852 APPENDIX A. THE THEORY OF THE RIEMANNN INTEGRAL∗

It is called the indicator function because it indicates whether x is in E according towhether it equals 1. For a function f ∈ R (Rn) and E a contented set, f XE ∈ R (Rn)by Corollary A.3.2. Then ∫

Ef dV ≡

∫f XEdV.

So what are examples of contented sets?

Theorem A.4.8 Suppose E is a bounded contented set in Rn and f ,g : E → R are twofunctions satisfying f (x) ≥ g(x) for all x ∈ E and f XE and gXE are both in R (Rn).Now define

P≡ {(x,xn+1) : x ∈ E and g(x)≤ xn+1 ≤ f (x)} .

Then P is a contented set in Rn+1.

Proof: Let G be a grid such that for k = f XE ,gXE ,

UG (k)−LG (k)< ε/4. (1.14)

Also let K ≥∑mj=1 vn (Q j) where the Q j are the boxes which intersect E. Let {ai}∞

i=−∞be a

sequence on R, ai < ai+1 for all i, which includes

MQ j ( f XE)+ε

4mK,MQ j ( f XE) ,MQ j (gXE) ,

mQ j ( f XE) ,mQ j (gXE) ,mQ j (gXE)−ε

4mK

for all j = 1, · · · ,m. Now define a grid on Rn+1 as follows.

G ′ ≡ {Q× [ai,ai+1] : Q ∈ G , i ∈ Z}

In words, this grid consists of all possible boxes of the form Q× [ai,ai+1] where Q ∈ Gand ai is a term of the sequence just described. It is necessary to verify that for P ∈ G ′,XP ∈R

(Rn+1

). This is done by showing that UG ′ (XP)−LG ′ (XP)< ε and then noting

that ε > 0 was arbitrary. For G ′ just described, denote by Q′ a box in G ′. Thus Q′ =Q× [ai,ai+1] for some i.

UG ′ (XP)−LG ′ (XP) ≡ ∑Q′∈G ′

(MQ′ (XP)−mQ′ (XP)

)vn+1

(Q′)

=∞

∑i=−∞

m

∑j=1

(MQ′j

(XP)−mQ′j(XP)

)vn (Q j)(ai+1−ai)

and all sums are bounded because the functions f and g are given to be bounded. Therefore,there are no limit considerations needed here. Thus

UG ′ (XP)−LG ′ (XP) =

m

∑j=1

vn (Q j)∞

∑i=−∞

(MQ j×[ai,ai+1] (XP)−mQ j×[ai,ai+1] (XP)

)(ai+1−ai) .

Consider the inside sum with the aid of the following picture.