12) Measure theory

Question 1

Def 12.1 σ-algebra

Answer 1

Let X be a set. A σ-algebra R on X is a collection of subsets
of X, written R ⊆ 2^X, (Power set of X)
such that
(i) X ∈ R;
(ii) if A, B ∈ R, then A \ B ∈ R;
(x ∈ R, so X\A ∈ R) ( Ø)∈ R,
(iii) if (An) is any sequence in R, then ∪nAn ∈ R. (countable unions)

e.g. if X={a,b,c,d} one possible σ-algebra on X is
={∅, {a,b}, {c,d}, {a,b,c,d}
In general, a finite algebra is always a σ-algebra.

for a sigma algebra we will also have closure under intersections and empty sets

Question 2

Whats wrong with Riemann integrals?

Answer 2

RI solitting using sub intervals and using areas of rectangles for an upper and lower RS to converge to the RI

However conisder us trying to measure the area of
f(x)=
{1 x in Q
{0 x not in Q

Q is dense everywhere
but total measure of all rationals =0

introduce lebesgue integral:

by slices of domain and consider what x’s give the function values in that range

Question 3

(Def set algebra)

Answer 3

3 in σ-algebra replaced:
if (A_n)1 ^ m is any finite family in R, then ∪{n=1,m} An ∈ R;

as sigma implies a countable property

Question 4

(Def set ring)

Answer 4

only 2 in σ-algebra
ii) if A, B ∈ R, then A \ B ∈ R;
(x ∈ R, so X\A ∈ R) ( Ø)∈ R,

In set theory, a ring of sets is a collection of sets that satisfies the following properties:

1. Closure under finite unions: If A and B are sets in the ring, then their union A ∪ B is also in the ring.
2. Closure under differences: If A and B are sets in the ring, then their difference A - B (set of elements in A but not in B) is also in the ring.

These properties make a ring of sets a useful concept in set theory and other areas of mathematics, particularly in measure theory and abstract algebra.

Question 5

The main difference between a sigma algebra and a ring of sets

Answer 5

The main difference between a sigma algebra and a ring of sets lies in their closure properties.

A sigma algebra of sets is a collection of sets that is closed under countable unions and complementation.

On the other hand, a ring of sets is a collection of sets that is closed under finite unions and differences.

In summary, the key distinction is that a sigma algebra is closed under countable unions and complementation, while a ring of sets is closed under finite unions and differences

Question 6

The ring generated by the sets {1,2} and {4,5} on the set X={1,2,3,4,5} would consist of the following sets:

Answer 6

1. The empty set ∅
2. {1,2}
3. {4,5}
4. {1,2,4,5}
5. {3}

These sets are obtained by taking finite unions and differences of the given sets {1,2} and {4,5}. This collection of sets forms the ring generated by the given sets on the set X={1,2,3,4,5}.

Question 7

The sigma algebra generated by the sets {1,2} and {4,5} on the set X={1,2,3,4,5} would consist of the following sets:

Answer 7

1. The empty set ∅
2. {1,2}
3. {4,5}
4. {1,2,3,4,5}
   {1,2,4,5}
   {3,4,5}
   {1,2,3}

{3}???

These sets are obtained by taking countable unions and complements of the given sets {1,2} and {4,5}. This collection of sets forms the sigma algebra generated by the given sets on the set X={1,2,3,4,5}.

Question 8

Def σ-ring

Answer 8

only 2 and 3 in σ-algebra
(ii) if A, B ∈ R, then A \ B ∈ R;
(x ∈ R, so X\A ∈ R) ( Ø)∈ R,
(iii) if (An) is any sequence in R, then ∪nAn ∈ R. (countable unions)

Question 9

Def R(s)

Answer 9

collection of finite disjoint sets is a Ring generated by semiring S:
ie minimal ring A ⊂ R(s) iff A = ⊔ A_k for k=1,…n

Question 10

Pairwise disjoint

Answer 10

written A ⊔B

A_n ∩ A_m = Ø for n≠ m
helpful to work with these - a notion of basis

Question 11

For a σ-algebra R and A, B ∈ R, we have show in R: A ∩ B =

Answer 11

A ∩ B
= X \ (X \ (A ∩ B))
= X \ ((X \ A) ∪ (X \ B)) ∈ R

ie intersections are also in the sigma algebra!

Question 12

**Def 12.3 Semiring
**

Answer 12

A semiring S of sets is a collection such that
(i) it is closed under intersection;
(ii) for A, B ∈ S we have
A \ B = C_1 ⊔ . . . ⊔ C_N with C_k ∈ S (finite)
(any non-empty semiring contain the empty set, no complement property, basis for a ring

it is a union of finite elements)

complement property relaxed, it doesn’t need to be in the set just be a finite union of things

Question 13

Def: σ-algebra generated by D

Answer 13

Given any collection D ⊆2^X ∃σ-algebra R S.t D ⊆ R and

st if S is any other σ-algebra with D ⊆ S then R ⊆ S

Called the σ-algebra generated by D

(to generate this we ensure it contains elements of D, is closed under complements, has the empty set, closed under unions. Closure under intersections will be true too.(

Question 14

2^x power set is

Answer 14

a sigma algebra which contains X

we have closure under complements
complements are finite unions of disjoint sets

Question 15

the intersection of a family of sigma algebras

Answer 15

is a sigma algebra itself

Question 16

Examples of semirings: collections of intervals [a,b] on R?

Answer 16

semiring but not a ring
Diagram:
[c,d) a<c<d<b
complement of set [a,b)∪[b,d) (can represent with unions C_k in S)
[c,d) c<a<d<b
complement of set [d,b) cant be as not all in [a,b}
(semiring means it has closure under intersections. which are collections of finite disjoin unions; typically do not have additive inverses (complements) that are also in the collection so is not a ring)

Question 17

Examples of semirings: collections of half open rectangles {a≤x <b, c≤ y<d }on a plane

Answer 17

semiring but not a ring

diagram: B not necessarily contained in A
consider A\B, which can be partitioned also by half open rectangles

Question 18

Let S be a semiring

What is R(S)

Answer 18

The collection of all finite disjoint unions ⊔{k=1,n} Ak, where Ak ∈ S, is a ring. We call it the ring R(S) generated by the semiring S.

(ii) Any ring containing S contains R(S) as well.

(iii) The collection of all finite (not necessarily disjoint!) unions ⊔_{k=1,n} Ak, where Ak ∈ S, coincides with R(S).

Question 19

to find the smallest sigma algebra containing a semiring

Answer 19

we form intersection of al sigma algebras containing that semiring

Question 20

Let S be a semiring. Show that
(i) The collection of all finite disjoint unions ⊔
k=1 to n Ak, where A_k ∈ S, is a
ring.

Answer 20

This is R(S) the ring generated by a semiring S

Found by takinng finite disjoing unions in A

We show it is a ring by showing: closed udner complements for some A= union a_k B= union b_k
A\B=(union a_k)(union b_k)

A_1(B_1 union B_2)= A_1\B_1 intersection A_1\B_1
A_2(B_1 union B_2)=…

Define c_kj=A_k\ Bj

thus defined A\B as the union of C_kj

Question 21

(iii) The collection of all finite (not necessarily disjoint!) unions ⊔ k=1 to n A_k,
where A_k ∈ S, coincides with R(S)

Answer 21

-

Question 22

(ii) Any ring containing S contains R(S) as well.-

Answer 22

-

Question 23

Def 12.6 Measure

Answer 23

A measure is a map µ : R → [0, ∞] defined on a (semi-)ring
(or σ-algebra) R, such that if
A = ⊔_nA_n for A ∈ R and a finite subset (A_n) of R, then
µ(A) = Σµ(A_n). This property is called additivity of a measure

Question 24

measure property for empty set

Answer 24

µ(∅) = 0.
equiv to
There is a set A ∈ R such that µ(A) < ∞.
exercise show that they are equiv

Question 25

Def 12.8 measure map is countably additive if

Answer 25

µ is countably
additive (or σ-additive) if for any countable infinite family (An) of pairwise disjoint sets from R such that
A = ⊔_nAn ∈ R we have
µ (A) = Σ_n µ(An).

(If the sum diverges, then as it will be the sum of positive numbers, we can, without problem, define it to be +∞)

Question 26

continuity property of an additive measure

Answer 26

µ(Lim{n→∞}( ⊔(k=1)^n A_k)
=Lim{n→∞} µ(⊔(k=1)^n A_k)
in general when we have inf series we cant reorder terms but

we can commute limits with inf unions of measures

Question 27

difference between additivity of a measure and sigma additivity

Answer 27

additivity of a measure:

for a disjoint union the measure s defined as the sum of measures

sigma additivity:continuity property of the measure
limit of measure commutes
(sigma additive we also have <

Question 28

Defn 12.10 measure µ is
finite if
σ-finite if

Answer 28

A measure µ is finite if µ(A) < ∞ for all A ∈ R.

A measure µ is σ-finite if X is a union of countable number of sets X_k, such that for any A ∈ R and any k ∈ N the intersection A ∩ X_k is in R and µ(A ∩X_k) < ∞.

Question 29

distribution laws (A∪B)∩C

Answer 29

(A∪B)∩C = (A∩C) ∪ (B∩C)

Question 30

A ⊆ B B = ?⊔?

Answer 30

A ⊆ B B = A ⊔ (B\A)

Question 31

Examples: f(x)=
{1 x ∈Q
{0 x ∉Q

defined on R

Answer 31

Every interval contains Q- dense

Question 32

Examples: Consider a semiring S
. Show that
A =⊔{k=1,..n} of A_k A_k ∈S
B =⊔{j=1,..n} of B_j B_j ∈S
A\B from the collection of all finite disjoint unions is a ring

Answer 32

Diagram: Consider A₁,A₂, B₁,B₂
s.t A’s disjoint and B’s disjoint
consider n=2
A₁ (B₁⊔B₂) = A₁ \B₁ ∩ A₁ \B₂
A₂(B₁⊔B₂) = A₂ \B₁ ∩ A₂\B₂
(A_k\B_j = C_1⊔…⊔C_N )
So A\B = (⊔A_k)(⊔B_j)
for k=1,….n and j=1,…n
= ⊔{k=1,..n}[∩{j=1,…,m} (A_k)(B_j)]
= ⊔{k=1,..n}[∩{j=1,…,m}⊔{l=1,..kj}C_lkj
=⊔C_lkj which is also a semiring

Question 33

Example S= {(a,b) :a<b} a,b in R
X=R is it a semiring?

Answer 33

1 yes Ø∈S whenever a<b there exists c∈R s.t. a<c<b 2

Question 34

Example fix a∈R define µ by:
µ(A)=1 if a∈A
µ(A)=0 otherwise
additive?

Answer 34

Yes, this is an example of a µ, additive measure

Question 35

Example: For the ring of intervals (of disjoint unions from semi ring S, The collection of intervals [a, b) on the real line;)
[a,b)

define µ(([a,b])=b-a on S
is it a measure? σ-additive?

Answer 35

Its a measure with σ-additivity, disjoint sets add length

Question 36

Example : For ring (collection of all finite disjoint unions ⊔k=1,..,n Ak, where Ak ∈ S, is a ring R(S) generated by the semiring S.) obtained from the semiring The collection of all rectangles {a ⩽ x < b, c ⩽ y < d} on the plane

define µ(V) = (b−a)(d−c) for the rectangle V = {a ⩽ x < b, c ⩽ y < d}
S.
σ-additive?

Answer 36

It will be again a σ-additive measure.

Question 37

Let X = N and R = 2^N, we define

µ(A) = 0 if A is a finite subset of X = N
and
µ(A) = +∞ otherwise

σ-additive.?

Answer 37

Let A_n = {n}, then
µ(A_n) = 0 and
µ(⊔_n A_n) = µ(N) = +∞ ̸=
Σ_n µ(A_n) = 0.

Thus, this measure is not
σ-additive and limit does not commute

Question 38

Example: Consider interval [0,1) and partitions which don’t coincide
[0,1) = A₁ ⊔A₂
[0,1) = B₁ ⊔B₂ ⊔B₃
What method do we use to help define a measure?

Answer 38

Define another partition
[0,1) = C₁₁ ⊔ C₁₂ ⊔C₂₂ ⊔C₂₃
HOW?
A⊂ R(s) ⇔ A = ⊔{k=1,..n} A_k
s.t. A_k ∈ S
But this is not unique:
A= ⊔{1,2} A_k = ⊔_{1,2,3} B_j

Let C_jk = A_j ∩B_k (by defn C_jk are disjoint)

C_jk ⊂ A _j and C_jk ⊂ B _k

and we can rebuild
A_j = ⊔{k=1,…,m} C_jk
B_k = ⊔{j=1,…,m} C_jk using these

Question 39

Exercise 12.11

Answer 39

Exercise 12.11. Modify the example 12.9(i) to obtain
(i) a measure which is not finite, but is σ-finite. (Hint: let the measure
count the number of integer points in a set).
(ii) a measure which is not σ-finite. (Hint: assign µ(A) = +∞ if a ∈ A.)

Question 40

Prop 12.12
Let µ be a σ-additive measure on a σ-algebra R. Then:

Answer 40

Let µ be a σ-additive measure on a σ-algebra R. Then:

1) If A, B ∈ R with A ⊆ B, then
µ(A) ⩽ µ(B) [
“monotonicity of a measure”];
2)If A, B ∈ R with A ⊆ B and µ(B) < ∞, then µ(B \ A) = µ(B) − µ(A);
3) If (An) is a sequence in R, with A1 ⊆ A2 ⊆ A3 ⊆ · · ·. Then
limn→∞ µ(An) = µ (∪nAn).

4)If (A_n) is a sequence in R, with A₁⊇ A₂ ⊇ A ₃ ⊇ · · ·.
If µ(A_m) < ∞
for some m, then
lim(n→∞) µ(A_n) = µ (∩_n A_n).

Question 41

PROOF 12.12
Let µ be a σ-additive measure on a σ-algebra R. Then:

1) If A, B ∈ R with A ⊆ B, then
µ(A) ⩽ µ(B) [
“monotonicity of a measure”];
2)If A, B ∈ R with A ⊆ B and µ(B) < ∞, then µ(B \ A) = µ(B) − µ(A);
3) If (An) is a sequence in R, with A1 ⊆ A2 ⊆ A3 ⊆ · · ·. Then
limn→∞ µ(An) = µ (∪nAn).

4)If (A_n) is a sequence in R, with A₁⊇ A₂ ⊇ A ₃ ⊇ · · ·.
If µ(A_m) < ∞
for some m, then
lim(n→∞) µ(A_n) = µ (∩_n A_n).

Answer 41

Proof. The two first properties are easy to see. For the third statement, define
A = ∪n A_n,
B₁= A₁ and
B_n = A_n \ A{n−1}, n > 1.
Then An = ⊔k=1,…n B_n and
A = ⊔∞ k=1B_n.

Using the σ-additivity of measures µ(A) = P∞
k=1 µ(Bk) and
µ(An) = Σk=1,…,n µ(Bk).
From the theorem in real analysis that any monotonic
sequence of real numbers converges (recall that we admit +∞ as limits’ value)
we have
µ(A) = Σ, k=1,…,∞ µ(Bk)
= limn→∞ Σk=1,…,n µ(B_k) = limn→∞ µ(A_n).
The last
statement can be shown similarly

Question 42

Exercise 12.13 Let a measure µ on N be defined by
µ(A) = 0 for finite A and
µ(A) = ∞ for infinite A
additive? σ-additive?

Answer 42

Check that µ is additive but not σ-additive. Therefore
give an example that µ does not satisfy 12.12 3)

If (An) is a sequence in R, with A1 ⊆ A2 ⊆ A3 ⊆ · · ·. Then
limn→∞ µ(An) = µ (∪nAn).

Question 43

12.2. Extension of Measures

Answer 43

From now on we consider only finite measures, an
extension to σ-finite measures will be done later.

Question 44

Prop 12.14 extending measure on S to R(S)

Answer 44

Any measure µ′ on a semiring S is uniquely extended to a
measure µ on the generated ring R(S).
If the initial measure was
σ-additive, then the extension is σ-additive as well.

Question 45

Proof: Prop 12.14 extending measure on S to R(S)

Answer 45

Written notes detailed

e.g. we might extend the measure on [a,b]= b-a

Say we have the semi ring and we construct the minimal ring generated by S R(S)

A in R(S) means we can always represent as a not necessarily unique but any collection of union of disjoint sets can be used to represent elements of the ring

but there is a unique way to extend the measure and keep sigma additivity if there are multiple ways to represent we make C_jk = intersections of the A_j and B_k
some are empty some not
the intersections form building blocks for A_j and A_l and other intersections make building blocks for B_j and B_k

Question 46

infinite sums of series

Answer 46

usually the sum of the sum of infinite series orderings cannot be swapped, e.g if we have negative terms -1+1/2 -1/3+1/4+….. not absolutely convergent but reordering terms may produce a sum

but as measures are positive then for a positive series convergence implies absolute convergence
absolute converges implies we can arbitrarily reorder terms

Question 47

**Def 12.15: Outer Measure

**

Answer 47

Let S be a semi-ring of subsets in X, and µ be a measure
defined on S.

An outer measure µ∗ on X is a map µ∗: 2^X → [0, ∞] defined by

µ∗(A) = inf [ Σ_k
µ(Ak), such that A ⊆ ∪_k A_k, Ak ∈ S ]

(a covering of A with A_k then the outer measure of A is smaller than the outer measure of the covering

(imagine a set dotted about, we form a covering and measure this instead

may not be additive e.g reals and outer measure on the irrationals might be 0 ) but a way to estimate/measure for any arbitrary set

Question 48

Is an outer measure a measure?

Answer 48

an outer measure may be not a measure in the sense of Defn. 12.6 due to a lack of additivity

outer measures dont have to be measures
but measures are outer measures

Question 49

The Lebesgue outer measure on R example 12.17

Answer 49

R is defined out of the measure
from ( µ([a, b)) = b − a)

or A ⊆ R, as
µ∗(A) = inf{

Σ from j =1 to ∞ of
[(bj − aj)]
: A ⊆ ∪ [aj, bj) (j=1 to ∞ for union)

}

We make this definition, as intuitively, the “length”, or measure, of the interval
[a, b) is (b − a)

For example, for outer Lebesgue measure we have µ∗(A) = 0 for any countable set, which follows, as clearly µ∗({x}) = 0 for any x ∈ R

Question 50

outer Lebesgue measure we have µ∗(A) = 0

Answer 50

or example, for outer Lebesgue measure we have µ∗(A) = 0 for any countable set, which follows, as clearly µ∗({x}) = 0 for any x ∈ R

Question 51

Lemma 12.18 µ∗([a, b])

Answer 51

Let a < b. Then
µ∗([a, b]) = b − a

Proof: For ϵ > 0, as [a, b] ⊆ [a, b + ϵ), we have that µ
∗
([a, b]) ⩽ (b − a) + ϵ. As
ϵ > 0, was arbitrary, µ
∗
([a, b]) ⩽ b − a.
To show the opposite inequality we observe that [a, b) ⊂ [a, b] and µ
∗
[a, b) =
b − a (because [a, b) is in the semi-ring) so µ
∗
[a, b] ⩾ b − a by 12.16(ii).

Question 52

symmetric difference of sets.

Answer 52

A △ B = (A ∪ B) \ (A ∩ B

Question 53

Def 12.19 Lebesgue measurable

Answer 53

Given an outer measure µ∗ defined by a measure µ on a semiring S, we define A ⊆ X to be Lebesgue measurable if for any ε > 0 there is a finite union B of elements in S (in other words: B ∈ R(S) by Lem. 12.5(iii)),
such that µ∗(A △ B) < ε.

Question 54

µ∗(A △ B) < ε.

Answer 54

used for lebesgue measurable set

symmetric difference

not common points :in union but not in intersection

means
can be approximated by other elements,
as this difference is as small as you like, as close as you like imagine estimating the area
these arent far from elements of the ring, which are sigma addiitive

Question 55

Straight away which elements are measurable S R(S)?

Answer 55

obviously all elements of S and R(S) are measurable.
R(S) is used as a sense space to approx. A

1)approximate elements A in R(S) with A themselves as they are in the ring!
the outer measure for A would equal the measure for A in this case

To show this wes need to use properties to show

µ(A) ≤ µ(A) due to inf defn
and
µ(A) ≤ µ(A) taking semiring decomposition into smaller parts

Question 56

elements in L: our algebra of measurable sets

Answer 56

R(S) and S elements within are lebesgue measurable as can be approx by themselves in R(S), in L

unions of elements in the set L also:
As their unions’ symmetric difference can be as small as you like

their outer measure is not additive, but we can use the sum of the symmetric difference for the outer measure as an upper bound (epsilon over 2)

Question 57

using unions
A_1 union A_2)
symmetric diff
(B_1 union B_2)

Answer 57

is a subset of
(A_1△ B_1) union( A_2 △ B_2)

thus outer measure of LHS less than or equal to outer measure RHS
less than our equal to epsilon as both lebesgue measurbable
epsilon/2

(first inequality from us taking complement related to symmetric difference we have containment)

Question 58

example define a function of pairs of elements A, B ∈ L as the
outer measure of the symmetric difference of A and B:
d(A, B) = µ∗(A △ B).

Show that d is a metric on the collection of cosets with respect to the equivalence relation: A ∼ B if d(A, B) = 0

Answer 58

Hint: to show the triangle
inequality use the inclusion:
A △ B ⊆ (A △ C) ∪ (C △ B)

doesnt seperate different sets d(A,B)=0 not held for different sets can’t distinguish

e.g [0,1) and [0,1]

symmetric difference
[0,1}△[0,1]={1}
outer measure of a sing point*=0

as it can be covered by arbitrarily small intervals inf=0
{1} in [1, epsilon)

Question 59

Let a sequence (εn) → 0 be monotonically decreasing. For a Lebesgue
measurable A there exists a sequence (An) ⊂ R(S) such that d(A, An) <
εn for each n. Show that (An) is a Cauchy sequence for the distance
d (12.2).

Answer 59

set is measurable if we can arbitrarily approximate with cauchy sequence of elements from the ring

set is lebesgue measurable if it is arbitrarily close to element from our ring

Question 60

Def 12.21 alt defn of a measurable set Carathéodory measurable

Answer 60

Given an outer measure µ∗
, we define E ⊆ X to be
Carathéodory measurable if
µ∗(A) = µ∗(A ∩ E) + µ∗(A \ E),

for any A ⊆ X

we cant cinsider all of P(X) to be measurable st we have sigma additivity but restrict to sets as above

Question 61

Carathéodory measurable equivalent:

Answer 61

As µ∗ is sub-additive, this is equivalent to

µ∗(A) ⩾ µ∗(A ∩ E) + µ∗(A \ E) (A ⊆ X),

as the other inequality is automatic.

Question 62

Lebesgue and Caratheodory

Answer 62

Exercise*:
Show that measurability by Lebesgue and Carathéodory are equivalent.

Question 63

THM 12.23 Lebesgue- supposing the ring R(S) is an algebra (that it contains the maximal element X) then the outer measure of any set is finite and…

Answer 63

(Lebesgue). Let** µ∗ be an outer measure** on** X** defined by a semiring S, and let L be the collection of all Lebesgue measurable sets for µ∗
. Then L is a σ-algebra, and if µ˜ is the restriction of µ ∗ to L, then µ˜ is a measure.

Furthermore,
µ˜ is σ-additive on L if µ is σ-additive on S.

Question 64

Proof theorem 12.23 (Lebesgue)

Answer 64

Sketch of proof notes

Diagram A₁ A₂⊂ L and approx these by elements of the ring (fuzzy picture of approximations B₁ and B₂
e.g.
(A₁∪A₂ ) △ (B₁∪B₂) ⊂ (A₁△B₁) ∪(A₂△B₂)
of x in LHS then either x∈ (A₁∪A₂) or x∈ (B₁∪B₂) only one.
Apply the outer measure which is monotonic and sub additive but not additive: µ( (A₁∪A₂ ) △ (B₁∪B₂)) ≤µ(A₁△B₁) + µ*(A₂△B₂) < ε/2 + ε/2

applied to general number k

Proof:
* R(S)⊂ L.
* Now show µ∗(A) = µ(A) for a set
A ∈ R(S)….
*Now we will show that R(S) with the distance d(A, B) = µ∗(A △ B) is an incomplete metric space with the measure µ being uniformly continuous functions….
*We can check that measurable sets form an algebra,,,
see proof in notes
lemma 12.24 used to show additivity and that additive on L, checking then countably additivity…

Question 65

Lemma 12.24 µ∗(A △ B)

Answer 65

|µ∗(A) − µ∗(B)| ⩽ µ∗(A △ B), that is µ∗ is uniformly continuous in the metric d(A, B)=µ∗(A △ B)

Proof: uses inclusions A ⊂ B∪(A △ B) and B ⊂ A∪(A △ B).

µ* can be extended by continuity to L, it is not additive but becomes thus

Question 66

Corollary 12.25. Lebesgue measurable when?

Answer 66

Let E ⊆ Reals be open or closed. Then E is Lebesgue measurable.

EVERY SUCH OPEN OR CLOSED SET OF THE REALS IS LEBESGUE MEASURABLE

( This is a common trick, using the density and the countability of the rationals. As σ-algebras are closed under taking complements, we need only show that open sets are Lebesgue measurable.
Intervals (a, b) are Lebesgue measurable by the very definition. Now let U ⊆ R be open. For each x ∈ U, there exists a_x < b_x with x ∈ (a_x, b_x) ⊆ U. By making a_x slightly larger, and bx slightly smaller, we can ensure that a_x, b_x ∈ Q. Thus
U = ∪_x(a_x, b_x). Each interval is measurable, and there are at most a countable number of them (endpoints make a countable set) thus U is the countable (or finite) union of Lebesgue measurable sets, and hence U is Lebesgue measurable itself.

Question 67

EVERY SUCH OPEN OR CLOSED SET OF THE REALS IS LEBESGUE MEASURABLE

Answer 67

Approximate open intervals with half open:
(c,d)
[C,d)
symmetric difference {c}
outer measure of {c}=0
(as outer measure is the imfinum of all coverings of the symmetric difference)

inf{ measure([C,C+epsilon}) =
inf epsilon =0

Every open set is a union of open balls centred at every point in the set
if this union is countable we will have L as a sigma algebra, countable unions which are measurable thus so is their union

Question 68

N Q R COUNTABLE?

Answer 68

Important to revise: countable unions of set might be referenced

N countable:
bijection with itself

Q countable dense subject of R
represent E= unio _alpha (a_alpha, B_alpha)

R uncountable

Question 69

REVISE TOO:
if x, y are countable then x X y

Answer 69

if x, y are countable then X X y (carteisan product)

= {(x,y) | x in X, y in Y}

countable as well

(Z x N is countable as well
Q= {m/n s.t m in Z and n in N}

Question 70

countable union of measurable sets

Answer 70

is measurable also

Question 71

Extending finite measure to σ-finite one:

Answer 71

Let
µ be a σ-additive and σ-finite measure defined on a semiring
in X = ⊔_k X_k, s.t. the restriction of µ to every X_k is finite.

Consider the Lebesgue extension µ_k of µ defined
within X_k.

A set A ⊂ X is measurable if every intersection A∩X_k is µ_k measurable.
For a such measurable set A we define its measure by the identity:
µ(A) = Σ_k µ_k(A ∩ X_k).

Question 72

measure µ defined on L complete if

Answer 72

a measure µ defined on L complete if whenever E ⊆ X is such that there
exists F ∈ L with µ(F) = 0 and E ⊆ F, we have that E ∈ L.

Measures constructed
from outer measures by the above theorem are always complete thm 12.23 Lebesgue

Question 73

more about null sets

Answer 73

On the example
sheet, we saw how to form a complete measure from a given measure. We call sets
like E null sets: complete measures are useful, because it is helpful to be able to
say that null sets are in our σ-algebra. Null sets can be quite complicated. For the
Lebesgue measure, all countable subsets of R are null, but then so is the Cantor set,
which is uncountable.

Question 74

Def 12.26 almost everywhere

Answer 74

If we have a property P(x) which is true except possibly
x ∈ A and µ(A) = 0, we say P(x) is almost everywhere or a.e.

Question 75

Cantors third set

Answer 75

Deletes middle third, 0 and end points of intervals will remain. n=1 deletes 1/3 n=2 deletes 2/9 n=3 deletes 4/27. Deletes a countable union of open sets- so deletes open set and this is measurable.
deleting: measures
(1/3) + (1/3)(2/3) + (1/3)(4/9) +….. = (1/3)/((1-(2/3)) = 1
(geometric series infinity aq^r from 0 = 1/(1-q)

Thus cantors third set
Lebesque measure=0 µ({x})=0
(lebsegue measure of a single poimt is 0, covered by small as you like intervals)

A=∪k=1 to ∞{a_k}
Then µ(A)= µ(∪k=1 to ∞{a_k}) =
Σk=1 ∞ µ({a_k})=0

by sigma additivity

so any countable set of points has measure 0
measure of the entire interval is 1
the measure of the cantor set =0
uncountable set here has measure 0
(turns out cantor points have bijection with binary decimals

Question 76

µ(Q∩[0,1])

Answer 76

µ(Q∩[0,1]) = 0 but Q is dense
thus size and dense not related and cantors set also shows this

Question 77

µ([0,b])

Answer 77

[0,b]= [0,a)∪([a,b]

µ([0,b])= µ([0,a))+ µ(([a,b])

Question 78

Consider the semiring S of intervals [a, b).
There is a simple description of all measures on it

Answer 78

For a measure µ define
F_µ(t) =
{µ([0, t)) if t > 0,
{0 if t = 0,
{−µ([t, 0)) if t < 0,

F_µ is monotonic and any monotonic function F defines a measure µ on S by the
µ([a, b)) = F(b) − F(a).

The correspondence is one-to-one with the additional
assumption F(0) = 0.

Question 79

Consider the semiring of half open intervals
S- {[a,b)]
measure of it:

Answer 79

µ([a,b)}=b-a
we have an extension of usual length to all measurable sets:

0<a<b

[0,b) = [0,a) union [a,b}
µ([0,b))= µ[[0,a))+ µ([a,b))
F(b)=F(a)+ µ([a,b))

µ([a,b))= F(b)-F(a)
we require psitive, thus F monotonic due to monotonicity of measure

Question 80

Thm 12.37 for
F_µ

Answer 80

The above measure µ is σ-additive on S if and only if F_µ is continuous from the left: F(t − 0) = F(t) for all t ∈ R.

—funct may have jumps but value at point a is attached to LHS
–we know additive but prove sigma additive

Proof: Diagram continuous from the left means discontinuity at point where function is defined coinciding with the left tail at that discontinuity
F(t − 0)= lim (ε>0 and ε→0)
of F(t-ε)

Intersection of all ∩[a- ε,a]=Ø

*F(t) − F(t − 0) = limε→0 µ([t − ε, t)) = µ(limε→0[t − ε, t)) = µ(∅) = 0, by the continuity of a σ-additive measure

assume [a, b) = ⊔k[ak, bk).
inequality inequality µ([a, b)) ⩾
k µ([ak, bk)) follows from additivity and monotonicity

opposite inequality: y take δk s.t. F(b) − F(b − δ) < ε and F(ak) − F(ak − δk) < ε/(2^k)
(use left continuity of F). Then the interval [a, b − δ] is covered by (ak −δk, bk), due to compactness of [a, b − δ] there is a finite subcovering. Thus
µ([a, b − δ)) ⩽
Σ(j=1 to N) ofµ([akj − δkj, bkj)) and µ([a, b)) ⩽Σ(j=1 to N) of µ([akj, bkj)) + 2ε

Question 81

µ(Q)

Answer 81

in R with lebesgue measure µ(Q)=0
so Q is a null set in R

Question 82

Example Take F(t)=t then the corresponding measure

Answer 82

take F(t) = t, then the corresponding measure is the
Lebesgue measure on R.

Question 83

take F(t) be the integer part of t,

Answer 83

take F(t) be the integer part of t, then µ counts the number of integer
within the set.

Question 84

Cantor function as follows α(x) = 1/2 on (1/3, 2/3); α(x) =
1/4 on (1/9, 2/9); α(x) = 3/4 on (7/9, 8/9), and so

Answer 84

diagram: ladder with different size steps
calc differences between end points=0 as funct is constant

This function
is monotonic and can be continued to [0, 1] by continuity, it is know as
Cantor ladder. The resulting measure has the following properties:
* The measure of the entire interval is 1.
µ_L(C)=0
µ[0,1])=1
complement C’=[0,1]\C
then µ_L(C’)=1 by additivity

• Measure of every point is zero.
• The measure of the Cantor set is 1, while its Lebesgue measure is 0??

µ_c(C’)=µ_c([1/3,2/3]∪[1/9,2/9]∪[7/9,8/9]∪…..)
=Σ_k µ_c ([a_k, b_k)
=Σ_k 0 = 0

counter? measure µ_c = 1
complimentary: any measure on R sum of cont functs?

Question 85

Def 12.40 product measure

Answer 85

.Let X and Y be spaces, and let S and T be semirings on
X and Y respectively. Then S × T is the semiring consisting of
{A × B :
A ∈ S, B ∈ T} (“generalised rectangles”).

Let µ and ν be measures on S
and T respectively. Define the product measure µ × ν on S × T by the rule**
(µ × ν)(A × B) = µ(A)ν(B)**

Question 86

example a measure is the product of 2 copies of pre-lebesgue measures

Answer 86

µ(v)=(b-a)(d-c) for half open rectangles is the product of the measure µ([a, b)) = b − a on semiring S

Question 87

general: what are measures

Answer 87

Measures are generalizations of length, area and volume, but are useful for much more abstract and irregular sets than intervals in {R} or balls in R^3 One might expect to define a generalized measuring function {R} that fulfills the following requirements:

1) an interval [a,b] has measure b-a
2)The measuring function µ is a non-negative extended real-valued function defined for all subsets of R
3)Translation invariance: For any set A and real x, set A +x={a+x: a in A } has the same measure
4) countable additivity for pairwise disjoint