68.3. A MULTIPLE INTEGRAL 2329

This says that when you consider such a function ∑i ciXAi1×···×Ainwith the Ak pairwise

disjoint, then to say that it equals 0 whenever some ti = t j is to say that it is really a sum over

all permutations of n indices taken from {1, · · · ,m} . Thus there are(

mn

)n! = P(m,n)

possible non zero terms in this sum.

Lemma 68.3.6 Consider the set of all ordered lists of n indices from {1,2, · · · ,m} . Thustwo lists are the same if they consist of the same numbers in the same positions. We denoteby i or j such an index, i from {1, · · · ,m} and j from {1, · · · ,q}. Also let

{A1, · · · ,Am} ,{

B1, · · · ,Bq}

are two lists of pairwise disjoint Borel sets from T having finite Lebesgue measure. Alsosuppose

∑i

ciXAi1×···×Ain= ∑

jdjXB j1×···×B jn

Then

∑i

ci

n

∏k=1

W(Aik

)= ∑

jdj

n

∏k=1

W(B jk

)Proof: Suppose that n = 1 first. Then you have

∑i

ciXAi = ∑j

d jXB j (68.3.9)

where the sets {Ai} and{

B j}

are disjoint. Clearly

Ai ⊇ ∪ jAi∩B j (68.3.10)

ConsiderciXAi ,∑

jciXAi∩B j (68.3.11)

If strict inequality holds in 68.3.10, then you must have a point in Ai\ ∪ jAi ∩B j wherethe left side of ∑i ciXAi equals ci but the right side would equal 0. Hence ci = 0 andso ∑ j ciXAi∩B j = 0 which shows that the two expressions in 68.3.11 are equal. If Ai =∪ jAi∩B j, it is also true that the two expressions in 68.3.11 are equal. Thus

∑i

ciXAi = ∑i

∑j

ciXAi∩B j

Similar considerations apply to the right side. Thus

∑i

∑j

ciXAi∩B j = ∑j∑

id jXAi∩B j

∑i, j(ci−d j)XAi∩B j = 0