- #1

fishturtle1

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- Homework Statement:
- Let ##(X, \mathcal{A})## be a measurable space and ##(A_n)_{n\in\mathbb{N}}## be a strictly increasing sequence of ##\sigma## algebras. Show $$\mathcal{A}_\infty := \bigcup_{n\in\mathbb{N}} A_n$$ is never a ##\sigma## algebra.

- Relevant Equations:
- A sigma algebra on ##X## is a subset of ##\mathcal{P}(X)## that contains the identity, is closed under complements, and closed under countable union.

For all ##n\in\mathbb{N}## we have ##\emptyset \in A_n##. Hence, ##\emptyset \in \mathcal{A}_\infty##. Let ##A \in \mathcal{A}_\infty##. Then ##A \in A_k## for some ##k\in\mathbb{N}##. So ##A^c \in A_k##. Hence, ##A^c \in \mathcal{A}_\infty##. Thus, ##\mathcal{A}_\infty## is closed under complements. So ##\mathcal{A}_\infty## must fail countable union.

We have ##A_1 \subsetneq A_2 \subsetneq A_3 \subsetneq \dots##. Let us define a sequence ##(B_n)_{n\in \mathbb{N}^{\ge 2}}## where ##B_n## is some set in ##A_n## but not in ##A_{n-1}##.

Also, if ##\mathcal{A}_\infty## is closed under countable union, then ##\mathcal{A}_\infty## is closed under countable intersection. So maybe that's how to get a contradiction?

Consider ##\bigcap_{n\ge 2} B_n##. Is this on the right track?...

We have ##A_1 \subsetneq A_2 \subsetneq A_3 \subsetneq \dots##. Let us define a sequence ##(B_n)_{n\in \mathbb{N}^{\ge 2}}## where ##B_n## is some set in ##A_n## but not in ##A_{n-1}##.

Also, if ##\mathcal{A}_\infty## is closed under countable union, then ##\mathcal{A}_\infty## is closed under countable intersection. So maybe that's how to get a contradiction?

Consider ##\bigcap_{n\ge 2} B_n##. Is this on the right track?...