In section 1.8 of Sidney Resnick's Adventures in Stochastic Processes, he writes
In general, suppose we have a probability space $(\Omega, \mathcal A, P)$, and an increasing family of $\sigma$-fields $\mathcal F_n,n\geqslant0$; i.e., $\mathcal F_n\subset F_{n+1}$. Define $$ \mathcal F_{\infty} = \bigvee_{n\geqslant 0}\mathcal F_n := \sigma\left(\bigcup_{n=0}^\infty\mathcal F_n\right).$$
What does $\bigvee_{n\geqslant 0}\mathcal F_n$ denote? Is it just shorthand for the expression on the right?
Usually, $\vee$ denotes the maximum; more generally,
$$\bigvee_{n \in \mathbb{N}} f_n = \sup\{f_n; n \in \mathbb{N}\}$$
is the supremum of a sequence of real numbers $(f_n)_{n \in \mathbb{N}}$, i.e. $\bigvee_{n \in \mathbb{N}} f_n$ is the smallest number which is larger or equal than $f_n$ for all $n \in \mathbb{N}$.
Here the situation is quite similar:
$$\bigvee_{n \in \mathbb{N}} \mathcal{F}_n = \sigma \left( \bigcup_{n \in \mathbb{N}} \mathcal{F}_n \right)$$
is the smallest $\sigma$-algebra which contains all $\sigma$-algebras $\mathcal{F}_n$. In this sense, we can see $\bigvee_{n \in \mathbb{N}} \mathcal{F}_n$ as a supremum.