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How do we know a sigma-algebra is closed under countable intersections?

A sigma-algebra is closed under countable intersections because any countable intersection can be rewritten in terms of complements and countable unions using De Morgan's law. Since a sigma-algebra is already closed under complements and countable unions, it must also be closed under countable intersections.

Conditions

  • Each Ak∈FA_k \in \mathcal{F}.
  • F\mathcal{F} is closed under complements and countable unions.

Reasoning, step by step

  1. Start with a countable sequence of sets A1,A2,…A_1, A_2, \ldots in F\mathcal{F}.
  2. Take the complement of each set: Akc∈FA_k^c \in \mathcal{F} (closure under complements).
  3. Form the countable union of these complements: ⋃k=1∞Akc∈F\bigcup_{k=1}^{\infty} A_k^c \in \mathcal{F} (closure under countable unions).
  4. Take the complement of that union: (⋃k=1∞Akc)c∈F\left(\bigcup_{k=1}^{\infty} A_k^c\right)^c \in \mathcal{F} (closure under complements).
  5. Apply De Morgan's law: ⋂k=1∞Ak=(⋃k=1∞Akc)c\bigcap_{k=1}^{\infty} A_k = \left(\bigcup_{k=1}^{\infty} A_k^c\right)^c, proving the intersection is in F\mathcal{F}.
  6. Conclude that F\mathcal{F} is closed under countable intersections.

Example

The video explicitly shows the identity ⋂k=1∞Ak=(⋃k=1∞Akc)c\bigcap_{k=1}^{\infty} A_k = \left(\bigcup_{k=1}^{\infty} A_k^c\right)^c to demonstrate this closure property.

Common misconceptions

  • Thinking that countable intersections must be added as a separate axiom to the definition of a sigma-algebra.
  • Confusing finite intersections with countable intersections.

Watch the explanation

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Answers are generated from source material and independently checked. Consult the original video or creator if something is unclear.