Extreme Points in the Balayage Order

August 12, 1997

Fix μ with σ-finite potential and let A denote the convex cone

{ν : νU ≤ μU}.

Using the integral representation theorem of [1], we give a short proof of the following result of Mokobodzki. Let A_{ex} denote the set of extreme points of A.

Theorem. A_{ex} = {μH_{B} : B ∈ E^{e}, B finely closed }.

Proof. (a) Let ν be an extreme point of A. By the main result of [1] there is a decreasing family {B(r) : 0 ≤ r ≤ 1} of finely closed E^{e}-measurable sets such that

ν = ∫_{0}^{1} μH_{B(r)} dr.

The extremality of ν and the monotonicity of r ↦ B(r) imply that r ↦ H_{B(r)} is constant on [0, 1]. Thus ν = μH_{B(1/2)}.

(b) Fix ν of the form μH_{B}, where B is E^{e}-measurable and finely closed. Suppose there are elements ν_{1} and ν_{2} of A such that ν = αν_{1} + (1 − α)ν_{2}, where 0 < α < 1. By Rost ’s theorem there are randomized stopping times T_{1} and T_{2} such that

ν_{i} = μP_{T_{i}} i = 1, 2.

Each ν_{i} is carried by B, so

ν_{i} = μP_{T_{i}} = μP_{T_{i}}H_{B} = μP_{D_{i}},

where D_{i} := T_{i} + D_{B}^{∘}θ_{T_{i}}. Thus

μH_{B} = ν = αν_{1} + (1 − α)ν_{2} = αμP_{D_{1}} + (1 − α)μP_{D_{2}},

which forces μH_{B} = μP_{D_{i}} (i = 1, 2) since we clearly have μP_{D_{i}} downstream from μH_{B}. Therefore D_{i} = D_{B}, P^{μ}-a.s. for i = 1, 2. Consequently, ν = μH_{B} = μP_{D_{i}} = ν_{i} (i = 1, 2), so ν is an extreme point of A.

Reference

[1] P.J. Fitzsimmons (1991): Skorokhod embedding by randomized hitting times,” In: Seminar on Stochastic Processes 1990, Birkhauser, Boston, pp. 183-191.