The treatment of Lebesgue measure given above is a special case of something called product
measure. You can take the product measure for any two finite or σ finite measures. A measure
space
(Ω, ℱ,μ)
is called σ finite if there are measurable subsets Ω_{n} such that μ
(Ωn)
< ∞ and
Ω = ∪_{n=1}^{∞}Ω_{n}. σ finite
Given two finite measure spaces,
(X,ℱ, μ)
and
(Y,S,ν)
, 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 defined on
this σ algebra such that
μ × ν(A ×B ) = μ (A)ν(B )
whenever A ∈ℱ and B ∈S.
Definition 22.1.1Let
(X,ℱ, μ)
and
(Y,S,ν)
be two measure spaces. Ameasurable rectangle is a set of the formA × B where A ∈ℱ and B ∈S.
With this lemma, it is easy to define product measure.
Let
(X,ℱ,μ )
and
(Y,S,ν)
be two finite measure spaces. Define K to be the set of
measurable rectangles, A × B, A ∈ℱ and B ∈S. Let
{ ∫ ∫ ∫ ∫ }
G ≡ E ⊆ X × Y : X dμdν = X dνdμ (22.1)
Y X E X Y E
(22.1)
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 E^{C}∈G. Observe X_{EC} = 1 −X_{E} and so
∫ ∫ ∫ ∫
XEC dμdν = (1− XE )dμdν
Y X ∫Y∫X
= (1− XE )dνdμ
∫X∫Y
= X Cdνdμ
X Y E
which shows that if E ∈G, then E^{C}∈G.
Next I want to show G is closed under countable unions of disjoint sets of G. Let
{Ai}
be
a sequence of disjoint sets from G. Then
∫ ∫ ∫ ∫ ∞ ∫ ∞ ∫
X ∞ dμdν = ∑ X dμdν = ∑ X dμdν
Y X ∪i=1Ai Y X i=1 Ai Y i=1 X Ai
∑∞ ∫ ∫ ∑∞ ∫ ∫
= XAidμdν = XAidνdμ
i=1 Y X i=1 X Y
∫ ∑∞ ∫ ∫ ∫ ∑∞
= XAidνdμ = XAidνdμ
∫X i∫=1 Y X Y i=1
∞
= X Y X∪i=1Aidνdμ, (22.2)
the interchanges between the summation and the integral depending on the monotone
convergence theorem. Thus G is closed with respect to countable disjoint unions.
one
can define a measure, denoted by μ × ν and that for every E ∈ σ
(K)
,
∫ ∫ ∫ ∫
(μ× ν)(E ) = XEd μdν = XEd νdμ. (22.3)
Y X X Y
(22.3)
Now here is Fubini’s theorem.
Theorem 22.1.2Let f : X × Y → [0,∞] be measurable with respect to the σalgebra, σ
(K )
just defined and let μ × ν be the product measure of 22.3where μ and ν arefinitemeasures on
(X, ℱ)
and
(Y,S )
respectively. Then
∫ ∫ ∫ ∫ ∫
X×Y fd(μ× ν) = Y X fdμdν = X Y fdνdμ.
Proof: Let
{sn}
be an increasing sequence of σ
(K )
measurable simple functions which
converges pointwise to f. The above equation holds for s_{n} in place of f from what was shown
above. The final result follows from passing to the limit and using the monotone convergence
theorem. ■
The symbol, ℱ×S denotes σ
(K)
.
Of course one can generalize right away to measures which are only σ finite.
Theorem 22.1.3Let f : X ×Y → [0,∞] be measurable with respect to the σ algebra,σ
(K )
just defined and let μ × ν be the product measure of 22.3where μ and ν are σ finitemeasures on
(X, ℱ)
and
(Y,S )
respectively. Then
∫ ∫ ∫ ∫ ∫
fd(μ× ν) = fdμdν = fdνdμ.
X×Y Y X X Y
Proof: Since the measures are σ finite, there exist increasing sequences of sets,
{Xn}
and
{Yn}
such that μ
(Xn )
< ∞ and ν
(Yn)
< ∞. Then μ and ν restricted to X_{n} and Y_{n}
respectively are finite. Then from Theorem 22.1.2,
∫ ∫ ∫ ∫
fdμd ν = fdνdμ
Yn Xn Xn Yn
Passing to the limit yields
∫ ∫ ∫ ∫
fdμd ν = fdνdμ
Y X X Y
whenever f is as above. In particular, you could take f = X_{E} where E ∈ℱ×S and
define
∫ ∫ ∫ ∫
(μ× ν)(E ) ≡ X dμdν = X dνdμ.
Y X E X Y E
Then just as in the proof of Theorem 22.1.2, the conclusion of this theorem is obtained.
■
It is also useful to note that all the above holds for ∏_{i=1}^{n}X_{i} in place of X × Y. You
would simply modify the definition of G in 22.1 including all permutations for the
iterated integrals and for K you would use sets of the form ∏_{i=1}^{n}A_{i} where A_{i}
is measurable. Everything goes through exactly as above. Thus the following is
obtained.
Theorem 22.1.4Let
{(Xi,ℱi,μi)}
_{i=1}^{n}be σ finite measure spaces and let∏_{i=1}^{n}ℱ_{i}denote the smallest σ algebra which contains the measurable boxes of the form∏_{i=1}^{n}A_{i}whereA_{i}∈ℱ_{i}. Then there exists a measure λ defined on∏_{i=1}^{n}ℱ_{i}such that if f : ∏_{i=1}^{n}X_{i}→ [0,∞] is∏_{i=1}^{n}ℱ_{i}measurable, and