The following proof is from Probability by Davar Khoshnevisan. There is a symbol $\vee$ in the third sentence of the proof. What does this symbol mean, please? There seems no definition about it in the book and I certainly do not recall seeing this symbol in previous studies. Thank you!
Kolmogorov's Zero-One Law. If $\{ X_i\}^\infty_{i=1}$ are independent random variables, then their tail $\sigma$-algebra $\mathscr{T}$ is trivial in the sense that for all $E\in\mathscr{T}$, $P(E)=0$ or $1$. Consequently, any $\mathscr{T}$-measurable random variable is a constant almost surely.
Proof. Our strategy is to prove that every $E\in\mathscr{T}$ is independent of itself, so that $P(E)=P(E\cap E)=P(E)P(E)$. Since $E\in\mathscr{T}$, it follows that $E$ is independent of $\sigma(\{ X_i\}^{n-1}_{i=1})$. Because this is true for each $n$, Lemma 6.15 (iii) ensures that $E$ is independent of $\vee^\infty_{n=1}\sigma(\{ X_i\}^{n-1}_{i=1})$, which is defined to be the smallest $\sigma$-algebra that contains $\cup^\infty_{n=1}\sigma(\{ X_i\}^{n-1}_{i=1})$. In other words, $E$ is independent of all the $X_i$'s, and hence of itself.
The context is not quite set theory, but Boolean algebras, in particular $\sigma$-algebras. The symbol $\lor$ in that context is the supremum of two elements, or any collection of elements when defined.
If we think about these algebras as algebras of sets then this is exactly the union of the two sets, or in the case of a $\sigma$-algebra countably many sets.