that d P0 d = d P0 d n j=0 c j Pj d n j=0 c

that d P0 dλ = d P0 d n j=0 c j Pj d n j=0 c j Pj dλ

is also A0-measurable. (ii): Let λ = ∞
j=1 c j Pθ j be equivalent to P. Then pairwise sufficiency of T implies for any θ0 that d Pθ0 /(d Pθ0 + dλ) and hence d Pθ0 /dλ is a measurable function of T .]