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Let $\mathcal F,\mathcal G$ be $\sigma$-algebras on a set $\Omega$. We define the join and meet of $\mathcal F$ and $\mathcal G$ to be $$\mathcal F\vee\mathcal G:=\sigma(\mathcal F\cup\mathcal G)$$ and $$\mathcal F\wedge\mathcal G:=\mathcal F\cap\mathcal G,$$ respectively.

However, how is $\mathcal F\cap\mathcal G$ defined? Is it $$\{A:A\in\mathcal F\text{ and }A\in\mathcal G\}\tag1$$ or $$\{A\cap B:A\in\mathcal F\text{ and }B\in\mathcal G\}\tag2?$$

In the theory of sets, I think it should be $(1)$. But wouldn't it make much more sense to define $\mathcal F\wedge\mathcal G$ to be $(2)$?

Note that $(2)$ is not a $\sigma$-algebra (or am I missing something?), but the $\sigma$-algebra generated by $(2)$ is actually $\mathcal F\vee\mathcal G$. On the other hand, $(1)$ is itself a $\sigma$-algebra.

0xbadf00d
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1 Answers1

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Notation from lattice theory ...

least upper bound: $\mathcal F\vee\mathcal G$ is the smallest $\sigma$-algebra $\mathcal A$ such that $\mathcal F \subseteq \mathcal A$ and $\mathcal G \subseteq A$.

greatest lower bound: $\mathcal F\wedge\mathcal G$ is the largest $\sigma$-algebra $\mathcal A$ such that $\mathcal A \subseteq \mathcal F$ and $\mathcal A \subseteq G$.

It turns out that the second one is merely the intersection, since the intersection of two $\sigma$-algebras (on the same set) is a $\sigma$-algebra. The first one is not (in general) the union, so it needs more study.

We can do similar things for: subgroups of a given group; topologies on a given set; real-valued functions on a given set; and many more.

GEdgar
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