measurable. Let An be a sequence of step functions, converging
uniformly to A except on an evanescent set. For each n, we have
u (An) = E(f 2AdXz). Since A is bounded, E(fA dXz) exists, as we
have shown. Moreover,
IE(f 2AdXz E(f 2AdX )I = E(J R2An-A)zdXzI
+ + +
5 E(f2 IAnAl zd lz)
5 (sup IA -Az )IXm 0
2
zcR2
as n m by uniform convergence. Thus E(fAndX ) + E(fA dX ). For
each n, the double integral is equal to iX (An), and Px(An) Nx(A)
(since X has finite variation, which we shall prove in a minute),
hence we have equality in passing to the limit, i.e.,
S(A) = E(R 2A dX ).
Now, we must show that (i) pX is a stochastic measure, and (ii)
uX has finite variation. The first is easy: if M E M is evanescent,
then M(w) 0 for almost all w, hence
S (w)z dX (w) = 0 P-a.s., so E(C(1M) dX ) = 0,
H