The intersection of an arbitrary family of Dynkin systems is itself a Dynkin system

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[math]\newcommand{\bigudot}{ \mathchoice{\mathop{\bigcup\mkern-15mu\cdot\mkern8mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}} }[/math][math]\newcommand{\udot}{\cup\mkern-12.5mu\cdot\mkern6.25mu\!}[/math][math]\require{AMScd}\newcommand{\d}[1][]{\mathrm{d}^{#1} }[/math]

Statement

Let [ilmath]\{\mathcal{D}_\alpha\}_{\alpha\in I} [/ilmath] be an arbitrary family of Dynkin systems on a set [ilmath]X[/ilmath], then[1] we claim:

  • [math]\mathcal{D}:\eq\bigcap_{\alpha\in I}\mathcal{D}_\alpha[/math] is itself a Dynkin system on [ilmath]X[/ilmath].

Proof

Recall the definition of a Dynkin system:

Given a set [ilmath]X[/ilmath] and a family of subsets of [ilmath]X[/ilmath], which we shall denote [ilmath]\mathcal{D}\subseteq\mathcal{P}(X)[/ilmath] is a Dynkin system[1] if:

  • [ilmath]X\in\mathcal{D} [/ilmath]
  • For any [ilmath]D\in\mathcal{D} [/ilmath] we have [ilmath]D^c\in\mathcal{D} [/ilmath]
  • For any [ilmath](D_n)_{n=1}^\infty\subseteq\mathcal{D}[/ilmath] is a sequence of pairwise disjoint sets we have [ilmath]\udot_{n=1}^\infty D_n\in\mathcal{D}[/ilmath]

Proof

  • There are 3 properties we must show, we will show them in the order above:
    1. Show that [ilmath]X\in\mathcal{D} [/ilmath]
      • [ilmath]X\in\mathcal{D}\iff\forall\alpha\in I[X\in\mathcal{D}_\alpha][/ilmath] by definition of intersection
        • Well let [ilmath]\alpha\in I[/ilmath] be given
          • [ilmath]X\in\mathcal{D}_\alpha[/ilmath] as [ilmath]\mathcal{D}_\alpha[/ilmath] is itself a Dynkin system on [ilmath]X[/ilmath].
        • Since [ilmath]\alpha\in I[/ilmath] was arbitrary we have shown it for all
      • Thus [ilmath]X\in\mathcal{D} [/ilmath] - as required
    2. Show: [ilmath]\forall D\in\mathcal{D}[X-D\in\mathcal{D}][/ilmath] (recall the set complement of a set in another set can be defined in terms of the relative complement as done here)
      • Let [ilmath]D\in\mathcal{D} [/ilmath] be given, we wish to show [ilmath]X-D\in\mathcal{D} [/ilmath]
        • [ilmath]X-D\in\mathcal{D}\iff\forall\alpha\in I[X-D\in\mathcal{D}_\alpha][/ilmath] by definition of intersection
          • We will show the RHS of this
          • Let [ilmath]\alpha\in I[/ilmath] be given
            • As [ilmath]D\in\mathcal{D} [/ilmath] by definition, we see that [ilmath]\forall\beta\in I[D\in\mathcal{D}_\beta][/ilmath] by definition of intersection again
            • Thus [ilmath]D\in\mathcal{D}_\alpha[/ilmath] for the given [ilmath]\alpha[/ilmath]
              • As [ilmath]D\in\mathcal{D}_\alpha[/ilmath] and [ilmath]\mathcal{D}_\alpha[/ilmath] is a Dynkin system on [ilmath]X[/ilmath]
                • [ilmath]X-D\in\mathcal{D}_\alpha[/ilmath]
          • Since [ilmath]\alpha\in I[/ilmath] was arbitrary we have shown it for all
        • Thus [ilmath]\forall\alpha\in I[X-D\in\mathcal{D}_\alpha][/ilmath]
      • This is the very definition of [ilmath]X-D\in\mathcal{D} [/ilmath] - as required
    3. Let [math](D_n)_{n\in\mathbb{N} }\subseteq\mathcal{D} [/math] be a family of pairwise disjoint sets in [ilmath]\mathcal{D} [/ilmath] - we must show [ilmath]\bigudot_{n\in\mathbb{N} }D_n\in\mathcal{D} [/ilmath]
      • Again [math]\left[\bigudot_{n\in\mathbb{N} }D_n\in\mathcal{D}\right]\iff\left[\forall\alpha\in I\left(\bigudot_{n\in\mathbb{N} }D_n\in\mathcal{D}_\alpha\right)\right][/math] - we will show the RHS
        • Let [ilmath]\alpha\in I[/ilmath] be given
          • By hypothesis [ilmath]\forall n\in\mathbb{N}[D_n\in\mathcal{D}][/ilmath], so:
          • [ilmath]\forall\beta\in I\forall n\in\mathbb{N}[D_n\in\mathcal{D}_\beta][/ilmath]
            • Specifically: let [ilmath]\beta:\eq\alpha[/ilmath], then:
            • [ilmath]\forall n\in\mathbb{N}[D_n\in\mathcal{D}_\beta][/ilmath] - otherwise written as [ilmath](D_n)_{n\in\mathbb{N} }\subseteq\mathcal{D}_\alpha[/ilmath]
              • As [ilmath]\mathcal{D}_\alpha[/ilmath] is itself a Dynkin system:
                • [ilmath]\bigudot_{n\in\mathbb{N} }D_n\in\mathcal{D}_\alpha[/ilmath]
        • Since [ilmath]\alpha\in I[/ilmath] was arbitrary we have shown it for all.
      • This completes the proof

References

  1. 1.0 1.1 Measures, Integrals and Martingales - René L. Schilling