A = {([s,s']x[t,t']) \ ((s]x[t,t'])} \ I(s,s']xIt)l.
(Note that the points (s,t'),(s',t) do not belong to A!) Then
m (A) = m ([s,s']x[t,t']) t,) m (s m ((s,s']x{t})
= f(s',t') f(s_,tD) f(s ,t_) + f(s_,t_) (f(s,t')
f(s_,t') f(s ,t_) + f(s_,t_)) (f(s',t) f(s;,t_)
f(s,t) + f(s,t_))
= f(s',t') f(s,t') f(s',t) + f(s,t),
which is how m (A) was originally defined. The measure of other
rectangles can be computed similarly using the parts already
explicitly given.
3.2 Functions Associated With Measures
In this section we consider the converse problem, namely, given
an E-valued measure m on R2 with finite variation, is it possible to
associate a function with finite variation such that m m in the
sense of Theorem 3.1.1? The following theorem provides a partial
answer to this question.
Theorem 3.2.1. Let m: B(R2) E be a measure with finite variation
Iml. There exists a function f: R2 + E with VarR(f) < on bounded
R