaces.
- III.5. Riemann Integrability
- III.6. Mode of Convergence
- IV. Product Measures
- IV.1. Product σ-algebras
- IV.2. Product Measures.
- IV.3. Monotone Class Lemma.
- IV.4. Fubini-Tonelli Theorem
- IV.5. Lebesgue measure on Rd
- V. Differentiation on Euclidean Space
- V.1. Hardy-Littlewood maximal function
- V.2. Lebesgue Differentiation Theorem.
- VI. Normed Vector Spaces
- VI.1. Metric Spaces and Normed Spaces
- VI.2. Lp spaces
- VI.3. Embedding Properties of Lpspaces
- VI.4. Banach Spaces Faye Jackson MATH 597 - TOC
- VI.5. Bounded Linear Transformations (BLTs)
- VI.6. Dual of Lp spaces
- VII. Signed and Complex Measures
- VII.1. Signed Measures
- VII.2. Absolutely Continuous Measures
- VII.3. Lebesgue Differentiation Theorem for regular Borel measures on Rd
- VII.4. Monotone Differentiation Theorem
- VII.5. Functions of bounded variation.
- VII.6. Absolutely Continuous Functions.
- VIII. Hilbert Spaces
- VIII.1. Inner Product Spaces
- VIII.2. Orthonormal Bases.
- IX. Intro to Fourier Analysis
- References
- Todo list
- Let X = { 0 , 1 , 2 , 3 }. Then there are measures μ : P (X) → [0, ∞] as below μ(A) = #A μ({ 0 , 1 }) = 2 μ(A) =
X
i∈A
2 i^ μ({ 0 , 1 }) = 3
- For X = N 0 := { 0 } ∪ N we have μ : P (N 0 ) → [0, ∞] via μ(A) = #A μ({ 2 , 4 , 6 ,.. .}) = ∞ μ(A) = e−^1
X
i∈A
i! μ(X) = 1
Using μ({i}) = ai and μ(A) =
P
i∈A ai^ is a good measurement of^ A. This works because the sums are always countable.
- X = R, want μ : P (R) → [0, ∞] μ(A) = #A not interesting μ(A) = length of A?. Here we’d have μ((a, b)) = b − a. Can we extend μ reasonably to all subsets of R? What if instead we take μ((a, b)) = eb^ − ea? We also can’t extend this to all subsets! Theorem II.1.1 (Banach-Tarski) In Rd, for d ≥ 3, one can divide a ball into finitely many subsets and put back into two balls of same radius. We will try to define a “measure” on X, that is μ : A → [0, ∞] for a “suitable” A ⊆ P (X).
II.2. σ-algebras
Definition II.2. Let X be a set. A collection A of subsets of X is called a σ-algebra on X provided that
- ∅ ∈ A.
- A is closed under complements, aka if E ∈ A then Ec^ ∈ A.
- A is closed under countable unions. that is if E 1 , E 2 ,... ∈ A then
S∞
i=1 Ei^ ∈ A.
These have some simple properties
- X = ∅c^ ∈ A.
- It’s closed under countable intersections because ^ ∞ i=
Ei =
[^ ∞
i=
Eci
!c
- SNi=1 Ei = E 1 ∪ · · · ∪ EN ∪ ∅ ∪ · · ·. So it is closed under finite unions (+ intersections).
- Closed under E \ F = E ∩ F c^ and E△F = (E ∪ F ) \ (E ∩ F ).
Announcements
- Canvas/Modules
- Lecture summary after each class
- Suggested reading
- HW1 will be posted today: due next Thursday 1/13. 9pm. Now lets look at examples of σ-algebras Example II.2. We have the followoing basic σ-algebras
- A = P (X), the power σ-algebra
- A = {∅, X}, the trivial σ-algebra
- Let B ⊆ X, B ̸= ∅, B ̸= X. Then A = {∅, B, Bc, X} is a σ-algebra. Lemma II.2. Let Ai where i ∈ I be a family of σ-algebras over a fixed set X. Then T i∈I Ai is a σ-algebra over X.
Proof. Clearly ∅ ∈
T
i∈I Ai^ because^ ∅ ∈ Ai^ for all^ i.^ Now if^ E^ ∈^
T
i∈I Ai, then^ Ec^ ∈ Ai^ for each^ i, so Ec^ ∈
T∞
i=1 Ai^ as desired. Now if E 1 , E 2 ,... ∈ T i∈I Ai, then of course S∞ j=1 Ej ∈ Ai for each i, so S∞ j=1 Ej ∈ T i∈I Ai. Great! Definition II.2. For E ⊆ P (X), let ⟨E⟩ be the intersection of all σ-algebras on X containing E. We call ⟨E⟩ the σ-algebra generated by E. Example II.2. {∅, B, Bc, X} = ⟨{B}⟩ = ⟨{∅, Bc}⟩. Remark II.2. ⟨E⟩ is the smallest σ-algebra containing E (under the subset relation), and this uniquely characterizes E. Lemma II.2. We have the following (a) Suppose E ⊆ P (X) and A is a σ-algebra containing A. Then ⟨E⟩ ⊆ A. (b) Suppose E ⊆ F ⊆ P (X). Then ⟨E⟩ ⊆ ⟨F⟩ because E ⊆ ⟨F⟩.
Proof. DIY
A measure is also necessarily finite additive Example II.3.1 (a) For any (X, A), μ(A) = #A is called the counting measure. (b) Let x 0 ∈ X. For any (X, A), the Dirac measure at x 0 is denoted by δx 0 and takes the values
δx 0 =
1 if x 0 ∈ A 0 if x 0 ̸∈ A (c) Note that measures are closed under pointwise scalar multiplication and pointwise addition. Thus for (N, P (N)) w eknow that μ(A) =
X
i∈A
ai
is a measure where ai ∈ [0, ∞) for i ∈ N.
Announcements
- Get to know you Video
- HW1 Due Thursday 9pm
- Office Hours (not today)
- M 12:30-1:30, T 1:30-2:30 in-person EH
- Thursday 1-2, online Recall: Definition II.3. Note: For A, B ∈ A, A ⊆ B, then μ(B \ A) + μ(A) = μ(B).
And thus μ(A) ≤ μ(B) and μ(B \ A) = μ(B) − μ(A) if μ(A) < ∞. We must always be careful with ∞ when we subtract, because ∞ − ∞ is not well-defined. Theorem II.3. Let (X, A, μ) be a measure space. Then we have the following properties (1) Monotonicity: A ⊆ B ∈ A =⇒ μ(A) ≤ μ(B). (2) Countable subadditivity: If A 1 , A 2 ,... ∈ A then
μ
[^ ∞
i=
Ai
X^ ∞
i=
μ(Ai).
(3) Continuity from below / Montone Convergence Theorem (MCT) for sets: Given A 1 , A 2 ,... ∈ A satisfying A 1 ⊆ A 2 ⊆ · · · then
μ
[^ ∞
i=
Ai
= (^) nlim→∞ μ(An)
(4) Continuity from above: Given A 1 , A 2 ,... ∈ Asatisfying A 1 ⊇ A 2 ⊇ · · · and μ(A 1 ) < ∞ then
μ
^ ∞
i=
Ai
= (^) nlim→∞ μ(An)
Proof. (1) and (2) DIY.
For part (3), let B 1 = A 1 and Bi = Ai \ Ai− 1 for i ≥ 2. Then we know that [^ ∞ i=
Ai =
G^ ∞
i=
Bi
μ
[^ ∞
i=
Ai
X^ ∞
i=
μ(Bi) = (^) nlim→∞
X^ n i=
μ(Bi)
= (^) nlim→∞ μ
G^ n i=
Bi
= (^) nlim→∞ μ(An).
For part (4), let Ei = A 1 \ Ai. Then E 1 ⊆ E 2 ⊆ · · ·. Then
[^ ∞ i=
Ei =
[^ ∞
i=
A 1 \ Ai = A 1 \
^ ∞
i=
Ai
Now note that
μ
[^ ∞
i=
Ei
≤ μ(A 1 ) < ∞.
Therefore we have that ^ ∞ i=
Ai = A 1 \
[^ ∞
i=
Ei
μ
^ ∞
i=
Ai
= μ(A 1 ) − μ
[^ ∞
i=
Ei
= μ(A 1 ) − (^) nlim→∞ μ(En) = μ(A 1 ) − (^) nlim→∞ μ(A 1 ) − μ(An) = (^) nlim→∞ μ(An).
Example II.3. TAke N, P(N) with the counting measure. Then let An = {n, n + 1, n + 2,.. .}. Then note that A 1 ⊇ A 2 ⊇ · · · and ^ ∞ i=
Ai = ∅ =⇒ μ
^ ∞
i=
Ai
But μ(An) = ∞ for each n. This shows that finiteness is necessary for part (4). Definition II.3. Let (X, A, μ) be a measure space. Then
- A ⊆ X is a μ-null set if A ∈ A and μ(A) = 0.
- A ⊆ X is a μ-subnull set if there exists a μ-null set B with A ⊆ B. Note: A is not necessarily A-measurable.
- (X, A, μ) is a complete measure space if every μ-subnull set is A-measurable.
(1) μ∗^ is well-defined: This is easy, since inf is taken over a non-empty set bounded below by zero. (2) μ∗(∅) = 0. Just take all the Ei = ∅ to get a minimum (3) A ⊆ B implies μ∗(A) ≤ μ∗(B) because every cover of B by elements of E also covers A. Next class: we will prove countable subadditivity. Recall II.4. Tonelli’s theorem for series. If aij ∈ [0, ∞] then X (i,j)∈N^2
aij =
X^ ∞
i=
X^ ∞
j=
aij =
X^ ∞
j=
X^ ∞
i=
aij.
Read [Tao11], specifically Thm 0.0.2.
Proof of Proposition II.4.1: Countable subadditivity. Let A 1 , A 2 ,... ⊆ X. We wish to show that
μ∗
[^ ∞
n=
An
X^ ∞
n=
μ∗(An).
If one of the μ∗(An) = ∞, the result holds. Thus it suffices to consider the case when all μ∗(An) < ∞. We will instead prove that for every ε > 0 we have that
μ∗
[^ ∞
n=
An
X^ ∞
n=
μ∗(An) + ε.
We can call this trick
Give yourself a room of ε > 0
For each n ∈ N, there exists En, 1 , En, 2 ,... ∈ E such that [^ ∞ k=
En,k ⊇ An μ∗(An) ≤
X^ ∞
k=
ρ(En,k) ≤ μ∗(An) + 2 εn
Useful because μ∗(An) < ∞. Here we have used the ε/ 2 n-trick so that we don’t accumulate infinite error. Then [^ ∞ n=
An ⊆
[^ ∞
n=
[^ ∞
k=
En,k =
[
(n,k)∈N^2
En,k
μ∗
[^ ∞
n=
An
X
(n,k)∈N^2
ρ(Ek,n) =
X^ ∞
n=
X^ ∞
k=
ρ(Ek,n)
X^ ∞
n=
μ∗(En) + 2 εn =
X^ ∞
n=
μ∗(An) + ε.
Here we have used Tonelli’s theorem, because each ρ(En,k) satisfies 0 ≤ ρ(En,k) < ∞. Perfect! This proves the result by taking ε → 0.
Definition II.4. [Carath´eodory measurable] Let μ∗^ be an outer measure on X. We say that A ⊆ X is Carath´eodory measurable (abbrev. C-measurable) with respect to μ∗^ provided that for every E ⊆ X, μ∗(E) = μ∗(E \ A) + μ∗(E ∩ A) Lemma II.4. Let μ∗^ be an outer measure on X. Suppose B 1 ,... , BN are disjoint C-measurable sets. Then for all E ⊆ X,
μ∗^ E ∩
[^ N
i=
Bi
X^ N
i=
μ∗(E ∩ Bi).
This also implies that μ∗^ is finitely additive on C-measurable sets by setting E = X.
Proof. We see that
μ∗^ E ∩
[^ N
i=
Bi
= μ∗(E ∩ B 1 ) + μ∗^ E ∩
[^ N
i=
Bi
X^ N
i=
μ∗(E ∩ Bi).
Announcements
- No class on Monday (MLK day)
- HW 2 posted
- Another assignment
- Get to Know Video review
- Solution consent form. Lemma II.4. Improved version of Lemma II.4. Let μ∗^ be an outer measure on X. Suppose B 1 , B 2 ,... are disjoint C-measurable sets. Then for all E ⊆ X,
μ∗^ E ∩
[^ ∞
i=
Bi
X^ ∞
i=
μ∗(E ∩ Bi).
This also implies that μ∗^ is countably additive on C-measurable sets by setting E = X.
Proof. By countable subadditivity of μ∗^ we have that
X^ ∞ n=
μ∗(E ∩ Bn) ≥ μ∗^ (E ∩ ∪∞ n=1Bn).
Now monotonicity and Lemma II.4.2 implies that
μ∗^ (E ∩ ∪∞ n=1Bn) ≥ μ∗^
E ∩ ∪Nn=1Bn
X^ N
n=
μ∗(E ∩ Bn)
The right hand side then becomes μ∗(E 1 ∪ E 2 ∪ E 3 ) + μ∗(E 4 ) = μ∗(E 1 ∪ E 2 ) + μ∗(E 3 ) + μ∗(E 4 ) = μ∗(E 1 ∪ E 2 ) + μ∗(E 3 ∪ E 4 ) = μ∗(E 1 ∪ E 2 ∪ E 3 ∪ E 4 ). (a4) We show A is closed under countable disjoint unions. Let A 1 , A 2 ,... ∈ A be disjoint. Fix E ⊆ X. We need to show that
μ∗(E) = μ∗^ E ∩
[^ ∞
n==
An
[^ ∞
n=
An
Because μ∗^ is countable subadditive we know that
μ∗(E) ≤ μ∗^ E ∩
[^ ∞
n==
An
[^ ∞
n=
An
We then just need to show the other direction of the inequality. So fix N ∈ N. We know by Item (a3) that SNn=1 An ∈ A, and so by Lemma II.4.2, monotonicity, and countable subadditivity
μ∗(E) = μ∗^ E ∩
[^ N
n=
An
[^ N
n=
X^ N
n=
μ∗(E ∩ An) + μ∗^ E \
[^ ∞
n=
An
By taking N → ∞ and applying the result of countable subadditivity. (a5) We claim that being closed under complement (a2), closed under finite unions (a3), and closed under countable disjoint unions (a4) suffices to show that A is closed under countable unions. To do this, fix A 1 , A 2 ,... ∈ A. Now let
Bn = An \
n[− 1 i=
Ai.
Then S n An = S n Bn, but the Bn are disjoint, and all in A because of (a2),(a3). (b) We know that μ(∅) = μ∗(∅) = 0, and countable additivity on A follows from Lemma II.4.3 with E = X. (c) On HW2!
Recall II.4. Recall Proposition II.4.1. That is let E ⊆ P (X) such that ∅, X ∈ E. Now let ρ : E → [0, ∞] such that ρ(∅) = 0. Then
μ∗(A) := inf
X
i=
ρ(Ei) | Ei ∈ E,
[^ ∞
i=
Ei ⊇ A
is an outer measure on X.
Then we have the following
(E, ρ) Proposition II.4.1(^ P (X), μ∗) Theorem II.4.4(C-measurable sets, μ) Question: Do we have E ⊆ A and μ (^) E = ρ? No! Definition II.4. Let A 0 be an algebra on X (that is contains ∅, closed under complement, and closed under finite union). We say μ 0 : A 0 → [0, ∞] is a pre-measure if (a) μ 0 (∅) = 0 (b) Finite additivity: If A 1 ,... , An ∈ A are disjoint then
μ 0
[^ N
i=
Ai
X^ N
i=
μ 0 (Ai)
(c) Countable additivity within A 0 : If A ∈ A 0 and A =
S∞
i=1 Ai^ for disjoint^ Ai^ ∈ A, then
μ 0
[^ ∞
i=
Ai
X^ ∞
i=
μ 0 (Ai)
In fact (a) + (c) imply (b) by taking empty sets. Notation: [Fol99] uses M for σ-algebra and A for algebra. We use A for σ-algebra and A 0 for algebra. Example II.4. By next Wednesday we will consider A 0 as finite disjoint unions of (a, b] and
μ 0
X^ N
i=
(ai, bi]
X^ N
i=
(bi − ai).
This will generate the Lebesgue measure on R. Lemma II.4. μ 0 is monotone
Proof. DIY
Theorem II.4.6 (Hahn-Kolmogorov Theorem) Let μ 0 be a premeasure on the algebra A 0 on X. Let μ∗^ be the induced outer measure from (A 0 , μ 0 ) via Proposition II.4.1. Let A and μ be the Carath´eodory σ-algebra and measure for μ∗. Then (A, μ) extends (A 0 , μ 0 ). In other words, A ⊇ A 0 and μ (^) A 0 = μ 0.
Proof. Let’s go!
(a) We wish to show A ⊇ A 0. Let A ∈ A 0. We need to show A ∈ A, that is we need to show A is C-measurable. Concretely, for E ⊆ X we need μ∗(E) =? μ∗(E ∩ A) + μ∗(E ∩ Ac).
- HW 1 solutions (by you) – Canvas/HW 1 page
- Piazza made for the class Definition II.4. We call (A, μ) the Hahn-Kolmogorov (HK) extension of (A 0 , μ 0 ) where A 0 is an algebra and μ 0 is a premeasure. Namely, we define μ∗^ : P(X) → [0, ∞]
μ∗(E) := inf
( X∞
i=
μ 0 (Bi) | Bi ∈ A 0 ,
[^ ∞
i=
Bi ⊇ E
A := {A ⊆ X | ∀E ⊆ X, μ∗(E) = μ∗(E ∩ A) + μ∗(E ∩ Ac)} μ := μ∗ A. We have by Theorem II.4.6 that A 0 ⊆ A and that μ (^) A 0 = μ 0.
Theorem II.4.7 (Uniqueness of HK extension) Let A 0 be an algebra on X, μ 0 a pre-measure on A 0. Let (A, μ) be the HK extension of (A 0 , μ 0 ). Let (A′, μ′) be some other extension of (A 0 , μ 0 ). If μ 0 is σ-finite (recall Definition II.3.4), then μ = μ′^ on A ∩ A′. Corollary II.4. Let μ 0 be a pre-measure on algebra A 0 on X. Suppose μ 0 is σ-finite. Then there exists a unique measure μ on ⟨A 0 ⟩ that extends μ 0. Furthermore, (i) the completion of (X, ⟨A 0 ⟩, μ) is the HK extension of (A 0 , μ 0 ) (HW) (ii) We have a formula for all A ∈ ⟨A⟩
μ(A) = inf
X
i=
μ 0 (Bi) | Bi ∈ A 0 ,
[^ ∞
i=
Bi ⊇ A
Proof of Theorem II.4.7. Let A ∈ A ∩ A′. We need to show that μ∗(A) = μ(A) = μ′(A). Again we prove two inequalities
(a) Show μ∗(A) ≥ μ′(A) (HW) (b) We will show μ∗(A) ≤ μ′(A). First (i) Assume μ∗(A) < ∞. Then fix ε > 0, then there exists Bi ∈ A 0 with B :=
S∞
i=1 Bi^ ⊇^ A^ so that
μ(A) + ε ≥
X
i=
μ 0 (Bi) =
X^ ∞
i=
μ(Bi)
≥ μ(B)
Faye Jackson January 21st, 2022 MATH 597 - II.
Then since A ⊆ B, μ(A) < ∞, we know that μ(B \ A) = μ(B) − μ(A) ≤ ε. On the other hand using continuity from below
μ(B) = (^) Nlim →∞ μ
[^ N
i=
Bi
= (^) Nlim →∞ μ′
[^ N
i=
Bi
= μ′(B)
Then we have by part (a) that μ(A) ≤ μ(B) = μ′(B) = μ′(A) + μ′(B \ A) ≤ μ′(A) + μ(B \ A) ≤ μ′(A) + ε Perfect! We win by taking ε → 0. (ii) Assume μ(A) = ∞. Because μ 0 is σ-finite we know X =
S∞
i=1 Xn^ for some^ Xn^ ∈ A^0 satisfying μ 0 (Xn) < ∞. Replacing Xn by X 1 ∪ · · · ∪ Xn ∈ A 0 , we may assume X 1 ⊆ X 2 ⊆ · · ·. Then note μ(A ∩ Xn) < ∞ so by part (i) we have μ(A ∩ Xn) ≤ μ′(A ∩ Xn). Now by continuity of the measure μ(A) = (^) nlim→∞ μ(A ∩ Xn) ≤ (^) nlim→∞ μ′(A ∩ Xn) = μ′(A).
This finishes the proof!
Announcements
II.5. Borel measures on R
Definition II.5. A function F : R → R is an inreasing function provided that for x ≤ y we have F (x) ≤ F (y). A function F : R → R which is increasing and right-continuous (that is limx→a+ F (x) = F (a) for all a) is called a distribution function. Example II.5. These functions are distributions
- F (x) = x
- F (x) = ex
- F (X) = 1 for x ≥ 0 and F (x) = 0 for x < 0.
Faye Jackson January 21st, 2022 MATH 597 - II.
ℓF ((−∞, ∞)) = F (∞) − F (−∞).
We now define μ 0 := μ 0 ,F : H → [0, ∞] by
μ 0 (A) =
X^ N
k=
ℓF (Ik)
if A may be written as a finite disjoint union
SN
k=1 Ik^ of^ h-intervals. Then μ 0 is well-defined and a pre-measure on H.
Proof. There are a few conditions to verify
(a) μ 0 is well-defined. This can be shown by taking a common “refinement” of two expressions I 1 ,... , IN and J 1 ,... , JM which both union to A ⊆ H. (b) μ 0 (∅) = 0 ✓. (c) μ 0 is finitely additive ✓. (d) μ 0 is countably additive within H. That is suppose A ∈ H and A = S∞ i=1 Ai, disjoint union, Ai ∈ H. These cases look something like
(0, 1] =
[^ ∞
i=
i + 1 ,^
i
It is enough to consider the case where A = I, Ak = Ik all h-intervals. (why?) Furthermore the statement is easy to extend to the infinite cases, so we focus on I = (a, b] (HW) Suppose that (a, b] =
S∞
n=1(an, bn], disjoint. We must check that
F (b) − F (a) =?
X^ ∞
n=
(F (bn) − F (an)).
We know that for all N
(a, b] ⊇
[^ N
n=
(an, bn]
F (b) − F (a) ≥
X^ N
n=
(F (bn) − F (an))
F (b) − F (a) ≥
X^ ∞
n=
(F (bn) − F (an)).
Fix ε > 0. Since F is right-continuous, there exists a′^ > a such that F (a′) − F (a) < ε. For each n ∈ N, ther eis a point b′ n > bn such that F (b′ n) − F (bn) < 2 εn. We then see that
[a′, b] ⊆
[^ ∞
n=
(an, b′ n)
[a′, b] ⊆
[^ N
n=
(an, b′ n)
(a′, b] ⊆
[^ N
n=
(an, b′ n]
F (b) − F (a′) ≤
X^ N
n=
F (b′ n) − F (an)
F (b) − F (a) ≤ F (b) = F (a′) + ε
≤ ε +
X^ ∞
n=
F (b′ n) − F (an)
≤ ε +
X^ ∞
n=
F (bn) − F (an) + 2 εn
= 2ε +
X^ ∞
n=
F (bn) − F (an).
taking ε → 0 yields the result.
Announcements
- Piazza.
- HW3 Q3 typo, B ⊆ Ac should be B ⊆ A.
- Office hour reminder
- M 12:30-1:30pm in-person.
- T 1:30-2:30pm in-person.
- Th 1-2pm online. Theorem II.5.4 (Locally finite Borel measures on R) Our work last time in fact classifies locally finite Borel measures on R (a) If F : R → R is a distribution function, then there exists a unique locally finite Borel measure μF on R satisfying μF ((a, b]) = F (b) − F (a) for every a < b in R. This essentially follows from Theorem II.4.7. (b) Suppose F, G : R → R are distribution functions. Then, μF = μG on B(R) if and only if F − G is a constant function. (HW) (c) Lemma II.5.1 implies that these are all of the locally finite Borel measures on R.
II.6. Lebesgue-Stieltjes measures on R
The general sketch of what is going on F dist. fn HK μF on Carath´eory σ-algebra AμF ⊇ B(R).
Then HW3 implies that (AμF , μF ) = (B(R), μF ) (the completion). Definition II.6.1 (Lebesgue-Stieltjes measure) For a distribution function F , we call μF on AμF the Lebesgue-Stieltjes measure corresponding to F. A special case, when F (x) = x, is called the Lebesgue measure m on the Lebesgue σ-algebra L. Example II.6.1 (Discrete Measures) (a) Write F (x−) = (^) alim→x− F (a) F (x+) = (^) alim→x+ F (a).
