be two complete measure spaces. In this section consider the
problem of defining a product measure, μ × ν which is defined on a σ algebra of sets of
X ×Y such that
-----
(μ× ν)
(E × F )
= μ
(E )
ν
(F)
whenever E ∈S and F ∈T . I found the
following approach to product measures in [?] and they say they got it from
[?].
Definition 10.9.1Let ℛ denote the set of countable unions of sets of the form A×B,where A ∈S and B ∈T (Sets of the form A × B are referred to as measurablerectangles) and also let
_{i=1}^{p}. Then
using the same procedure, replace each of
C^
_{k}×
^D
_{k} with finitely many disjoint rectangles
such that none of these intersect A_{p+1}× B_{p+1} while preserving the union of all the
sets involved. The process stops when you have gotten to n. This proves the
lemma.
Lemma 10.9.3If Q = ∪_{i=1}^{∞}A_{i}× B_{i}∈ℛ, then there exist disjoint sets, of the formA_{i}^{′}×B_{i}^{′}such that Q = ∪_{i=1}^{∞}A_{i}^{′}×B_{i}^{′}, each A_{i}^{′}×B_{i}^{′}is a subset of some A_{i}×B_{i}, andA_{i}^{′}∈S while B_{i}^{′}∈T . Also, the intersection of finitely many sets of ℛ is a set of ℛ. Forρ defined in 10.9.31, it follows that 10.9.32and 10.9.33hold for any element of ℛ.Furthermore,
∑ ∑
ρ (Q ) = μ (A ′i) ν(B′i) = ρ(A ′i × B′i).
i i
Proof:Let Q be given as above. Let A_{1}^{′}× B_{1}^{′} = A_{1}× B_{1}. By Lemma 10.9.2, it is
possible to replace A_{2}×B_{2} with finitely many disjoint rectangles,
′ ′
{Ai × B i}
_{i=2}^{m2} such
that none of these rectangles intersect A_{1}^{′}× B_{1}^{′}, each is a subset of A_{2}× B_{2},
and
∞ m2 ′ ′ ∞
∪i=1Ai × Bi = (∪ i=1A i ×B i)∪(∪k=3Ak × Bk)
Now suppose disjoint rectangles,
{A ′i × B′i}
_{i=1}^{mp} have been obtained such that each
rectangle is a subset of A_{k}× B_{k} for some k ≤ p and
∪∞ Ai × Bi = (∪mp A′× B ′)∪ (∪∞ Ak × Bk).
i=1 i=1 i i k=p+1
By Lemma 10.9.2 again, there exist disjoint rectangles
′ ′
{A i × Bi}
_{i=mp+1}^{mp+1} such that
each is contained in A_{p+1}× B_{p+1}, none have intersection with any of
′ ′
{Ai × Bi}
_{i=1}^{mp}
and
∞ ( mp+1 ′ ′) ( ∞ )
∪i=1Ai × Bi = ∪i=1 A i × Bi ∪ ∪k=p+2Ak ×Bk .
Note that no change is made in
{A ′i × B′i}
_{i=1}^{mp}. Continuing this way proves the
existence of the desired sequence of disjoint rectangles, each of which is a subset of at
least one of the original rectangles and such that
Q = ∪∞i=1A′i × Bi′.
It remains to verify x →X_{Q}
(x,y)
is μ measurable for all y and
∫
y → X (x,y)dμ
Q
is ν measurable whenever Q ∈ℛ. Let Q ≡∪_{i=1}^{∞}A_{i}×B_{i}∈ℛ. Then by the first part of
this lemma, there exists
{A ′× B′}
i i
_{i=1}^{∞} such that the sets are disjoint and
∪_{i=1}^{∞}A_{i}^{′}× B_{i}^{′} = Q. Therefore, since the sets are disjoint,
∑∞ ∞∑
XQ (x,y) = XA ′×B ′(x,y) = XA′(x)XB ′(y).
i=1 i i i=1 i i
It follows x →X_{Q}
(x,y)
is measurable. Now by the monotone convergence theorem,
∫ ∫ ∑∞
XQ (x,y)dμ = XA ′i (x)XB ′i (y)dμ
∞ i=1 ∫
= ∑ X ′(y) X ′(x)dμ
i=1 Bi Ai
∞∑
= XB′(y)μ (A ′i) .
i=1 i
It follows y →∫X_{Q}
(x,y)
dμ is measurable and so by the monotone convergence theorem
again,
∫ ∫ ∫ ∞∑ ′
XQ (x,y)dμdν = XB′i (y)μ (A i)dν
i=1∫
∞∑ ′
= XB′i (y)μ (A i)dν
i=∞1
= ∑ ν(B′)μ(A′). (10.9.34)
i=1 i i
This shows the measurability conditions, 10.9.32 and 10.9.33 hold for Q ∈ℛ and also
establishes the formula for ρ
If ∪_{i}A_{i}× B_{i} and ∪_{j}C_{j}× D_{j} are two sets of ℛ, then their intersection is
∪i ∪j (Ai ∩Cj) ×(Bi ∩Dj )
a countable union of measurable rectangles. Thus finite intersections of sets of ℛ are in
ℛ. This proves the lemma.
Now note that from the definition of ℛ if you have a sequence of elements of ℛ then
their union is also in ℛ. The next lemma will enable the definition of an outer
measure.
Lemma 10.9.4Suppose
{Ri}
_{i=1}^{∞}is a sequence of sets of ℛ then
∑∞
ρ (∪ ∞i=1Ri) ≤ ρ(Ri).
i=1
Proof: Let R_{i} = ∪_{j=1}^{∞}A_{j}^{i}×B_{j}^{i}. Using Lemma 10.9.3, let
′ ′
{Am × Bm }
_{m=1}^{∞} be a
sequence of disjoint rectangles each of which is contained in some A_{j}^{i}×B_{j}^{i} for some i,j
such that
∪ ∞i=1Ri = ∪∞m=1A ′m × B′m.
Now define
Si ≡ {m : A ′ ×B ′ ⊆ Ai ×Bi for some j}.
m m j j
It is not important to consider whether some m might be in more than one S_{i}. The
important thing to notice is that
∑
ρ(∪∞i=1Ri) = ρ(A′m × B ′m)
m
∑∞ ∑ ′ ′
≤ ρ (A m ×B m)
i=1m∈Si
∑∞ ′ ′ ∞∑
≤ ρ(∪m ∈SiA m × Bm) ≤ ρ(Ri).
i=1 i=1
This proves the lemma.
So far, there is no measure and no σ algebra. However, the next step is to define
an outer measure which will lead to a measure on a σ algebra of measurable
sets from the Caratheodory procedure. When this is done, it will be shown
that this measure can be computed using ρ which implies the important Fubini
theorem.
Now it is possible to define an outer measure.
Definition 10.9.5For S ⊆ X × Y, define
(μ×-ν)(S) ≡ inf{ρ(R ) : S ⊆ R, R ∈ ℛ }. (10.9.35)
(10.9.35)
The following proposition is interesting but is not needed in the development which
follows. It gives a different description of
-----
(μ× ν)
.
Proposition 10.9.6
-----
(μ × ν)
(S)
= inf
∑ ∞
{ i=1μ (Ai)ν (Bi) : S ⊆ ∪∞i=1Ai × Bi}
Proof:Let λ
(S)
≡ inf
∑ ∞ ∞
{ i=1μ (Ai)ν(Bi) : S ⊆ ∪i=1Ai × Bi}
. Suppose
S ⊆∪_{i=1}^{∞}A_{i}× B_{i}≡ Q ∈ℛ. Then by Lemma 10.9.3, Q = ∪_{i}A_{i}^{′}× B_{i}^{′} where these
rectangles are disjoint. Thus by this lemma, ρ
(Q )
= ∑_{i=1}^{∞}μ
′
(Ai)
ν
′
(B i)
≥ λ
(S)
and so
λ
(S )
≤
-----
(μ × ν)
(S)
. If λ
(S )
= ∞, this shows λ
(S)
=
-----
(μ× ν)
(S )
. Suppose then that
λ
(S )
< ∞ and λ
(S )
+ ε >∑_{i=1}^{∞}μ
(Ai)
ν
(Bi)
where Q = ∪_{i=1}^{∞}A_{i}×B_{i}⊇ S. Then by
Lemma 10.9.3 again, ∪_{i=1}^{∞}A_{i}× B_{i} = ∪_{i=1}^{∞}A_{i}^{′}× B_{i}^{′} where the primed rectangles are
disjoint, each is a subset of some A_{i}× B_{i} and so