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I have been reading J L Doob Stochastic Processes. I was going through the supplement section and found this theorem:

$\Omega$ is the space of points $\omega$. Let $\mathcal{F}_o$ be a field of $\omega$ sets. Then, if $\mathcal{G}$ is class of $\omega$ sets such that $\mathbb{B}(\mathcal{F}_o) \subset \mathcal{G}$ (Where, $\mathbb{B}(\mathcal{F}_o)$ is Borel/$\sigma-$field of $\mathcal{F}_o$ sets). $(i) \mathcal{F}_o \subset \mathcal{G}$, and $(ii)$ if $\Lambda_j \in \mathcal{G},\ j \ge 1$ and if either $\Lambda_1 \subset \Lambda_2 \subset \Lambda_3 \cdots, \ \bigcup_1^\infty \Lambda_j = \Lambda$ or $\Lambda_1 \supset \Lambda_2 \cdots, \ \bigcap_1^\infty \Lambda_j = \Lambda$ then $\Lambda \in \mathcal{G}$.

The item $(i)$ is easy to see. The second one has left me clueless. If the $\Lambda_j \in \mathcal{F}_o$ or $\mathbb{B}(\mathcal{F}_o)$ it is fairly straight forward since $\mathbb{B}(\mathcal{F}_o) \subset \mathcal{G}$. If $\Lambda_j \notin \mathcal{F}_o$ or $\mathbb{B}(\mathcal{F}_o)$, not sure how to proceed. Understandably, $\mathcal{G}$ is one of the many sets that contains $\mathbb{B}(\mathcal{F}_o)$. Not sure how I go about showing the result. Any help would be appreciated and thanks in advance.

On thinking some more about the problem, I seem to have a counter example. Let $X$ be a set of Integers. Then $\mathcal{F}_o = \{\emptyset, X\}$ is a field. Let $A_n = \{\text{set of all even numbers including 2n}\}$. Let $\mathcal{G} = \{ \emptyset, X\} \bigcup \{A_n : n \in \mathbb{N} \}$. Now $\mathcal{G}$ is a counter example. Since $\bigcup_1^n A_n$ is the set of all even integers and does not belong to $\mathcal{G}$. Can someone confirm my conclusion, and explain what I am missing from the original theorem.

Doob Theorem 1.2

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    I don't understand what $\Bbb B(\mathcal F_0)$ is supposed to be: is $\mathcal{F}_0$ already a $\sigma$-field or just a field? – 2017-01-23
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    @Cett $\mathcal{F}_o$ is just a Field, $\mathbb{B}(\mathcal{F}_o)$ is the sigma-field of $\mathcal{F}_o$ sets. In essence it is the $\sigma$-field generated by $\mathcal{F}_o$. – 2017-01-23
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    (i) should read $\mathcal F_0\subset\mathcal G$, not your $\mathcal F_0\in\mathcal G$. In (ii), are you sure the hypothesis is not that $\Lambda_j\in\mathcal F_0$ instead of your $\Lambda_j\in\mathcal G$? – 2017-02-23
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    @Did I have made the correction you were right it should be a subset and not belong to. I actually took an image and posted it. I would love to understand how to prove the theorem. – 2017-02-24
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    You are simply taking some hypotheses of the theorem for its conclusion and vice versa. The hypotheses are that $\mathcal F_0$ is a field, that $\mathcal F_0\subset\mathcal G$, and that (i) and (ii) hold. The conclusion is that $\mathcal B(\mathcal F_0)\subset\mathcal G$. – 2017-02-24
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    @Did Are you serious? That would make sense. But that is not the way it has been stated though. I would have actually put $\mathcal{B({F_o})}$ at the end. – 2017-02-24
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    @Did Are you serious? That would make sense. But that is not the way it has been stated though (would you agree?). I interpreted it as if $\mathcal{B({F_o})} \subset \mathcal{G}$ then show that $\Lambda \in \mathcal{G}$, and hence my counter example. I would have expected the statement to read: Let $\mathcal{F}_o$ be a field. Let $\mathcal{G}$ be a class of $\omega$ sets such that (i) and (ii) are true implies $\Lambda \in \mathcal{G}$, then $\mathcal{B({F}_o)} \subset \mathcal{G}$. I wasted hours on it, Boy do I feel Stupid. – 2017-02-24
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    This (the version in my comment) is exactly how the theorem is stated, yes. I just had to read. – 2017-02-24
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    @Did, You mean the theorem is stated correctly? – 2017-02-24
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    Yes it is. $ $ $ $ – 2017-02-24
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    @Did Fair enough.. I should have thought a bit more I guess. Thanks I appreciate the help. Had I interpreted it right, the proof is fairly trivial to show. – 2017-02-24

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