10.12.1 General Theory
Given two finite measure spaces,
there is a way to define a σ
algebra of subsets of X ×Y
, denoted by ℱ×S
and a measure, denoted by μ×ν
on this σ
algebra such that
whenever A ∈ℱ and B ∈S. This is naturally related to the concept of iterated integrals
similar to what is used in calculus to evaluate a multiple integral. The approach is based
on something called a π system, [?].
Definition 10.12.1 Let
be two measure spaces. A
measurable rectangle is a set of the form A × B where A ∈ℱ and B ∈S.
Definition 10.12.2 Let Ω be a set and let K be a collection of subsets of Ω. Then
K is called a π system if ∅,Ω ∈K and whenever A,B ∈K, it follows A ∩ B ∈K.
Obviously an example of a π system is the set of measurable rectangles
The following is the fundamental lemma which shows these π systems are useful. This
lemma is due to Dynkin.
Lemma 10.12.3 Let K be a π system of subsets of Ω, a set. Also let G be a
collection of subsets of Ω which satisfies the following three properties.
- If A ∈G, then AC ∈G
i=1∞ is a sequence of disjoint sets from G then ∪i=1∞Ai ∈G.
Then G⊇ σ
, where σ
is the smallest σ algebra which contains K.
Proof: First note that if
then ∩ℋ yields a collection of sets which also satisfies 1 - 3. Therefore, I will assume in
the argument that G is the smallest collection satisfying 1 - 3. Let A ∈K and
I want to show GA satisfies 1 - 3 because then it must equal G since G is the smallest
collection of subsets of Ω which satisfies 1 - 3. This will give the conclusion that for
A ∈K and B ∈G, A ∩ B ∈G. This information will then be used to show that
if A,B ∈G then A ∩ B ∈G. From this it will follow very easily that G is a
σ algebra which will imply it contains σ
. Now here are the details of the
Since K is given to be a π system, K⊆GA. Property 3 is obvious because if
sequence of disjoint sets in
because A ∩ Bi ∈G and the property 3 of G.
It remains to verify Property 2 so let B ∈GA. I need to verify that BC ∈GA. In other
words, I need to show that A ∩ BC ∈G. However,
Here is why. Since B ∈GA, A ∩ B ∈G and since A ∈K⊆G it follows AC ∈G by
assumption 2. It follows from assumption 3 the union of the disjoint sets, AC and
and then from 2
the complement of their union is in G
. Thus GA
and this implies since G
is the smallest such, that GA ⊇G
. However, GA
constructed as a subset of G
. This proves that for every B ∈G
and A ∈K
, A ∩ B ∈G
Now pick B ∈G
I just proved K⊆GB. The other arguments are identical to show GB satisfies 1 -
3 and is therefore equal to G. This shows that whenever A,B ∈G it follows
A ∩ B ∈G.
This implies G is a σ algebra. To show this, all that is left is to verify G is closed
under countable unions because then it follows G is a σ algebra. Let
. Then let
because finite intersections of sets of G
are in G
. Since the Ai′
are disjoint, it
Therefore, G⊇ σ
and this proves the Lemma.
With this lemma, it is easy to define product measure.
be two finite measure spaces. Define
to be the set of
measurable rectangles, A × B
, A ∈ℱ
and B ∈S
where in the above, part of the requirement is for all integrals to make sense.
Then K⊆G. This is obvious.
Next I want to show that if E ∈G then EC ∈G. Observe XEC = 1 −XE and so
which shows that if E ∈G
, then EC ∈G
Next I want to show G is closed under countable unions of disjoint sets of G. Let
be a sequence of disjoint sets from
the interchanges between the summation and the integral depending on the
monotone convergence theorem. Thus G
is closed with respect to countable disjoint
From Lemma 10.12.3, G⊇ σ
Also the computation in 10.12.55
that on σ
one can define a measure, denoted by
μ × ν
and that for every
E ∈ σ
Now here is Fubini’s theorem.
Theorem 10.12.4 Let f : X ×Y → [0,∞] be measurable with respect to the σ algebra,
just defined and let μ × ν be the product measure of 10.12.56 where μ and ν are
finite measures on
be an increasing sequence of
measurable simple functions
which converges pointwise to
The above equation holds for sn
in place of f
was shown above. The final result follows from passing to the limit and using the
monotone convergence theorem. ■
The symbol, ℱ×S denotes σ
Of course one can generalize right away to measures which are only σ finite.
Theorem 10.12.5 Let f : X ×Y → [0,∞] be measurable with respect to the σ algebra,
just defined and let μ×ν be the product measure of 10.12.56 where μ and ν are σ
finite measures on
Proof: Since the measures are σ finite, there exist increasing sequences of sets,
. Then μ
restricted to Xn
and Y n
respectively are finite. Then from Theorem 10.12.4
Passing to the limit yields
whenever f is as above. In particular, you could take f = XE where E ∈ℱ×S and
Then just as in the proof of Theorem 10.12.4, the conclusion of this theorem is obtained.
This proves the theorem.
It is also useful to note that all the above holds for ∏
i=1nXi in place of X ×Y. You
would simply modify the definition of G in 10.12.54 including all permutations for the
iterated integrals and for K you would use sets of the form ∏
i=1nAi where Ai is
measurable. Everything goes through exactly as above. Thus the following is
Theorem 10.12.6 Let
i=1n be σ finite measure spaces and let ∏
denote the smallest σ algebra which contains the measurable boxes of the form ∏
where Ai ∈ℱi. Then there exists a measure, λ defined on ∏
i=1nℱi such that if
] is ∏
i=1nℱi measurable, and
is any permutation of