CONTENTS

Solution: Intersection of Sigma-Algebras

Exercise: Intersection of Sigma-Algebras

Part 1: G∩H\mathcal{G} \cap \mathcal{H} is a σ\sigma-algebra

Let K=G∩H\mathcal{K} = \mathcal{G} \cap \mathcal{H}. Verify each axiom.

Axiom 1. Ω∈G\Omega \in \mathcal{G} (because G\mathcal{G} is a σ\sigma-algebra) and Ω∈H\Omega \in \mathcal{H} similarly. Hence Ω∈G∩H=K\Omega \in \mathcal{G} \cap \mathcal{H} = \mathcal{K}.
Axiom 2. Suppose A∈KA \in \mathcal{K}. Then A∈GA \in \mathcal{G}, so Ac∈GA^c \in \mathcal{G}. Also A∈HA \in \mathcal{H}, so Ac∈HA^c \in \mathcal{H}. Combining: Ac∈G∩H=KA^c \in \mathcal{G} \cap \mathcal{H} = \mathcal{K}.
Axiom 3. Suppose A1,A2,…∈KA_1, A_2, \ldots \in \mathcal{K}. Then each An∈GA_n \in \mathcal{G}, so ⋃nAn∈G\bigcup_n A_n \in \mathcal{G}. Each An∈HA_n \in \mathcal{H}, so ⋃nAn∈H\bigcup_n A_n \in \mathcal{H}. Hence ⋃nAn∈K\bigcup_n A_n \in \mathcal{K}.

All three axioms hold, so K\mathcal{K} is a σ\sigma-algebra. □\square

Remark. The same proof shows that the intersection of any family of σ\sigma-algebras (even an uncountable family) is a σ\sigma-algebra — this is exactly what makes the definition σ(C)=⋂{F:C⊆F}\sigma(\mathcal{C}) = \bigcap \{\mathcal{F} : \mathcal{C} \subseteq \mathcal{F}\} sensible.

Part 2: G∪H\mathcal{G} \cup \mathcal{H} can fail

Take Ω={1,2,3,4}\Omega = \{1, 2, 3, 4\} and:

G={∅, {1,2}, {3,4}, Ω}\mathcal{G} = \{\emptyset,\, \{1, 2\},\, \{3, 4\},\, \Omega\} H={∅, {1,3}, {2,4}, Ω}\mathcal{H} = \{\emptyset,\, \{1, 3\},\, \{2, 4\},\, \Omega\}

Each is a σ\sigma-algebra on Ω\Omega (direct check: closed under complements — {1,2}c={3,4}\{1,2\}^c = \{3,4\} — and unions).

G∪H={∅, {1,2}, {3,4}, {1,3}, {2,4}, Ω}\mathcal{G} \cup \mathcal{H} = \{\emptyset,\, \{1,2\},\, \{3,4\},\, \{1,3\},\, \{2,4\},\, \Omega\}
Check axiom 3. The union {1,2}∪{1,3}={1,2,3}\{1, 2\} \cup \{1, 3\} = \{1, 2, 3\} is not in G∪H\mathcal{G} \cup \mathcal{H}. So G∪H\mathcal{G} \cup \mathcal{H} is not a σ\sigma-algebra — the failing axiom is closure under unions, and {1,2,3}\{1, 2, 3\} is the missing set. (Several other unions also fail to land in G∪H\mathcal{G} \cup \mathcal{H}; any one of them suffices as a witness.)

Part 3: the generated σ\sigma-algebra

To compute σ(G∪H)\sigma(\mathcal{G} \cup \mathcal{H}), close the collection under unions, intersections, and complements:

  • Intersections: {1,2}∩{1,3}={1}\{1,2\} \cap \{1,3\} = \{1\}, {1,2}∩{2,4}={2}\{1,2\} \cap \{2,4\} = \{2\}, {3,4}∩{1,3}={3}\{3,4\} \cap \{1,3\} = \{3\}, {3,4}∩{2,4}={4}\{3,4\} \cap \{2,4\} = \{4\}.

So every singleton {1},{2},{3},{4}\{1\}, \{2\}, \{3\}, \{4\} is in σ(G∪H)\sigma(\mathcal{G} \cup \mathcal{H}). As in the previous exercise, isolating all singletons on a finite Ω\Omega forces the generated σ\sigma-algebra to be the full power set:

σ(G∪H)=2Ω(all 16 subsets).\sigma(\mathcal{G} \cup \mathcal{H}) = 2^\Omega \quad \text{(all $16$ subsets).}
Interpretation. G\mathcal{G} answers the question "is the outcome in {1,2}\{1, 2\}?" — it tells you a coarse binary split of Ω\Omega. H\mathcal{H} answers the orthogonal question "is the outcome in {1,3}\{1, 3\}?". Individually, each gives a two-way partition. Together they give the four-way partition into singletons — complete information.
This is the discrete version of what happens continuously: two independent sources of information (e.g. the path of Wt(1)W_t^{(1)} and the path of Wt(2)W_t^{(2)}) generate σ\sigma-algebras whose union needs closure to become a valid σ\sigma-algebra — the join σ(Ft(1)∪Ft(2))\sigma(\mathcal{F}_t^{(1)} \cup \mathcal{F}_t^{(2)}) captures the combined information.

Takeaways

  • Intersections preserve the σ\sigma-algebra property, unions do not. This asymmetry is the reason the "smallest σ\sigma-algebra containing C\mathcal{C}" is defined via intersection but information combination requires the generated σ\sigma-algebra.
  • Combining information is not set-union. Two observers with different information pool their knowledge via the join σ(G∪H)\sigma(\mathcal{G} \cup \mathcal{H}), not the raw union.
  • Orthogonal binary partitions can generate the full power set. Two well-chosen two-element binary splits of a four-point space already contain all the information.
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Solution — Intersection of Sigma-Algebras | q4quant.studio