854 APPENDIX A. THE THEORY OF THE RIEMANNN INTEGRAL∗

=(MQ j (gXE)−mQ j (gXE)

)+(MQ j ( f XE)−mQ j ( f XE)

)+

ε

2m

(m

∑j=1

v(Q j)

)−1

.

Therefore, by (1.14),UG ′ (XP)−LG ′ (XP)≤

m

∑j=1

vn (Q j)[(

MQ j (gXE)−mQ j (gXE))+(MQ j ( f XE)−mQ j ( f XE)

)]+

m

∑j=1

v(Q j)ε

2m

(m

∑j=1

v(Q j)

)−1

= UG ( f )−LG ( f )+UG (g)−LG (g)+ε

2

<ε

4+

ε

4+

ε

2= ε.

Since ε > 0 is arbitrary, this proves the theorem. ■

Corollary A.4.9 Suppose f and g are continuous functions defined on E, a contented setin Rn and that g(x)≤ f (x) for all x ∈ E. Then

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

is a contented set in Rn.

Proof: Since E is contented, meaning XE is integrable, it follows from Theorem A.4.6the set of discontinuities of XE has Jordan content 0. But the set of discontinuities of XEis ∂E defined as those points x such that B(x,r) contains points of E and points of EC

for every r > 0. Extend f and g to equal 0 off E. Then the set of discontinuities of theseextended functions still denoted as f ,g is ∂E which has Jordan content 0. This reduces tothe situation of Theorem A.4.8. ■

As an example of how this can be applied, it is obvious a closed interval is a contentedset in R. Therefore, if f ,g are two continuous functions with f (x) ≥ g(x) for x ∈ [a,b], itfollows from the above theorem or its corollary that the set

P1 ≡ {(x,y) : g(x)≤ y≤ f (x)}

is a contented set in R2. Now using the theorem and corollary again, suppose f1 (x,y) ≥g1 (x,y) for (x,y) ∈ P1 and f ,g are continuous. Then the set

P2 ≡ {(x,y,z) : g1 (x,y)≤ z≤ f1 (x,y)}

is a contented set in R3. Clearly you can continue this way obtaining examples of contentedsets. ■

Note that as a special case, it follows that every box is a contented set. Therefore, if Biis a box, functions of the form

m

∑i=1

aiXBi

are integrable. These functions are called step functions.The following theorem is analogous to the fact that in one dimension, when you inte-

grate over a point, the answer is 0.