236 CHAPTER 11. ABSTRACT MEASURE AND INTEGRATION

Lemma 11.3.2 If {An} is an increasing sequence in [−∞,∞], then sup{An}= limn→∞ An.

The following lemma is useful also and this is a good place to put it. First{

b j}∞

j=1 isan enumeration of the ai j if

∪∞j=1{

b j}= ∪i, j

{ai j}.

In other words, the countable set,{

ai j}∞

i, j=1 is listed as b1,b2, · · · .

Lemma 11.3.3 Let ai j ≥ 0. Then ∑∞i=1 ∑

∞j=1 ai j = ∑

∞j=1 ∑

∞i=1 ai j. Also if

{b j}∞

j=1 is anyenumeration of the ai j, then ∑

∞j=1 b j = ∑

∞i=1 ∑

∞j=1 ai j.

Proof: First note there is no trouble in defining these sums because the ai j are allnonnegative. If a sum diverges, it only diverges to ∞ and so ∞ is written as the answer.

∞

∑j=1

∞

∑i=1

ai j ≥ supn

∞

∑j=1

n

∑i=1

ai j = supn

limm→∞

m

∑j=1

n

∑i=1

ai j

= supn

limm→∞

n

∑i=1

m

∑j=1

ai j = supn

n

∑i=1

∞

∑j=1

ai j =∞

∑i=1

∞

∑j=1

ai j. (11.3.10)

Interchanging the i and j in the above argument the first part of the lemma is proved.Finally, note that for all p,

p

∑j=1

b j ≤∞

∑i=1

∞

∑j=1

ai j

and so ∑∞j=1 b j ≤ ∑

∞i=1 ∑

∞j=1 ai j. Now let m,n > 1 be given. Then

m

∑i=1

n

∑j=1

ai j ≤p

∑j=1

b j

where p is chosen large enough that{

b1, · · · ,bp}⊇{

ai j : i≤ m and j ≤ n}. Therefore,

since such a p exists for any choice of m,n,it follows that for any m,n,

m

∑i=1

n

∑j=1

ai j ≤∞

∑j=1

b j.

Therefore, taking the limit as n→ ∞,

m

∑i=1

∞

∑j=1

ai j ≤∞

∑j=1

b j

and finally, taking the limit as m→ ∞,

∞

∑i=1

∞

∑j=1

ai j ≤∞

∑j=1

b j

proving the lemma.