It is easy to give a generalization to yield a theorem for the completion of product spaces. To
do this, here is a useful little theorem which is by now not surprising.
Theorem 22.2.1Let
(Ω,ℱ, μ)
be a complete measure space and let f ≤ g ≤ hbe functions having values in
[0,∞ ]
. Suppose also that f
(ω)
= h
(ω)
a.e. ω and thatf and h are measurable. Then g is also measurable. If
(Ω,G,λ)
is the completion ofa σ finite measure space
(Ω,ℱ, μ)
as described above in Theorem 9.1.4, then if f ismeasurable with respect to Ghaving values in
[0,∞ ]
, it follows there exist functionsh, g measurable with respectto ℱ, h ≤ f ≤ g, such that g = h a.e. Also if f hascomplex values and is G measurable, there exist h,g which are ℱ measurable such that
|h|
≤
|f|
≤
|g|
and h = f = g a.e.
Proof: Consider the first claim. g = f + X_{}
[g>f]
(g− f)
. f is given to be measurable. The
second term on the right is also measurable because
[ ]
X[g>f](g − f) > α
= Ω if α < 0 while if
α ≥ 0, the set is contained in
[h − f > 0]
, which is a measurable set of measure
zero.
Now consider the last assertion. By Theorem 7.1.6 on Page 445 there exists an increasing
sequence of nonnegative simple functions,
{sn}
measurable with respect to G which converges
pointwise to f. Letting
m∑n
sn(ω) = cnkXEnk (ω ) (22.4)
k=1
(22.4)
be one of these simple functions, it follows from Theorem 9.1.4, there exist sets, F_{k}^{n}∈ℱ such
that F_{k}^{n}⊆ E_{k}^{n}⊆ G_{k}^{n} and μ
Hence b_{n}≤ s_{n}≤ t_{n}, and t_{n} = b_{n} a.e. Let
g (ω ) ≡ lim nsu→p∞ tn(ω)
and let
h(ω) = lim nin→f∞ bn(ω).
Then h,g are ℱ measurable, h ≤ f ≤ g, and both h,g equal f off N ≡∪_{n}
[tn − bn > 0]
.
Because off this set of measure zero, both functions equal s_{n} which converges to
f.
The last claim follows from considering the positive and negative parts of real and
imaginary parts. ■
Theorem 22.2.2Let
{(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 if
(i1,⋅⋅⋅,in)
is any permutation of
(1,⋅⋅⋅,n )
,then
∫ ∫ ∫
fdλ = ⋅⋅⋅ fdμi1 ⋅⋅⋅dμin
Xin Xi1
Let
(∏n ∏n------)
i=1 Xi, i=1ℱi,λ
denote the completion of this product measure space andlet
∏n
f : Xi → [0,∞ ]
i=1
be ∏_{i=1}^{n}ℱ_{i}measurable. Then there exists N ∈∏_{i=1}^{n}ℱ_{i}such that λ
(N )
= 0 and anonnegative function, f_{1}measurable with respect to∏_{i=1}^{n}ℱ_{i}such that f_{1} = f off N and if
(i1,⋅⋅⋅,in)
is any permutation of
(1,⋅⋅⋅,n)
, then
∫ ∫ ∫
fdλ = ⋅⋅⋅ f1dμi ⋅⋅⋅dμi .
Xin Xi1 1 n
Furthermore, f_{1}may be chosen to satisfy either f_{1}≤ f or f_{1}≥ f.
Proof: This follows immediately from Theorem 22.1.4 and Theorem 22.2.1.
By the second theorem, there exists a function f_{1}≥ f such that f_{1} = f for all
(x1,⋅⋅⋅,xn )
∕∈
N, a set of ∏_{i=1}^{n}ℱ_{i} having measure zero. Then by Theorem 9.1.4 and
Theorem 22.1.4