i.e., 1x(M) = 0. As for (ii), let M E M, (Mi) i = 1,2,...,n disjoint
n1
sets form M, with UMi C M. We have then IZ bX(MI) [ =
1 =1-1
n n n
I E( 2 1M )dX dx) E(f 2(1M zdlXz) =E( Z 2 zd X z )
i=1 R I i=1 R I i=1 R i
E(f 2 (U M zdlX)) since all sections are disjoint = E(f (1M) zdII ).
The last integral is independent of the family (Mi), so by taking
supremum we get luvx(M) S E(f 2M) zdlX|z) 5 E(IX|I) < =, so pX has
finite variation (in particular IpX is bounded by E(IXI[)).
Uniqueness of p is evident: this completes one half of the
correspondence. Next we prove the converse.
2) Let p be a stochastic measure with finite variation p|l.
Assume, first, pO (then Mfl = p). We will associate an increasing
process X (increasing in both senses) satisfying (4.4.1); then the
final result is an easy consequence of the decomposition of measures
with finite variation.
2
For each bounded r.v. Y and u R consider the raw process
Yu(w) Y(w)-I,](z). The map A defined by A (Y) = p(Yu)
z Luuj u u
p(Y-I[ ,u) is a bounded, positive measure on (2,F). In fact: for
B E F, A (B) p(l -1 ou ) = p(Bx[O,u]) S u(nx[0,u]). Also, u is P-
u B [0,u] u
absolutely continuous; if P(B) = 0, then A (B) = p(Bx[O,u]) = 0 since
p is a stochastic measure. Then A has a density au with respect to
u u