rectangles R such that m is the measure associated with f by Theorem

3.1.1, i.e., such that for all rectangles R = ((s,t),(s',t')] we have


 m(R) = AR(f) = f(s',t') f(s',t) f(s,t') + f(s,t).


 Proof. Define f: R2 + E by f(s,t) = m((-C,(s,t)])* for

 (s,t) R 2. We show first of all that AR(f) = m(R) for bounded

 rectangles R. We have, denoting R = ((s,t),(s',t')]:


 AR(f) = f(s',t') f(s',t) (f(s,t') f(s,t))


 = m((- (s',t')]) m((-',(s',t)]) [m((-=,(s,t')])


 m((-m,(s,t)])]


 = m((-",(s',t')] \ (-",(s',t)]) m((-=,(s,t')] \ (-C,(s,t)])


 = m({(-<,(s',t')] \ (-=,(s',t)]} \ {(-",(s,t')] \ (-,(s,t)]})


 = m(R).


 We now show that f has finite variation on bounded rectangles,

i.e., that Var (f) < = for bounded rectangles R = [(s,t),(s',t')].

Assume note: suppose there exists R [(s,t),(s',t')] such that

VarR(f) = + =. Denote by R the half-open rectangle ((s,t),(s',t')].

Let o: s = sO < s < ... < s s' be a partition of [s,s'],

T: t = t < tI < ... < tn t' be a partition of [t,t'], and let

P = oXT be the corresponding partition of R (of. Prop. 2.2.5).



* Note: (--,(s,t)] = z e R2: z 6 (s,t)J.