Probability Tidbits 3 - Pi Systems

Suppose we have a set S, I is a π-system on S if I is a set of subsets of S satisfying:

I≠∅I1,I2∈I⟹I1∩I2∈I or in other words, the subsets of S in I are closed under intersection.

This simple system is visually intuitive; we can often work with just an underlying π-system of a σ-algebra to prove certain properties about the measure space. I know that’s really vague, so let’s formalize that a bit.

Useful Tool: λ Systems

A λ-system (also called Dynkin system) D is a set of subsets of S satisfying:

S∈DA,B∈D,A⊂B⟹B−A∈D(An)n∈D,∪nAn=A,Ai∩Aj=∅∀i≠j⟹A∈D

This looks really similar to a σ-algebra, but isn’t exactly it. Here, we can only make countable union statements on disjoint sets of the space which is a big difference. We can also swap the third statement of a λ-system with the equivalent statement that:

Gn∈Dlimn→∞Gn=G∈D

This is true by monotone convergence theorem.

σ-algebra and λ,π -systems

A π-system I of some set S extended to a λ-system is denoted λ(I) (or d(I) if we use the Dynkin notation). This λ(I) is the smallest λ-system containing I. A set Σ is both a π-system and a λ-system iff it’s a σ-algebra.

If a set Σ is a σ-algebra, then we have that it’s closed under countable intersections via DeMorgan’s law:

∪nAn=A∈ΣAc∈Σ⟹(∪nAn)c=∩nAcn∈Σ

This is a stronger statement than what π-system states, so it should be one as well. This set should also be a λ-system because you can just set each An in the countable union to be disjoint.

In the other direction, if a set Σ is both a π-system and a λ-system then we’ll show it’s also a σ-algebra. Specifically, we’d like to show that countable unions of not necessarily disjoint sets (An)n is in the set (since all the other properties are basically inherited from λ-systems). Note that:

A∪B=S−(Ac∩Bc)∈Σ

The complement, set subtraction is inherited from λ-system (and since S is the whole set, that satisfies the subset requirement). The intersection is inherited from π-system. This allows us to union two sets even if they’re not disjoint! To finish, we construct a set Gn:=∪i≤nAi and its limit G∈Σ because it’s a λ-system.

Surprisingly, λ-systems generated from a π-system are π-systems themselves. This result, paired with our previous result, gives us that the generated λ-system contains a σ-algebra.

Suppose we have the set:

D={B∈λ(I):B∩C∈λ(I),∀C∈λ(I)}

Is D a π-system? It should be, since we have by definition that this set is closed under intersection. This set should also be nonempty since S∩C=C,∀C∈λ(I) so the whole set S must belong in D.

Is D a λ-system? It should be as well. We already know from the above result that S∈λ(I). For two sets B⊂A∈D, we have to show A−B∈D. For all C∈λ(I), (A−B)∩C must be in D for A−B∈D. Thankfully,

(A−B)∩C=(A∩C)−(B∩C)(distributivity)A∩C∈λ(I),B∩C∈λ(I)(by definition)A∩C⊂B∩C(intersection preserves subset)⟹(A∩C)−(B∩C)∈λ(I)⟹(A−B)∩C∈D

Finally, for disjoint countable unions of An∈D, each An∩C∈λ(I)∀i,n, so by definition of a λ-system we have (∪nAn)∩C∈λ(I). Now recall λ(I) is defined as the smallest λ-system containing I. Here, D is a subset of λ(I) but it also is a λ-system containing I, so D=λ(I)! This means any π-system extended to a λ-system is also a π-system (since D is that λ-system), and thus a σ-algebra is always contained within this extension.

The Punchline

Why did we mention all this stuff about π and λ-systems? Because two measures that agree on the same π-system will be the same measure on the σ-algebra generated by that π-system!

To be more specific, suppose we have a π-system I on a set S and we have two measures μ1,μ2 on σ(I), the generated σ-algebra of the π-system. If μ1(S)=μ2(S)<∞ and μ1(A)=μ2(A),∀A∈I, then μ1(A)=μ2(A),∀A∈σ(I).

To do this, we need to loop in a λ-system constructed as:

D={A∈σ(I):μ1(A)=μ2(A)}

Let’s show that it is indeed a λ-system. Obviously, S∈σ(I) for any I, and we have that the measures are equal on the whole set. To show that B−A∈D,∀A⊂B∈D, note that

μ1(B−A)=μ1(B)−μ1(A)=μ2(B)−μ2(A)=μ2(B−A)

Since measures of disjoint sets are additive. In fact, measures of disjoint sets are countably additive, which also conveniently proves ∪nAn∈D, if An∈D and disjoint (hint: use monotone convergence theorem again). Since we just proved that D is a λ-system, we know it contains a σ-algebra within it by our previous result. However, D⊂σ(I) by construction, yet it contains σ(I), which means D=σ(I)! This means the measures μ1,μ2 will agree on the generated σ-algebra (which turns out to be D).

Borel σ-algebra

Constructing a π-system is often pretty easy. For reals, we can create a π-system as:

π(R)={(−∞,x]:x∈R}

Compare this to B, the borel σ-algebra generated by the topology on reals, this is much easier to reason with. With the previous result, we now know that if two probability measures agree on π(R) they will also agree on the generated σ(π(R))=σ(R)=B.