Kenneth Kuttler

A.5. ITERATED INTEGRALS 857

and therefore, the only if part of the equivalence is obvious.Conversely, let G be a grid such that (1.17) holds with ε replaced with ε

2 . It is necessaryto show that there is a grid such that (1.17) holds with no primes on the Q. Let F be arefinement of G obtained by adding the points α i

k +ηk where ηk ≤ η and is also chosenso small that for each i = 1, · · · ,n,

αik +ηk < α

ik+1.

You only need to have ηk > 0 for the finitely many boxes of G which intersect the boundedset where f is not zero. Then for

Q≡n

∏i=1

[α

iki,α i

ki+1]∈ G ,

Let

Q̂≡n

∏i=1

[α

iki+ηki

,α iki+1]

and denote by Ĝ the collection of these smaller boxes. For each set Q in G there is thesmaller set Q̂ along with n boxes, Bk,k = 1, · · · ,n, one of whose sides is of length ηk andthe remainder of whose sides are shorter than the diameter of Q such that the set Q is theunion of Q̂ and these sets Bk. Now suppose f equals zero off the ball B

(0,R

). Then

without loss of generality, you may assume the diameter of every box in G which hasnonempty intersection with B(0,R) is smaller than R

3 . (If this is not so, simply refine Gto make it so, such a refinement leaving (1.17) valid because refinements do not increasethe difference between upper and lower sums in this context either.) Suppose there are Psets of G contained in B(0,R) (So these are the only sets of G which could have nonemptyintersection with the set where f is nonzero.) and suppose that for all x, | f (x)| < C/2.Then

∑Q∈F

(MQ ( f )−mQ ( f ))v(Q)≤ ∑Q̂∈Ĝ

(MQ̂ ( f )−mQ̂ ( f )

)v(Q)

+ ∑Q∈F\Ĝ

(MQ ( f )−mQ ( f ))v(Q)

The first term on the right of the inequality in the above is no larger than ε/2 becauseMQ̂ ( f )−mQ̂ ( f )≤MQ′ ( f )−mQ′ ( f ) for each Q. Therefore, the above is dominated by

≤ ε/2+CPnRn−1η < ε

whenever η is small enough. Since ε is arbitrary, f ∈R (Rn) as claimed. ■

A.5 Iterated IntegralsTo evaluate an n dimensional Riemannn integral, one uses iterated integrals. Formally, aniterated integral is defined as follows. For f a function defined on Rn+m,

y→ f (x,y)

A.5. ITERATED INTEGRALS 857and therefore, the only if part of the equivalence is obvious.Conversely, let Y be a grid such that (1.17) holds with € replaced with 5. It is necessaryto show that there is a grid such that (1.17) holds with no primes on the Q. Let ¥ bearefinement of ¥ obtained by adding the points ati, +1, where 1; < 1 and is also chosenso small that for each i = 1,--- ,n,Oe +N, < Oey.You only need to have 7, > 0 for the finitely many boxes of Y which intersect the boundedset where f is not zero. Then forno=]][ Oj,» Uh, 41 EY,i=eeLet3O= J] lou, +5, 0441]i=land denote by G the collection of these smaller boxes. For each set Q in & there is thesmaller set Q along with n boxes, By,k = 1,--- ,n, one of whose sides is of length n;, andthe remainder of whose sides are shorter than the diameter of Q such that the set Q is theunion of Q and these sets B,. Now suppose f equals zero off the ball B (0,%). Thenwithout loss of generality, you may assume the diameter of every box in Y which hasnonempty intersection with B(0,R) is smaller than & . (If this is not so, simply refine ¥to make it so, such a refinement leaving (1.17) valid “because refinements do not increasethe difference between upper and lower sums in this context either.) Suppose there are Psets of Y contained in B(0,R) (So these are the only sets of Y which could have nonemptyintersection with the set where f is nonzero.) and suppose that for all a, | f(x)| < C/2.ThenY (Mo(f)—mo(f))v(@) < Y (Mg (A) —ma(f)) ¥()OCF 0G+ yp (Mo (f) —mo(F)) v(Q)QEF\GThe first term on the right of the inequality in the above is no larger than €/2 becauseMo (Sf) — m6 (f) < Ma (f) — mq (f) for each Q. Therefore, the above is dominated by<e/2+CPnR"'n <ewhenever 1) is small enough. Since € is arbitrary, f € & (IR") as claimed. HiA.5 Iterated IntegralsTo evaluate an n dimensional Riemannn integral, one uses iterated integrals. Formally, aniterated integral is defined as follows. For f a function defined on R"*”,y— f(x,y)