# What does the meet of two sigma algebras mean?

I came across this notation of which I am unfamiliar;

$$\mathscr{F}=\mathscr{G}_{1}\vee \mathscr{G}_{1}$$

where $$\mathscr{G}_{1}$$ and $$\mathscr{G}_{2}$$ are both sigma-fields of subsets of $$\Omega$$. It is claimed $$\mathscr{F}$$ is larger than both of $$\mathscr{G}_{1}$$ and $$\mathscr{G}_{2}$$ suggesting to me that

$$\mathscr{F}=\mathscr{G}_{1}\vee \mathscr{G}_{1}=\mathscr{G}_{1}\cup \mathscr{G}_{1}$$

but I know this last union is not always a sigma-field so perhaps this is not the meaning here?

Staying on this subject, would the notation $$\mathscr{G}_{1}\subset\mathscr{G}_{2}$$ mean the same as $$\mathscr{G}_{1}\leq\mathscr{G}_{2}$$, the latter statement which (I assume anyway) means $$\mathscr{G}_{2}$$ is finer than $$\mathscr{G}_{1}$$?

For some reason I cannot seem to find a clear definition of this for sigma-fields - sets, partitions etc yes, but not sigma fields - for example here;

https://math.stackexchange.com/questions/1345598/does-meet-of-two-partitions-of-a-set-always-exist

Any help as ever appreciated.

• Offhand, I would expect this to be defined within the context of a lattice, as explained at Wikipedia. The natural lattice in this context would be that of all sigma-fields defined on $\Omega;$ the (partial) ordering would be that inherited from the power set $\mathcal{P}(\Omega).$ In other words, the meet would be the coarsest sigma-field of which both components are subfields. That's not usually their union--it will be bigger. It would help to know the context in which you came across this notation in case some other lattice is understood.
– whuber
Commented Dec 11, 2018 at 23:22
• Thank you @whuber. As I responded to Jonas I believe his answer is correct given the context - I think you are describing the same thing in the sense that $\sigma(\mathcal{G}_{1}\cup \mathcal{G}_{2})$ is this coarsest sigma-field you mention? The context I found this in is "coarsening at random" - essentially where loss of information occurs with censoring or missing values in repeated measures. In this context $\mathcal{G}_{1}$ represents the sigma-field generated by the indicator variable denoting if a random variable is observed, and $\mathcal{G}_{2}$ is for the random variable itself. Commented Dec 12, 2018 at 8:28
• Yes, he is describing the same thing in a more basic way. Commented Dec 12, 2018 at 8:30

Indeed, $$\mathcal{G_1} \cup \mathcal{G_2}$$ is in general no $$\sigma$$-field. Therefore, one may set $$\mathcal{G_1} \vee \mathcal{G_2} = \sigma( \mathcal{G_1} \cup \mathcal{G_2}),$$ where $$\sigma(C)$$ denotes the $$\sigma$$-field that is generated by the set system $$C \subseteq 2^\Omega$$. This function is defined by $$\sigma(C) = \bigcap \left\lbrace \mathcal{G} : \mathcal{G} \supseteq C, \mathcal{G} \text{ is \sigma-field on \Omega}\right\rbrace.$$
• Thank you @Jonas for your answer - this makes sense to me given the context I encountered the notation. So essentially $\mathcal{F}=\mathcal{G}_{1}\vee \mathcal{G}_{2}$ is the smallest sigma-field containing $\mathcal{G}_{1}\cup \mathcal{G}_{2}$ in the sense that if $\mathcal{F}'$ is another sigma-field $\mathcal{G}_{1}\cup \mathcal{G}_{2}\subset\mathcal{F}'$ then $\mathcal{F}'\subset\mathcal{F}$? Commented Dec 12, 2018 at 8:14
• In the last line, I would say $\mathcal{F}' \supseteq \mathcal{F}$, since $\mathcal{F}$ is indeed the smallest $\sigma$-field that contains $\mathcal{G_1}\cup \mathcal{G_2}$. Commented Dec 12, 2018 at 8:26