Chapter II Measure and Integral

Chapter II

Measure and Integral

1 Classes of Sets

Given: a non-empty set Ω and a class A⊂P(Ω) of subsets. Put A⁺=n[ⁿ

i=1

A_i :n ∈N∧A₁, . . . , A_n ∈Apairwise disjointo . Definition 1.

(i) A closed w.r.t. intersections or∩–closed iff A, B ∈A⇒A∩B ∈A.

(ii) A closed w.r.t. unions or ∪–closed iff A, B ∈A⇒A∪B ∈A.

(iii) A closed w.r.t. complements or^c–closed iff A∈A⇒A^c := Ω\A∈A.

(iv) A semi-algebra (in Ω) if (a) Ω∈A,

(b) A∩–closed,

(c) A∈A⇒A^c ∈A⁺. (v) A algebra (in Ω) if

(a) Ω∈A, (b) A∩–closed,

(c) A^c–closed.

(vi) A σ-algebra (in Ω) if (a) Ω∈A,

(b) A₁, A₂, . . . ∈A⇒S∞

n=1A_n∈A, (c) A^c–closed.

Remark 1. Let A denote a σ-algebra in Ω. Recall that a probability measure P on (Ω,A) is a mapping

P :A→[0,1]

such thatP(Ω) = 1 and

A₁, A₂, . . .∈A pairwise disjoint ⇒ P[^∞

i=1

A_i

=

∞

X

i=1

P(A_i).

Moreover, (Ω,A, P) is called a probability space, and P(A) is the probability of the event A∈A.

Remark 2.

(i) A σ-algebra ⇒ Aalgebra ⇒ Asemi-algebra.

(ii) A closed w.r.t. intersections ⇒A⁺ closed w.r.t. intersections.

(iii) A algebra and A₁, A₂ ∈A ⇒A₁∪A₂, A₁\A₂, A₁ MA₂ ∈A.

(iv) A σ-algebra andA1, A2,· · · ∈A ⇒ T∞

n=1An ∈A.

Example 1.

(i) Let Ω =R and consider the class of intervals

A={]a, b] :a, b∈R∧a < b} ∪ {]−∞, b] :b∈R} ∪ {]a,∞[ :a∈R} ∪ {R,∅}.

Then A is a semi-algebra, but not an algebra.

(ii) {A∈P(Ω) :A finite or A^c finite} is an algebra, but not aσ-algebra in general.

(iii) {A∈P(Ω) :A countable or A^c countable} is a σ-algebra.

(iv) P(Ω) is the largest σ-algebra in Ω, {∅,Ω} is the smallestσ-algebra in Ω.

Definition 2.

(i) A monotone class (in Ω) if

(a) A₁, A₂, . . . ∈A∧A_n ↑A¹ ⇒A∈A, (b) A₁, A₂, . . . ∈A∧A_n ↓A² ⇒A∈A.

(ii) A Dynkin class (in Ω) if (a) Ω∈A,

(b) A₁, A₂ ∈A∧A₁ ⊂A₂ ⇒A₂\A₁ ∈A, (c) A₁, A₂, . . . ∈A pairwise disjoint ⇒S∞

n=1A_n∈A.

Remark 3. A σ-algebra ⇒A monotone class and Dynkin class.

1I.e.,An ⊆An+1 for allnandA=S

nAn 2I.e.,An+1⊆An for allnandA=T

nAn

Theorem 1.

(i) For every algebraA

A σ-algebra ⇔ Amonotone class.

(ii) For every Dynkin class A

A σ-algebra ⇔ A closed w.r.t. intersections.

Proof. Ad (i), ‘⇐’: Let A₁, A₂, . . . ∈ A and put B_m = Sm

n=1A_n and B = S∞ n=1A_n. Then B_m ↑B. Furthermore, B_m ∈A since A is an algebra. Thus B ∈A since A is a monotone class.

Ad (ii), ‘⇐’: For A ∈ A we have A^c = Ω\A ∈ A since A is a Dynkin class. For A, B ∈A we have

A∪B =A∪(B\(A∩B))∈A

since A is also closed w.r.t. intersections. Thus, for A₁, A₂, . . . ∈A and B_m as previ- ously we getB_m ∈Aand

∞

[

n=1

A_n=

∞

[

m=1

(B_m\Bm−1)∈A, whereB₀ =∅.

Remark 4. Consider σ-algebras (algebras, monotone classes, Dynkin classes)A_i for i∈I 6=∅. Then T

i∈IAi is a σ-algebra (algebra, monotone class, Dynkin class), too.

Given: a classE⊂P(Ω).

Definition 3. The σ-algebra generated by E σ(E) = \

{A:A σ-algebra in Ω∧E⊂A}.

Analogously, α(E), m(E), δ(E) the algebra, monotone class, Dynkin class, respec- tively,generated by E.

Remark 5. For γ ∈ {σ, α, m, δ} and E,E₁,E₂ ⊂P(Ω) (i) γ(E) is the smallest ‘γ-class’ that contains E, (ii) E1 ⊂E2 ⇒γ(E₁)⊂γ(E₂),

(iii) γ(γ(E)) =γ(E).

Example 2. Let Ω =N and E={{n}:n∈N}. Then

α(E) = {A∈P(Ω) :A finite or A^c finite}=:A.

Proof: A is an algebra, see Example 1, and E ⊂ A. Thus α(E) ⊂ A. On the other hand, for every finite set A ⊂ Ω we have A = S

n∈A{n} ∈ α(E), and for every set A⊂Ω with finite complement we have A= (A^c)^c ∈α(E). Thus A⊂α(E).

Moreover,

σ(E) =P( ), m(E) =E, δ(E) =P( ).

Theorem 2. [Monotone class theorem, set version]

(i) E closed w.r.t. intersections ⇒ σ(E) =δ(E).

(ii) E algebra⇒ σ(E) =m(E).

Proof. Ad (i): Remark 3 implies

δ(E)⊂σ(E).

We claim that

δ(E) is closed w.r.t. intersections. (1) Then, by Theorem 1.(ii),

σ(E)⊂δ(E).

Put

CB ={C ⊂Ω :C∩B ∈δ(E)}, B ∈δ(E), so that (1) is equivalent to

∀B ∈δ(E) :δ(E)⊂CB. (2)

It is straightforward to verify that

∀B ∈δ(E) :C_B Dynkin class. (3) Moreover, since Eis closed w.r.t. intersections,

∀E ∈E:E⊂C_E. Therefore

∀E ∈E:δ(E)⊂C_E, i.e., for allE ∈E, B ∈δ(E), E∩B ∈δ(E); hence

∀B ∈δ(E) :E⊂C_B. Since C_B is a Dynkin system, δ(B)⊂C_B.

Ad (ii): Obviously, m(E) ⊂ σ(E). By Part (ii) of Theorem 1, it is enough to show that m(E) is an algebra. This amounts to the claim that

m(E) is ^c–closed and∩–closed. (4) First, the class

C:={A ∈m(E) : A^c ∈m(E)}

is monotone, contains E by assumption, and thus equals m(E). Second, in complete analogy to Part (i), for B ∈m(E) it follows that the set

CB ={C ⊂Ω :C∩B ∈m(E)}

is a monotone class containingEand thusm(E), so thatm(E) is indeed∩–closed.

Lemma 1. E semi-algebra ⇒α(E) = E⁺.

Proof. Clearly E⊂E⁺ ⊂α(E). It remains to show that E⁺ is an algebra. For A=

n

[

i=1

Ai ∈E⁺, Ai ∈E disjoint,

B =

n

[

i=1

B_i ∈E⁺, B_i ∈E disjoint, A∩B = [

i≤n j≤m

(A_i∩B_j), (A_i∩B_j)∈Edisjoint.

Hence E⁺ is ∩–stable. For A=

n

[

i=1

A_i ∈E⁺, A_i ∈E disjoint, with

A^c_i = [

j≤n_i

B_jⁱ, B_jⁱ ∈E disjoint, we have

A^c = \

i≤n

[

j≤ni

B_jⁱ

= [

(j1,...,jn) ji≤n_i

\ⁿ

i=1

Bⁱ_ji

| {z }

∈E disjoint

.

Hence A^c ∈E⁺, and E⁺ is an algebra.

Put

R=R∪ {−∞,∞},

and equip this with the metricd(x, y) :=|arctan(x)−arctan(y)|. ThenRis a complete, compact, separable, order complete metric space. For a∈R set

(±∞) + (±∞) =a+ (±∞) = (±∞) +a=±∞, a/_±∞ = 0, a·(±∞) = (±∞)·a=







±∞ if a >0 0 if a = 0

∓∞ if a <0 as well as−∞< a < ∞.

Recall that (Ω,G) is a topological space iff G⊂P(Ω) satisfies (i) ∅,Ω∈G,

(ii) G is closed w.r.t. to intersections,

(iii) for every family (G_i)i∈I with G_i ∈G we have S

i∈IG_i ∈G.

G is the set of open subsets of Ω, and the complements of open sets are the closed subsets of Ω. K ⊂Ω is compact iff for every family (G_i)i∈I with G_i ∈G and

K ⊂[

i∈I

Gi

there is a finite setI₀ ⊂I such that

K ⊂ [

i∈I₀

G_i.

For Ω =R^k and Ω =R^k, we consider the natural (product) topologies G_k,G_k. Definition 4. For every topological space (Ω,G)

B(Ω) =σ(G) is the Borel-σ-algebra (in Ω w.r.t. G). We shorten

B=B(R), B=B(R), B_k=B(R^k),B_k=B(R^k), Remark 6. We have

B_k =σ({F ⊂R^k :F closed}) = σ({K ⊂R^k :K compact})

=σ({]−∞, a] :a∈R^k}) =σ({]−∞, a] :a∈Q^k}) and

B={B ⊂R:B∩R∈B}. (5) One can prove that #B_k = #R^k, and thus

Bk P(R^k) see Billingsley (1979, Exercise 2.21).

Definition 5. For any σ-algebra Ain Ω and Ωe ⊂Ω Ae ={eΩ∩A:A∈A}

is the trace-σ-algebra of A in Ω, sometimes denoted bye Ωe∩A.

Remark 7.

(i) Ae is a σ-algebra in Ω.e

(ii) Ae 6⊂A in general, but if Ωe ∈A, then Ae ={A ∈A:A⊂Ω}.e (iii) A=σ(E)⇒Ae =σ({eΩ∩E :E ∈E}).

(iv) B_k =R^k∩B_k, see (5) for k= 1.

(v) [a, b[∩B_k =σ({[a, c[ :a≤c≤b}), see (iii).

2 Measurable Mappings

Definition 1. (Ω,A) is calledmeasurable space iff Ω6=∅ and A is a σ-algebra in Ω.

Elements A∈A are called (A–)measurable sets.

In the sequel, (Ωi,Ai) are measurable spaces for i= 1,2,3.

Remark 1. Let f : Ω₁ →Ω₂. For B ∈A₂, we set in short

{f ∈B}=f⁻¹(B) ={ω ∈Ω₁ : f(ω)∈B} ⊂Ω₁ (i) f⁻¹(A₂) ={f⁻¹(A) :A∈A₂} is a σ-algebra in Ω₁.

(ii) {A⊂Ω₂ :f⁻¹(A)∈A₁} is a σ-algebra in Ω₂.

Definition 2. f : Ω1 → Ω2 is A1-A2-measurable iff f⁻¹(A2) ⊂ A1. i.e., iff for all A∈A₂ we have {f ∈A} ∈A₁.

How can we prove measurability of a given mapping?

Theorem 1. If f : Ω₁ → Ω₂ is A₁-A₂-measurable and g : Ω₂ → Ω₃ is A₂-A₃- measurable, theng◦f : Ω₁ →Ω₃ is A₁-A₃-measurable.

Proof. (Compare Bemerkung 5.4,(i), Analysis IV)

(g◦f)⁻¹(A₃) =f⁻¹(g⁻¹(A₃))⊂f⁻¹(A₂)⊂A₁ .

Lemma 1. For f : Ω1 →Ω2 and E⊂P(Ω2)

f⁻¹(σ(E)) =σ(f⁻¹(E)).

Proof. Byf⁻¹(E)⊂f⁻¹(σ(E)) and Remark 1.(i) we get σ(f⁻¹(E))⊂f⁻¹(σ(E)).

Let F = {A ⊂ Ω₂ : f⁻¹(A) ∈ σ(f⁻¹(E))}. Then E ⊂ F and F is a σ-algebra, see Remark 1.(ii). Thus we getσ(E)⊂F, i.e., f⁻¹(σ(E))⊂σ(f⁻¹(E)).

Theorem 2. If A₂ =σ(E) with E⊂P(Ω₂), then

f isA₁-A₂-measurable ⇔ f⁻¹(E)⊂A₁ . Proof. (compare Lemma 5.2, Analysis IV) ‘⇒’ is trivial,

‘⇐’:Assume thatf⁻¹(E)⊂A₁. By Lemma 1,

f⁻¹(A2) = f⁻¹(σ(E)) =σ(f⁻¹(E))⊂σ(A1) =A1.

Corollary 1. Let (Ω_i,G_i) be topological spaces. Then every continuousf : Ω₁ →Ω₂ isB(Ω )-B(Ω )-measurable.

Proof. (Compare Korollar 5.3, Analysis IV) For continuous f we have f⁻¹(G₂)⊂G₁ ⊂σ(G₁) =B(Ω₁).

Theorem 2 shows the claim.

Given: measurable spaces (Ω_i,A_i) for i ∈I 6=∅, mappings f_i : Ω →Ω_i for i∈ I and some non-empty set Ω.

Definition 3. The σ-algebra generated by (f_i)i∈I (and (A_i)i∈I) σ({f_i :i∈I}) =σ[

i∈I

f_i⁻¹(A_i)

. Moreover, setσ(f) = σ({f}).

Remark 2. σ({f_i :i∈I}) is the smallest σ-algebra A in Ω such that all mappings fi are A-Ai-measurable.

Theorem 3. For every measurable space (Ω,e A) and every mappinge g :Ωe →Ω, g is A-σ({fe _i :i∈I})-measurable ⇔ ∀i∈I :f_i◦g isA-Ae _i-measurable.

Proof. Use Lemma 1 to obtain g⁻¹(σ({f_i :i∈I})) = σ

g⁻¹[

i∈I

f_i⁻¹(A_i)

=σ[

i∈I

(f_i◦g)⁻¹(A_i) . Therefore

g⁻¹(σ({f_i :i∈I}))⊂Ae ⇔ ∀i∈I :f_i◦g_i is A-Ae _i-measurable.

Now we turn to the particular case of functions with values inRorR, and we consider the Borel σ-algebra in R or R, respectively. For any measurable space (Ω,A) we use the following notation

Z(Ω,A) = {f : Ω→R:f isA-B-measurable}, Z₊(Ω,A) = {f ∈Z(Ω,A) :f ≥0},

Z(Ω,A) =

f : Ω→R:f is A-B-measurable , Z₊(Ω,A) =

f ∈Z(Ω,A) :f ≥0 .

Every functionf : Ω→Rmay also be considered as a function with values in R, and in this case f ∈Z(Ω,A) iff f ∈Z(Ω,A).

Corollary 2. For≺ ∈ {≤, <,≥, >} and f : Ω→R,

f ∈Z(Ω,A) ⇔ ∀a∈R:{f ≺a} ∈A.

Proof. (Compare Satz 5.6, Bem.5.7, Analysis IV) For instance,B=σ({[−∞, a] :a∈R}) and

{f ≤a}=f⁻¹([−∞, a])

and B=σ({[−∞, a] :a ∈R}), see Remark 1.6. It remains to apply Theorem 2.

Theorem 4. For f, g∈Z(Ω,A) and ≺ ∈ {≤, <,≥, >,=,6=}, {ω∈Ω :f(ω)≺g(ω)} ∈A.

Proof. For instance, Corollary 2 yields {ω∈Ω :f(ω)< g(ω)}= [

q∈Q

{f < q < g}

= [

q∈Q

({f < q} ∩ {g > q})∈A.

Theorem 5. For every sequence f1, f2, . . . ∈Z(Ω,A), (i) infn∈Nf_n, sup_n∈Nf_n ∈Z(Ω,A),

(ii) lim inf_n→∞f_n, lim sup_n→∞f_n∈Z(Ω,A),

(iii) if (f_n)n∈N converges at every point ω∈Ω, then limn→∞f_n∈Z(Ω,A).

Proof. (Compare Satz 5.8, 5.9, Analysis IV) Fora∈R

n∈infN

f_n< a

= [

n∈N

{f_n< a},

sup

n∈N

f_n≤a

= \

n∈N

{f_n≤a}. Hence, Corollary 2 yields (i). Since

lim sup

n→∞

f_n = inf

m∈N

sup

n≥m

f_n, lim inf

n→∞ f_n = sup

m∈N

n≥minf f_n, we obtain (ii) from (i). Finally, (iii) follows from (ii).

By

f⁺= max(0, f), f⁻ = max(0,−f)

we denote the positive part and the negative part, respectively, of f : Ω→R. Remark 3. For f ∈Z(Ω,A) we have f⁺, f⁻,|f| ∈Z₊(Ω,A).

Theorem 6. For f, g∈Z(Ω,A),

f±g, f ·g, f /g∈Z(Ω,A), provided that these functions are well defined.

Proof. (Compare Folgerung 5.5, Analysis IV) The proof is again based on Corollary 2. For simplicity we only consider the case that f and g are real-valued. Clearly g ∈Z(Ω,A) implies −g ∈Z(Ω,A), too. Furthermore, for every a∈R,

{f +g < a}= [

q∈Q

{f < q} ∩ {g < a−q},

and therefore f ±g ∈ Z(Ω,A). Clearly f ·g ∈ Z(Ω,A) if f is constant. Moreover, x7→x² defines a B-B-measurable function, see Corollary 1, and

f ·g = 1/4· (f +g)²−(f −g)²

We apply Theorem 1 to obtain f·g ∈Z(Ω,A) in general. Finally, it is easy to show that g ∈Z(Ω,A) implies 1/g ∈Z(Ω,A).

Definition 4. f ∈Z(Ω,A) is called simple function if |f(Ω)|<∞. Put Σ(Ω,A) = {f ∈Z(Ω,A) :f simple},

Σ₊(Ω,A) = {f ∈Σ(Ω,A) :f ≥0}. Remark 4. f ∈Σ(Ω,A) iff

f =

n

X

i=1

α_i·1_Ai

with α₁, . . . α_n ∈R pairwise different and A₁, . . . , A_n∈ A pairwise disjoint such that Sn

i=1A_i = Ω.

Theorem 7. (Compare Theorem 5.11, Analysis IV) For every (bounded) function f ∈ Z₊(Ω,A) there exists a sequence f₁, f₂,· · · ∈ Σ₊(Ω,A) such that f_n ↑ f (with uniform convergence).

Proof. Letn ∈N and put f_n =

n·2ⁿ

X

k=1

k−1

2ⁿ ·1_An,k +n·1_Bn where

An,k ={(k−1)/(2ⁿ)≤f < k/(2ⁿ)}, Bn={f ≥n}.

Now we consider a mappingT : Ω₁ →Ω₂ and a σ-algebra A₂ in Ω₂. We characterize measurability of functions with respect toσ(T) =T⁻¹(A2).

Theorem 8 (Factorization Lemma). For every function f : Ω₁ →R f ∈Z(Ω₁, σ(T)) ⇔ ∃g ∈Z(Ω₂,A₂) :f =g◦T.

Proof. ‘⇐’ is trivially satisfied. ‘⇒’: First, assume that f ∈Σ+(Ω1, σ(T)), i.e., f =

n

X

i=1

α_i·1_Ai

with pairwise disjoint setsA₁, . . . , A_n ∈σ(T). Take pairwise disjoint setsB₁, . . . , B_n∈ A₂ such thatA_i =T⁻¹(B_i) and put

g =

n

X

i=1

α_i·1_Bi. Clearly f =g◦T and g ∈Z(Ω₂,A₂).

Now, assume that f ∈ Z₊(Ω1, σ(T)). Take a sequence (fn)n∈N in Σ+(Ω1, σ(T)) ac- cording to Theorem 7. We already know thatf_n=g_n◦T for suitable g_n∈Z(Ω₂,A₂).

Hence

f = sup

n

f_n= sup

n

(g_n◦T) = (sup

n

g_n)◦T =g◦T whereg = sup_ng_n ∈Z(Ω₂,A₂).

In the general case, we already know that

f⁺ =g₁◦T, f⁻=g₂ ◦T for suitable g1, g2 ∈Z(Ω2,A2). Put

C ={g₁ =g₂ =∞} ∈A₂,

and observe that T(Ω₁)∩C = ∅ since f = f⁺ −f⁻. We conclude that f = g ◦T where

g =g1·1D−g2·1D ∈Z(Ω2,A2) with D=C^c.

Our method of proof for Theorem 8 is sometimes called algebraic induction.

3 Product Spaces

Example 1. A stochastic model for coin tossing. For a single trial,

Ω ={0,1}, A=P(Ω), ∀ω ∈Ω :P({ω}) = 1/2. (1) Forn ‘independent’ trials, (1) serves as a building-block,

Ω_i ={0,1}, A_i =P(Ω_i), ∀ω_i ∈Ω_i :P_i({ω_i}) = 1/2, and we define

Ω =

n

Y

i=1

Ω_i, A=P(Ω), ∀A ∈A:P(A) = |A|

|Ω|. Then

P(A₁× · · · ×A_n) = P₁(A₁)· · · · ·P_n(A_n) for all A_i ∈A_i.

Question: How to model an infinite sequence of trials? To this end, Ω =

∞

Y

i=1

Ω_i.

How to choose aσ-algebra Ain Ω and a probability measureP on (Ω,A)? A reason- able requirement is

∀n ∈N ∀A_i ∈A_i :

P(A₁× · · · ×A_n×Ω_n+1×Ω_n+2. . .) =P₁(A₁)· · · · ·P_n(A_n). (2) Unfortunately,

A=P(Ω)

is too large, since there exists no probability measure on (Ω,P(Ω)) such that (2) holds.

The latter fact follows from a theorem by Banach and Kuratowski, which relies on the continuum hypothesis, see Dudley (2002, p. 526). On the other hand,

A={A₁ × · · · ×A_n×Ω_n+1×Ω_n+2· · ·:n ∈N, A_i ∈A_i for i= 1, . . . , n} (3) is not aσ-algebra.

Given: a non-empty setI and measurable spaces (Ω_i,A_i) for i∈I. Put Y =[

i∈I

Ωi

and define

Y

i∈I

Ω_i ={ω ∈Y^I :ω(i)∈Ω_i for i∈I}.

Notation: ω = (ω_i)_i∈I for ω∈Q

i∈IΩ_i. Moreover, let

P0(I) ={J ⊂I :J non-empty, finite}.

The following definition is motivated by (3).

Definition 1.

(i) Measurable rectangle

A =Y

j∈J

Aj × Y

i∈I\J

Ωi

with J ∈ P₀(I) and A_j ∈ A_j for j ∈ J. Notation: R class of measurable rectangles.

(ii) Product (measurable) space (Ω,A) with components (Ω_i,A_i),i∈I, Ω =Y

i∈I

Ω_i, A=σ(R).

Notation: A=N

i∈IAi, product σ-algebra.

Remark 1. The class Ris a semi-algebra, but not an algebra in general. See ¨Ubung 2.3.

Example 2. Obviously, (2) only makes sense if A contains the product σ-algebra Nn

i=1A_i. We will show that there exists a uniquely determined probability measure P on the product space (Q∞

i=1{0,1},N∞

i=1P({0,1})) that satisfies (2), see Remark 4.3.(ii). The corresponding probability space yields a stochastic model for the simple case of gambling, which was mentioned in the introductory Example I.2.

We study several classes of mappings or subsets that generate the productσ-algebra.

Moreover, we characterize measurability of mappings that take values in a product space.

Put Ω =Q

i∈IΩi. For any ∅ 6=S ⊂I let π^I_S : Ω→Y

i∈S

Ω_i, (ω_i)i∈I 7→(ω_i)i∈S

denote the projection of Ω onto Q

i∈SΩ_i (restriction of mappings ω). In particular, for i ∈ I the i-th projection is given by π_{i}^I . Sometimes we simply write π_S instead of π^I_S and πi instead ofπ{i}.

Theorem 1.

(i) N

i∈IA_i =σ({π_i :i∈I}).

(ii) ∀i∈I :A_i =σ(E_i) ⇒ N

i∈IA_i =σ S

i∈Iπ_i⁻¹(E_i) .

Proof. Ad (i), ‘⊃’: We show that every projection π_i : Ω → Ω_i is N

i∈IA_i -A_i- measurable. For A_i ∈A_i

π_i⁻¹(A_i) =A_i× Y

i∈I\{i}

Ω_i ∈R.

Ad (i), ‘⊂’: We show that R ⊂ σ({π_i : i ∈ I}). For J ∈ P₀(I) and A_j ∈ A_j with j ∈J

Y

j∈J

A_j× Y

i∈I\J

Ω_i =\

j∈J

π⁻¹_j (A_j).

Ad (ii): By Lemma 2.1 and (i) O

i∈I

A_i =σ[

i∈I

π_i⁻¹(A_i)

=σ[

i∈I

σ(π_i⁻¹(E_i))

=σ[

i∈I

π_i⁻¹(E_i) .

Corollary 1.

(i) For every measurable space (eΩ,A) and every mappinge g :Ωe →Ω g is A-e O

i∈I

A_i-measurable ⇔ ∀i∈I :π_i◦g is A-Ae _i-measurable.

(ii) For every ∅ 6=S⊂I the projection π_S^I is N

i∈IA_i-N

i∈SA_i-measurable.

Proof. Ad (i): Follows immediately from Theorem 2.3 and Theorem 1.(i).

Ad (ii): Note that π_{i}^S ◦π^I_S =π_i^I and use (i).

Remark 2. From Theorem 1.(i) and Corollary 1 we get O

i∈I

Ai =σ({π_S^I :S ∈P0(I)}).

The sets

π^I_S⁻¹

(B) =B × Y

i∈I\S

Ω_i withS ∈P₀(I) andB ∈N

i∈SA_iare calledcylinder sets. Notation: Cclass of cylinder sets. The classC is an algebra in Ω, but not aσ-algebra in general. Moreover,

R⊂α(R)⊂C⊂σ(R), where equality does not hold in general.

Every product measurable set is countably determined in the following sense.

Theorem 2. For everyA∈ ⊗i∈IA_i there exists a non-empty countable setS ⊂I and a set B ∈ ⊗i∈SA_i such that

A= π_S^I−1 (B).

Proof. Put Ae =n

A∈O

i∈I

A_i :∃S ⊂I non-empty, countable ∃B ∈O

i∈S

A_i :A= π_S^I−1

(B)o .

By definition,Ae contains every cylinder set andAe ⊂N

i∈IA_i. It remains to show that Ae is a σ-algebra. Obviously, Ω∈A, and ife A = (π_S^I)⁻¹(B),A^c = (π_S^I)⁻¹(B^c). Finally, if An = (π^I_Sn)⁻¹(Bn), we define S =S

nSn and Ben = (π_S^S_n)⁻¹(Bn) =Bn×Q

i∈S\Bn ∈ N

i∈SA_i (see Corollary 1, (ii)); then

\

n

A_n=\

n

(π_S^I)⁻¹(Be_n) = ((π^I)_S)⁻¹ \

n

Be_n

! ,

hence T

nAn ∈A.e

Now we study products of Borel-σ-algebras.

Theorem 3.

B_k =

k

O

i=1

B, B_k=

k

O

i=1

B.

Proof. By Remark 16, B_k=σnY^k

i=1

]−∞, a_i] :a_i ∈R for i= 1, . . . , ko

⊂

k

O

i=1

B.

On the other hand, πi : R^k → R is continuous, hence it remains to apply Corollary 2.1 and Theorem 1.(i). Analogously,B_k =Nk

i=1Bfollows.

Remark 3. Consider a measurable space (Ω,e A) and a mappinge f = (f₁, . . . , f_k) :Ωe →R^k.

Then, according to Theorem 3, f is A-Be _k-measurable iff all functions f_i are A-B-e measurable.

4 Construction of (Probability) Measures

Given: Ω6=∅ and ∅ 6=A⊂P(Ω).

Definition 1. µ:A→R⁺∪ {∞}is called (i) additive if:

A, B ∈A∧A∩B =∅ ∧A∪B ∈A ⇒ µ(A∪B) = µ(A) +µ(B), (ii) σ-additive if

A1, A2, . . . ∈Apairwise disjoint ∧

∞

[

i=1

Ai ∈A ⇒ µ [^∞

i=1

Ai

=

∞

X

i=1

µ(Ai), (iii) content (on A) if

A algebra ∧ µadditive ∧ µ(∅) = 0, (iv) pre-measure (on A) if

A semi-algebra ∧ µ σ-additive ∧ µ(∅) = 0, (v) measure (on A) if

A σ-algebra ∧ µpre-measure, (vi) probability measure (on A) if

µmeasure ∧ µ(Ω) = 1.

Definition 2. (Ω,A, µ) is called a

(i) measure space, if µis a measure on the σ-algebra A in Ω,

(ii) probability space, ifµ is a probability measure on the σ-algebra Ain Ω.

Example 1.

(i) k–dimensional Lebesgue pre-measure λ_k, e.g., on cartesian products of intervals.

(ii) For any semi-algebraA in Ω and ω ∈Ω

δ_ω(A) = 1_A(ω), A∈A,

defines a pre-measure. If A is a σ-algebra, then δ_ω is called the Dirac measure at the point ω.

More generally: take sequences (ω_n)_n∈R in Ω and (α_n)_n∈N in R⁺ such that P∞

n=1α_n= 1. Then

µ(A) =

∞

X

n=1

α_n·1_A(ω_n), A∈A,

defines a discrete probability measure on any σ-algebra A in Ω. Note that µ = P∞

n=1α_n·ε_ωn.

(iii) Counting measure on a σ-algebra A

µ(A) = |A|, A ∈A.

Uniform distribution in the case |Ω|<∞ and A=P(Ω) µ(A) = |A|

|Ω|, A⊂Ω.

(iv) On the algebraA={A⊂Ω :A finite or A^c finite} let µ(A) =

(0 if |A|<∞

∞ if |A|=∞.

Then µis a content but not a pre-measure in general.

(v) For the semi-algebra of measurable rectangles in Example 3.1 andA_i ⊂ {0,1}

µ(A₁× · · · ×A_n×Ω_n+1× · · ·) = |A₁×. . .×A_n|

| {0,1}ⁿ| is well defined and yields a pre-measure µwith µ {0,1}^N

= 1.

Remark 1. For every content µonA and A, B ∈A

(i) A⊂B ⇒µ(A)≤µ(A∩B) +µ(A^c ∩B) =µ(B) (monotonicity), (ii) µ(A∪B) +µ(A∩B) = µ(A) +µ(B\A) +µ(A∩B) =µ(A) +µ(B), (iii) A⊂B ∧µ(A)<∞ ⇒µ(B\A) =µ(B)−µ(A),

(iv) µ(A)<∞ ∧µ(B)<∞ ⇒ |µ(A)−µ(B)| ≤µ(AMB),

(v) µ(A∪B) =µ(A) +µ(B∩A^c)≤µ(A) +µ(B) (subadditivity).

Theorem 1. Consider the following properties for a content µ onA:

(i) µ pre-measure, (ii) A₁, A₂, . . .∈A∧S∞

i=1A_i ∈A⇒µ S∞ i=1A_i

≤P∞

i=1µ(A_i) (σ-subadditivity), (iii) A₁, A₂, . . . ∈ A∧ A_n ↑ A ∈ A ⇒ lim_n→∞µ(A_n) = µ(A) (σ-continuity from

below),

(iv) A₁, A₂, . . . ∈ A ∧A_n ↓ A ∈ A∧ µ(A₁) < ∞ ⇒ limn→∞µ(A_n) = µ(A) (σ- continuity from above),

(v) A₁, A₂, . . .∈A∧A_n↓ ∅ ∧µ(A₁)<∞ ⇒limn→∞µ(A_n) = 0 (σ-continuity at ∅).

Then

(i)⇔ (ii) ⇔ (iii) ⇒ (iv) ⇔(v).

Ifµ(Ω)<∞, then (iii) ⇔(iv).

Proof. ‘(i) ⇒ (ii)’: PutBm =Sm

i=1Ai and B0 =∅. Then

∞

[

i=1

A_i =

∞

[

m=1

(B_m\B_m−1)

with pairwise disjoint sets Bm \Bm−1 ∈ A. Clearly Bm \Bm−1 ⊂ Am. Hence, by Remark 1.(i),

µ [^∞

i=1

A_i

=

∞

X

m=1

µ(B_m\Bm−1)≤

∞

X

m=1

µ(A_m).

‘(ii) ⇒ (i)’: LetA₁, A₂, . . .∈A be pairwise disjoint with S∞

i=1A_i ∈A. Then µ[^∞

i=1

A_i

≥µ[ⁿ

i=1

A_i

=

n

X

i=1

µ(A_i), and therefore

∞

X

i=1

µ(A_i)≤µ[^∞

i=1

A_i . The reverse estimate holds by assumption.

‘(i) ⇒ (iii)’: Put A₀ =∅ and B_m =A_m\Am−1. Then µ[^∞

i=1

A_i

=

∞

X

m=1

µ(B_m) = lim

n→∞

n

X

m=1

µ(B_m) = lim

n→∞µ[ⁿ

m=1

B_m

= lim

n→∞µ(A_n).

‘(iii) ⇒ (i)’: Let A1, A2, . . . ∈ A be pairwise disjoint with S∞

i=1Ai ∈ A, and put B_m =Sm

i=1A_i. Then B_m ↑S∞

i=1A_i and µ[^∞

i=1

A_i

= lim

m→∞µ(B_m) =

∞

X

i=1

µ(A_i).

‘(iv)⇒ (v)’ trivially holds.

‘(v)⇒ (iv)’: Use B_n=A_n\A↓ ∅.

‘(i)’ ⇒ (v)’: Note that µ(A₁) = P∞

i=1µ(A_i\A_i+1). Hence 0 = lim

k→∞

∞

X

i=k

µ(A_i\A_i+1) = lim

k→∞µ(A_k).

‘(iv)∧ µ(Ω)<∞ ⇒(iii)’: Clearly A_n ↑A impliesA^c_n ↓A^c. Thus µ(A) = µ(Ω)−µ(A^c) = lim

n→∞(µ(Ω)−µ(A^c_n)) = lim

n→∞µ(A_n).

Theorem 2 (Extension: semi-algebra algebra). For every semi-algebra A and every additive mapping µ:A→R⁺∪ {∞} with µ(∅) = 0

∃1µb content onα(A) : bµ|_A =µ.

Moreover, if µis σ-additive then bµis σ-additive, too.

Proof. We haveα(A) = A⁺, see Lemma 1.1. Necessarily

bµ[ⁿ

i=1

A_i

=

n

X

i=1

µ(A_i) (1)

for A₁, . . . , A_n ∈ A pairwise disjoint. Use (1) to obtain a well-defined extension of µ ontoα(A). It easily follows that µis additive or even σ-additive.

Example 2. For the semi-algebra Ain Example 1.(v)α(A) is the algebra of cylinder sets, and

µ(Ab ×Ω_n+1× · · ·) = |A|

| {0,1}ⁿ|, A ⊂ {0,1}ⁿ. Letµ be a pre–measure onA. The outer measure generated byµis

µ^∗(A) := inf ( _∞

X

i=1

µ(A_i) : A_i ∈A, A⊆ [

i=1

∞A_i )

,

It is straightforward that µ^∗(∅= 0) and that µ^∗ is monotone and σ–subadditive.

Theorem 3 (Extension: algebra σ-algebra, Carath´eodory). For every pre- measureµ on an algebra A,

(a) the class

A_µ^∗ :=n

A ⊆Ω : µ^∗(B) = µ^∗(A∩B) +µ^∗(A^c∩B)∀B ⊆Ωo is a σ–algebra, and µ^∗ is a measure on Aµ^∗.

(b) A ⊆ A_µ^∗, and µ = µ^∗ on A. In particular, there exists a measure µ^∗ on σ(A) extendingµ.

Proof. We will start with part (b), i.e., we show that (i) µ^∗|_A=µ,

(ii) ∀A∈A ∀B ∈P(Ω) : µ^∗(B) = µ^∗(B ∩A) +µ^∗(B∩A^c).

Ad (i): ForA∈A

µ^∗(A)≤µ(A) +

∞

X

i=2

µ(∅) =µ(A), and for A_i ∈A with A⊂S∞

i=1A_i µ(A) =µ[^∞

i=1

(A_i∩A)

≤

∞

X

i=1

µ(A_i∩A)≤

∞

X

i=1

µ(A_i)

Ad (ii): ‘≤’ holds due to sub-additivity of µ^∗; if B ⊆

∞

[

i=1

A_i with A_i ∈A, then A_i∩A, A_i∩A^c ∈A and

B∩A ⊆

∞

[

i=1

Ai∩A, B ∩A^c ⊆

∞

[

i=1

Ai∩A^c . This directly implies ‘≥’.

Now we prove (a); to this end, we claim first that

(iii) A_µ^∗ is ∩–closed, ∀A₁, A₂ ∈ A_µ^∗ ∀B ∈ P(Ω) : µ^∗(B) = µ^∗(B ∩(A₁ ∩A₂)) + µ^∗(B∩(A₁∩A₂)^c).

(iv) Aµ^∗ c–closed, i.e., A is an algebra.

Ad (iii): We have

µ^∗(B) =µ^∗(B∩A₁) +µ^∗(B∩A^c₁)

=µ^∗(B∩A₁∩A₂) +µ^∗(B∩A₁∩A^c₂) +µ^∗(B∩A^c₁) and

µ^∗(B∩(A₁∩A₂)^c) = µ^∗(B ∩A^c₁∪B∩A^c₂) = µ^∗(B∩A^c₂∩A₁) +µ^∗(B∩A^c₁).

Ad (iv): Obvious.

Next we claim thatµ^∗ is additive onA^∗, and even more,

(v) ∀A₁, A₂ ∈A_µ^∗ disjoint∀B ∈P(Ω) : µ^∗(B∩(A₁∪A₂)) =µ^∗(B∩A₁) +µ^∗(B∩ A₂).

In fact, sinceA₁∩A₂ =∅,

µ^∗(B ∩(A1∪A2)) =µ^∗(B∩A1) +µ^∗(B∩A2∩A^c₁) =µ^∗(B∩A1) +µ^∗(B∩A2).

At last, we claim that A^∗ is a Dynkin class and µ^∗ is σ–additive on A^∗, i.e., (vi) ∀A₁, A₂, . . .∈A_µ^∗ pairwise disjoint

∞

[

i=1

A_i ∈A_µ^∗ ∧ µ^∗[^∞

i=1

A_i

=

∞

X

i=1

µ^∗(A_i).

LetB ∈P(Ω). By (iv), (v), and monotonicity of µ^∗ µ^∗(B) = µ^∗

B∩

n

[

i=1

A_i +µ^∗

B∩[ⁿ

i=1

A_ic

≥

n

X

i=1

µ^∗(B∩A_i) +µ^∗

B ∩[^∞

i=1

A_ic . Use σ-subadditivity of µ^∗ to get

µ^∗(B)≥

∞

X

i=1

µ^∗(B ∩Ai) +µ^∗

B∩[^∞

i=1

Ai

c

≥µ^∗ B ∩

∞

[

i=1

A_i +µ^∗

B∩[^∞

i=1

A_ic

≥µ^∗(B).

Hence S∞

i=1Ai ∈Aµ^∗. Take B =S∞

i=1Ai to obtain σ-additivity of µ^∗|A_µ∗. Conclusions:

• A_µ^∗ is a Dynkin class and ∩–closed ((iv), (vi)), and hence a σ-algebra, see Theorem 1.1.(ii),

• A⊂A_µ^∗ by (ii), hence σ(A)⊂A_µ^∗.

• µ^∗|_Aµ∗ is a measure withµ^∗|_A =µ, see (vi) and (i).

Remark 2. The extension from Theorem 3 is non-unique, in general. For instance, on Ω =R, the pre–measure

µ(A) = ∞ ·#A =

(0 if A=∅

∞ otherwise, A∈α(J₁)

on the algebra generated by intervals (see Ex.1) has the extensions µ₁(A) = #A (counting measure) and µ₂(A) = ∞ ·#A to B.

Definition 3. µ:A→R+∪ {∞}is called (i) σ-finite, if

∃B₁, B₂, . . . ∈Apairwise disjoint : Ω =

∞

[

i=1

B_i∧ ∀i∈N:µ(B_i)<∞,

(ii) finite, if Ω∈A and µ(Ω) <∞.

Theorem 4 (Uniqueness). A₀ be ∩–closed, µ₁, µ₂ be measures on A = σ(A₀). If µ | is σ–finite and µ | =µ | , then µ =µ .

Proof. TakeBi according to Definition 3, with A0 instead of A, and put D_i ={A ∈A:µ₁(A∩B_i) = µ₂(A∩B_i)}.

Obviously, D_i is a Dynkin class andA₀ ⊂D_i. Theorem 1.2.(i) yields D_i ⊂A=σ(A₀) = δ(A₀)⊂D_i.

ThusA=D_i and for A∈A, µ1(A) =

∞

X

i=1

µ1(A∩Bi) =

∞

X

i=1

µ2(A∩Bi) = µ2(A).

Corollary 1. For every semi-algebraAand every pre-measureµonAthat isσ-finite

∃1µ^∗ measure on σ(A) : µ^∗|_A =µ.

Proof. Use Theorems 2, 3, and 4.

Remark 3. Applications of Corollary 1:

(i) For Ω =R^kand the Lebesgue pre-measureλ_konJ_kwe get the Lebesgue measure on B_k. Notation for the latter: λ_k.

(ii) In Example 1.(v) there exists a uniquely determined probability measure P on N∞

i=1P({0,1}) such that

P(A1× · · · ×An× {0,1} ×. . .) = |A₁× · · · ×A_n|

|{0,1}ⁿ|

for A₁, . . . , A_n ⊂ {0,1}. We will study the general construction of product measures in Section 8.

5 Integration

For the proofs, see Analysis IV or Elstrodt (1996, Kap. VI).

Fixed in this section: A measure space (Ω,A, µ). Notation:

• Σ₊ = Σ₊(Ω,A) (nonnegative simple functions),

• Z₊ =Z₊(Ω,A) (nonnegative A-B-measurable functions), Definition 1. Integral Let f ∈Σ₊,

f =

n

X

i=1

α_i·1_Ai , α_i ∈R, A_i ∈A. Then define its Integral w.r.t. µas

Z

f dµ=

n

X

i=1

α_i·µ(A_i). Lemma 1. The mapping R

·dµ: Σ₊ →R₊ is (i) positive–linear: R

(αf+βg)dµ=αR

f dµ+βR

g dµ,f, g ∈Σ₊, α, β ∈R+, (ii) monotone: f ≤g ⇒R

f dµ ≤R

g dµ (monotonicity).

Definition 2. Integral of f ∈Z₊ w.r.t. µ Z

f dµ= supnZ

g dµ:g ∈Σ₊∧g ≤fo .

Theorem 1 (Monotone convergence, Beppo Levi). (e.g., Thm.6.4, Analysis IV, SS06) Let f_n∈Z₊ such that

∀n∈N:f_n≤f_n+1.

Then Z

sup

n

f_ndµ= sup

n

Z

f_ndµ.

Remark 1. For everyf ∈Z₊ there exists a sequence of functions f_n ∈Σ₊ such that f_n ↑f, see Theorem 2.7.

Example 1. Consider

f_n= 1 n ·1_[0,n]

on (R,B, λ₁). Then

Z

f_ndλ₁ = 1, lim

n→∞f_n= 0.

Lemma 2. The mapping R

·dµ : Z →R is still positive–linear and monotone.

Theorem 2 (Fatou’s Lemma). (See, e.g., Lemma 6.6, Ananlysis IV, SS06) For every sequence (f_n)_n inZ₊

Z

lim inf

n→∞ f_ndµ≤lim inf

n→∞

Z

f_ndµ.

Proof. For g_n = infk≥nf_k we have g_n ∈ Z₊ and g_n ↑ lim inf_nf_n. By Theorem 1 and Lemma 1.(iii)

Z

lim inf

n f_ndµ= lim

n→∞

Z

g_ndµ≤lim inf

n→∞

Z

f_ndµ.

Theorem 3. Let f ∈Z₊. Then Z

f dµ= 0⇔µ({f > 0}) = 0.

Definition 3. A property Π holds µ-almost everywhere (µ-a.e., a.e.), if

∃A∈A:{ω∈Ω : Π does not hold for ω} ⊂A∧µ(A) = 0.

In case of a probability measure we say: µ-almost surely,µ-a.s., with probability one.

Notation: Z=Z(Ω,A) is the class ofA-B-measurable functions.

Definition 4. f ∈Zquasi-µ-integrable if Z

f₊dµ <∞ ∨ Z

f₋dµ < ∞.

In this case: integral of f (w.r.t. µ) Z

f dµ= Z

f+dµ− Z

f−dµ.

f ∈Z µ-integrable if Z

f₊dµ <∞ ∧ Z

f−dµ < ∞.

Theorem 4.

(i) f µ-integrable ⇒µ({|f|=∞}) = 0,

(ii) f µ-integrable ∧g ∈Z ∧ f =g µ-a.e. ⇒ g µ-integrable ∧ R

f dµ=R g dµ.

(iii) equivalent properties for f ∈Z:

(a) f µ-integrable, (b) |f| µ-integrable,

(c) ∃g :g µ-integrable∧ |f| ≤g µ-a.e.,

(iv) forf and g µ-integrable and c∈R

(a) f+gwell-definedµ-a.e. andµ-integrable withR

(f+g)dµ=R

f dµ+R g dµ, (b) c·f µ-integrable with R

(cf)dµ=c·R f dµ, (c) f ≤g µ-a.e. ⇒ R

f dµ≤R g dµ.

Theorem 5 (Dominated convergence, Lebesgue). Assume that (i) fn∈Z for n∈N,

(ii) ∃g µ-integrable ∀n∈N:|f_n| ≤g µ-a.e., (iii) f ∈Zsuch that limn→∞f_n=f µ-a.e.

Then f is µ-integrable and Z

f dµ= lim

n→∞

Z

fndµ.

Example 2. Consider

fn =n·1_]0,1/n[

on (R,B, λ₁). Then Z

f_ndλ₁ = 1, lim

n→∞f_n= 0.

6 L

-Spaces

Given: a measure space (Ω,A, µ) and 1≤p <∞. Put Z=Z(Ω,A).

Definition 1.

L^p =L^p(Ω,A, µ) = n

f ∈Z: Z

|f|^pdµ < ∞o .

In particular, for p = 1: integrable functions and L = L¹, and for p = 2: square- integrable functions. Put

kfk_p = Z

|f|^pdµ 1/p

, f ∈L^p.

Theorem 1 (H¨older inequality). Let 1< p, q < ∞ such that 1/p+ 1/q = 1 and letf ∈L^p,g ∈L^q. Then

Z

|f ·g|dµ≤ kfk_p· kgk_q. In particular, forp=q= 2: Cauchy-Schwarz inequality.

Proof. See Analysis III or Elstrodt (1996,§VI.1) as well as Theorem 5.3.

Theorem 2. L^p is a vector space and k · k_p is a semi-norm on L^p. Furthermore, kfkp = 0 ⇔ f = 0 µ-a.e.

Proof. See Analysis III or Elstrodt (1996,§VI.2).

Definition 2. Letf, f_n∈L^p forn∈N. (f_n)_n converges to f in L^p (in mean of order p) if

n→∞lim kf −f_nk_p = 0.

In particular, for p = 1: convergence in mean, and for p = 2: mean-square conver- gence. Notation:

f_n −→^Lp f.

Remark 1. Let f, f_n∈Z for n∈N. Recall (define) that (f_n)_n converges to f µ-a.e.

if

µ(A^c) = 0 for

A=

n→∞lim f_n=f =

lim sup

n→∞

f_n = lim inf

n→∞ f_n ∩

lim sup

n→∞

f_n=f ∈A.

Notation:

f_n ^µ-a.e.−→ f.

Lemma 1. Let f, g, f_n∈L^p for n∈N such that f_n−→^Lp f. Then f_n−→^Lp g ⇔ f =g µ-a.e.

Analogously for convergence almost everywhere.

Proof. For convergence in L^p: ‘⇐’ follows from Theorem 5.4.(ii). Use kf −gk_p ≤ kf−f_nk_p+kf_n−gk_p

to verify ‘⇒’.

For convergence almost everywhere: ‘⇐’ trivially holds. Use lim

n→∞f_n =f ∩

n→∞lim f_n =g ⊂ {f =g}

to verify ‘⇒’.

Theorem 3 (Fischer-Riesz). Consider a sequence (fn)n in L^p. Then (i) (f_n)_n Cauchy sequence ⇒ ∃f ∈L^p : f_n−→^Lp f (completeness), (ii) f_n−→^Lp f ⇒ ∃subsequence (f_nk)_k :f_nk ^µ-a.e.−→ f.

Proof. Ad (i): Consider a Cauchy sequence (f_n)_n and a subsequence (f_nk)_k such that

∀k ∈N ∀m≥n_k :kf_m−f_nkk_p ≤2^−k. For

g_k =f_nk+1−f_nk ∈L^p we have

k

X

`=1

|g_`| p

≤

k

X

`=1

kg_`k_p ≤

k

X

`=1

2^−` ≤1.

Putg =P∞

`=1|g<sub>`</sub>| ∈Z₊. By Theorem 5.1 Z

g^pdµ= Z

sup

k

X^k

`=1

|g`|p

dµ= sup

k

Z X^k

`=1

|g`|p

dµ≤1. (1)

Thus, in particular,P∞

`=1|g<sub>`</sub>|and P∞

`=1g<sub>`</sub> convergeµ-a.e., see Theorem 5.4.(i). Since f_nk+1 =

k

X

`=1

g_`+f_n1, we have

f = lim

k→∞f_nk µ-a.e.

for some f ∈Z. Furthermore,

|f −f_nk| ≤

∞

X

`=k

|g_`| ≤g µ-a.e., so that, by Theorem 5.5 and (1),

k→∞lim Z

|f −f_nk|^pdµ= 0.

It follows that

n→∞lim kf −f_nk_p = 0, too. Finally, by Theorem 2, f ∈L^p.

Ad (ii): Assume that

fn L^p

→f.

According to the proof of (i) there existsfe∈L^p and a subsequence (f_nk)_k such that f_nk ^µ-a.e.−→ fe∧f_nk −→^Lp f .e

Use Lemma 1.