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Existence of a locally finite Poisson point process

Statement

Assume countable choice and dependent choice. Let S be locally compact, second countable, and Hausdorff, and let ν be a locally finite, sigma-finite Borel measure on S. Then a locally finite Poisson point process with intensity ν exists. This existence assertion uses countable choice and dependent choice only through the countable product of probability spaces; it does not assume full choice.

Facts & Assumptions

Given: A locally compact, second-countable Hausdorff space S and a locally finite, sigma-finite Borel measure ν on S; sigma-finite means that S is a countable union of Borel sets of finite ν-measure (Finite, sigma-finite, and semifinite measures).

[F1]

There is an increasing compact exhaustion (Kn) covering S, with every compact subset contained in some Kn. (Second-countable locally compact Hausdorff spaces admit compact exhaustions)

[F2]

For each finite Borel measure ρ, the explicit mixed-binomial construction on ∐n≥0Sn gives a Poisson point process of intensity ρ, and finite disjoint counts have the independent Poisson laws. (Finite-intensity Poisson-process existence and mixed-binomial law)

[F3]

The countable product of the coordinate probability spaces exists under countable choice and dependent choice. (Assuming countable and dependent choice, countable products of arbitrary probability spaces)

[F4]

A countable sum of measures is a measure; increasing limits pass through nonnegative integrals. (Nonnegative scalar multiples and countable weighted sums of measures are measures, Monotone convergence for the integral, Measures on sigma-algebras)

[F5]

The scalar Poisson law has mass e−aak/k!, probability generating function ea(s−1), and is normalized for a≥0. (The scalar Poisson count law, Normalization, moments, and generating function of the Poisson count law)

[A1]

Countable choice is assumed only to invoke [F3]. (The Axiom of Countable Choice (ACω))

Proof

technique · direct
1.1givenF1F4F6construct

Put K0=∅, A1=K1, and Ai=Ki∖Ki−1 for i≥2. By [F6], each Ki is closed, so the disjoint sets Ai are Borel and cover S. Since Ai⊆Ki, countable additivity of the measure gives ν(Ai)≤ν(Ki)<∞, hence νi(B):=ν(B∩Ai) is finite. If i>m, then Ai∩Km=∅.

2.1givenstep 1.1F2F3A1A2construct

For each i, let Ωi be the explicit finite-intensity mixed-binomial space from [F2] for νi (a one-point space when νi=0), with law Pi and coordinate counting measure ηi. The countable product theorem [F3], under [A1]-[A2], supplies a probability space on which all these coordinates are independent. Each ηi is supported on Ai almost surely: its count on S∖Ai is Poisson with mean zero by [F2], hence zero almost surely.

3.1givenF1F2F4step 1.1step 2.1construct

Let E=⋂i≥1{ηi(S∖Ai)=0}. Each event in this intersection is measurable by the coordinate evaluation maps and has probability one by step 2.1; countable subadditivity applied to their null complements gives P(E)=1. For Borel B, define η(B)=∑i≥1ηi(B) on E and η(B)=0 off E. On E the countable sum is a measure by [F4], while off E it is the zero measure; its evaluations are measurable as the restriction to the measurable event E of increasing limits of finite sums. On E it is a counting measure because the coordinate construction lists finitely many point locations per cell and the countable union of those lists is countable; off E the empty configuration is a counting measure. On E, if C is compact, [F1] gives C⊆Km for some m; then η(C)=∑i=1mηi(C)<∞, because later cells avoid Km and each of the first m coordinates contains finitely many points. Off E all compact counts are zero. Thus η is a measurable point process with locally finite realizations everywhere, and agrees almost surely with the unmodified superposition, preserving all its count laws.

3.2givenF2F3F5A1A2step 2.1algebra

Fix pairwise-disjoint Borel sets B1,…,Br and write Nij=ηi(Bj) and aij=νi(Bj). By [F2], within each row i the Nij are independent Poisson variables of means aij; rows are independent by [F3] and [A1]-[A2], so all finitely many variables in the first n rows are independent. For independent X∼Pois⁡(a) and Y∼Pois⁡(b), direct convolution gives P(X+Y=k)=e−(a+b)∑l=0kalbk−l/(l!(k−l)!)=e−(a+b)(a+b)k/k! by the binomial theorem. Iterating this identity shows that Nj(n)=∑i=1nNij are independent across j with laws Pois⁡(aj(n)), where aj(n)=∑i=1naij.

4.1step 3.1step 3.2F4F5algebra

On the probability-one event E from step 3.1, the counts Nj(n) increase to Nj=η(Bj) and aj(n)↑aj:=ν(Bj). For finite aj, the events {Nj(n)≤kj for every j} decrease to {Nj≤kj for every j}; continuity from above follows by applying [F4] to their complementary increasing indicators. Their probabilities, computed from step 3.2, converge to ∏j∑k=0kje−ajajk/k!. Inclusion-exclusion over the finitely many lower rectangles then gives the product of Poisson point masses for every count vector, proving independence and the desired laws when all aj<∞.

5.1step 3.2step 4.1F5algebra

If aj=∞, then for each integer M≥0, continuity from above gives P(Nj≤M)=lim⁡ne−aj(n)∑k=0M(aj(n))k/k!=0. The limit is zero because e−ttk→0 for each fixed k, as follows from et≥tk+1/(k+1)!. Hence Nj=∞ almost surely. Applying the finite-coordinate argument of step 4.1 to the coordinates with finite aj, while the infinite-mean coordinates are almost surely constant at +∞, gives the required extended independent-count convention for the whole family.

6.1step 2.1step 3.1step 3.2step 4.1step 5.1A1A2∎

Steps 3.1, 3.2, 4.1, and 5.1 prove local finiteness and the independent Poisson count laws. The only choice principles used are [A1]-[A2] for the countable product in step 2.1.

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