Measure theory class notes - 2 August 2010, class 1
1
Introduction
We would like to define length for subsets of R. Is it possible to do this for all subsets of R? We
want length to have the following properties:
• The length of any interval (a, b] (with a ≤ b) is b − a.
• The length of an at most countable disjoint union of sets is the sum of the individual lengths.
Unfortunately we cannot define a length function on 2R satisfying these properties (proof omitted).
So we do the best that we can.
By default, intervals will mean intervals in R, and (a, ∞] will mean (a, ∞), [−∞, a) will mean
(−∞, a), and so on. If we want to talk of intervals in R̄ = R ∪ {∞, −∞}, we will use a R̄ subscript,
like [0, ∞]R̄ = [0, ∞] ∪ {∞}.
Intervals, semi-fields
Let S be the collection of all intervals (a, b], where −∞ ≤ a ≤ b ≤ ∞. This is a possible starting
point for defining lengths. What properties does it have?
Definition. Let Ω be a set. A semi-field over Ω is a subset S of 2Ω satisfying the following:
• Ω∈S
• A, B ∈ S =⇒ A ∩ B ∈ S
• A ∈ S =⇒ Ω \ A is a finite disjoint union of elements of S
S as defined above is a semi-feld over R. In fact, the complement of any element of S can be
expressed as a disjoint union of at most two elements of S .
Definition. A measure on a semi-field S over Ω is a function µ : S → [0, ∞]R̄ such that
• µ(∅) = 0
• If A ⊆ S is an at most countable collection of disjoint sets, and
[ X
µ
µ(A)
A =
S
A ∈ S , then
A∈A
This property is known as countable additivity (on the semi-field).
Theorem. The natural length function on the semi-field of intervals S is a measure.
(proof omitted)
As another example, consider an F : R → [0, 1] which is right continuous and nondecreasing. Take
the same semi-field of intervals, but define the length of (a, b] as F (b) − F (a). This also gives a
measure (proof omitted).
Measure theory class notes - 2 August 2010, class 1
2
The first extension : fields
Definition. Let Ω be a set. A field over Ω is a subset F of 2Ω satisfying the following:
• Ω∈F
• A, B ∈ F =⇒ A ∩ B ∈ F
• A ∈ F =⇒ Ω \ A ∈ F
Theorem. Let S be a semi-field over Ω. Let F be the collection of all finite disjoint unions of
elements of S . Then F is a field over Ω. This is known as the field generated by S .
Proof. By definition of a semi-field, Ω ∈ S . So Ω ∈ F . Let A, B ∈ F . Then we have A ⊆ S , a
S
finite collection of disjoint sets such that A = A. Similarly we have B ⊆ S , a finite collection
S
of disjoint sets such that B = B. By the distributive law,
[
(Ã ∩ B̃)
A∩B =
Ã∈A,B̃∈B
Since A and B are each a collection of disjoint sets, so is à ∩ B̃ : à ∈ A, B̃ ∈ B . By definition
of a semi-field, each of these sets also belongs to S . A ∩ B is a finite disjoint union of elements of
S , and so belongs to F .
Let A ∈ F . We want to show that Ω \ A ∈ F . As above we have A ⊆ S , a finite collection of
disjoint sets whose union is A. By De Morgan’s laws,
\
(Ω \ Ã)
Ω\A=
Ã∈A
By the definition of a semi-field, each Ω \ Ã is a finite disjoint union of elements of S , and so
belongs to F . By repeated application of the previous part, a finite intersection of elements of F
belongs to F . So Ω \ A ∈ F .
Note that by De Morgan’s laws, since a field is closed under complementation and finite intersections, it is also closed under finite unions.
Theorem. Suppose µ is a measure on a semifield S on the set Ω. Let F be the field generated
S
by S . Define µ̃ : F → [0, ∞]R̄ as follows: for A ∈ F , write A = A where A is a finite collection
of disjoint elements of S . Define
X
µ̃(A) =
µ(Ã)
Ã∈A
µ̃ is well-defined and agrees with µ on S . Also, µ̃ is countably additive on the field: If B ⊆ F is
a countable collection of disjoint sets whose union belongs to F , then
[ X
µ̃
B =
µ̃(B̃)
B̃∈B
Measure theory class notes - 2 August 2010, class 1
3
Proof. Suppose we have two ways of expressing A as a finite disjoint union of elements of S ,
S
S
S
A = A = B. For each à ∈ A, we have (since à ⊆ B)
[
à =
(Ã ∩ B̃)
B̃∈B
The sets on the RHS are disjoint and all in S , so
X
µ(Ã) =
µ(Ã ∩ B̃)
B̃∈B
Similarly, for each B̃ ∈ B,
µ(B̃) =
X
µ(Ã ∩ B̃)
Ã∈A
So
X
µ(Ã) =
Ã∈A
XX
µ(Ã ∩ B̃) =
Ã∈A B̃∈B
XX
µ(Ã ∩ B̃) =
B̃∈B Ã∈A
X
µ(B̃)
B̃∈B
This shows that µ̃ is well-defined.
Clearly µ̃ agrees with µ on S .
We now show countable additivity. Let A ⊆ F be an at most countable collection of disjoint sets
whose union A belongs to F .
[
A=
C
C∈A
For each C ∈ A, we have BC , a finite collection of disjoint elements of S whose union is C. Also,
we have B, a finite collection of disjoint elements of S whose union is A.
X
X
µ(E) (for all C ∈ A)
µ(D), µ̃(C) =
µ̃(A) =
E∈BC
D∈B
Take any particular D ∈ B : it is a subset of A =
so D is the disjoint union of the collection
S
C∈A
S
E∈BC
E (and all these E’s are disjoint),
{D ∩ E : E ∈ BC , C ∈ A}
All these sets are in S and µ is countably additive on S , so
X
µ(D ∩ E)
µ(D) =
C∈A,E∈BC
µ̃(A) =
X
µ(D)
D∈B
=
X
X
µ(D ∩ E)
D∈B C∈A,E∈BC
=
X X X
µ(D ∩ E)
(nonnegative values, can rearrange)
C∈A E∈BC D∈B
=
X X
µ(E)
C∈A E∈BC
=
X
µ̃(C)
C∈A
This completes the proof.
(similar argument as before, since E ⊆
[
B = A)