A.4. WHICH FUNCTIONS ARE INTEGRABLE? 851

Now let C denote all intersections of the form Q̃∩B(x,rx) such that x ∈ B(0,R) so thatC is an open cover of the compact set B(0,R). Let δ be a Lebesgue number for this opencover of B(0,R) and let F be a refinement of G such that every box in F has diameterless than δ . Now let F1 consist of those boxes of F which have nonempty intersectionwith B(0,R/2). Thus all boxes of F1 are contained in B(0,R) and each one is contained insome set of C. Let CW be those open sets of C, Q̃∩B(x,rx), for which x ∈W . Thus each

of these sets is contained in some ˜̃Q where Q ∈ GW . Let FW be those sets of F1 which aresubsets of some set of CW . Thus

∑Q∈FW

v(Q)< ε. (1.13)

because each Q in FW is contained in a set ˜̃Q described above and the sum of the volumesof these is less than ε by (1.11). Then

UF ( f )−LF ( f ) = ∑Q∈FW

(MQ ( f )−mQ ( f ))v(Q)

+ ∑Q∈F1\FW

(MQ ( f )−mQ ( f ))v(Q) .

If Q ∈F1 \FW , then Q must be a subset of some set of C \CW since it is not in any set ofCW . Say Q⊆ Q̃1∩B(x,rx) where x /∈W . Therefore, from (1.12) and the observation thatx /∈W , it follows ω f (x) = 0 and so

MQ ( f )−mQ ( f )≤ ε.

Therefore, from (1.13) and the estimate on f ,

UF ( f )−LF ( f )≤ ∑Q∈FW

Cv(Q)+ ∑Q∈F1\FW

εv(Q)

≤Cε + ε (2R)n ,

the estimate of the second sum coming from the fact that

B(0,R)⊆n

∏i=1

[−R,R] .

Since ε is arbitrary, this proves the theorem.1 ■

Definition A.4.7 A bounded set E is a Jordan set in Rn, also called a contented set in Rn

if XE ∈R (Rn). The symbol XE means

XE (x) =

{1 if x ∈ E0 if x /∈ E

1In fact one cannot do any better. It can be shown that if a function is Riemann integrable, then it must bethe case that for all ε > 0, (1.10) is satisfied for some grid G . This along with what was just shown is known asLebesgue’s theorem after Lebesgue who discovered it in the early years of the twentieth century. Actually, he alsoinvented a far superior integral which made the Riemann integral which is the topic of this appendix obsolete.