The $\sigma$-algebra generated by multiple random variables

measure-theoryprobabilityprobability theoryrandom variablesterminology

The Measure Theory textbook that I am studying from (by Donald Cohn) presents the following definition of the tail $\sigma$-algebra:

Let $X_n$ be a sequence of random variables and define $\tau _n =\sigma(X_{n+1}, X_{n+2}, …)$, then we call $\tau := \cap _{n=1}^{\infty}\tau _n$ the tail $\sigma$-algebra of $X_n$ which contain events that only depend on the limiting behaviour of the sequence.

My confusion comes by the way that we define $\tau _n$. The textbook defines the sigma algebras generated by a set and the sigma algebra generated by a function, however, this is an infinite sequence of random variables.

Since random variables are functions, I am tempted to extrapolate something from the definition of the sigma algebra generated by a function (which I know would be fine if it was just one random variable). Although, the fact that we have more than one makes me hesitant to do this.

My idea was that since $$\sigma(X_i) = \{X_i^{-1}(A) : A \in \mathbb{B}(\mathbb{R}) \}$$perhaps we have the following:

$$ \sigma(X_{n+1}, X_{n+2}, . . .) = \{(X_{n+1}^{-1}(A), X_{n+2}^{-1}(A) . . . ) : A \in \mathbb{R} \}$$

However, with this being a guess (at best), I'm looking for some clarification here.


To give some additional background, this is a precursor to a discussion on Kolmogorov's Zero-One Law. However, I haven't been able to understand this without this issue being clarified first.

Best Answer

$ \sigma(X_{n+1}, X_{n+2}, . . .) = \{(X_{n+1}^{-1}(A), X_{n+2}^{-1}(A) . . . ) : A \in \mathbb{R} \}$ is not correct. The right hand side is not a sigma algebra.

$ \sigma(X_{n+1}, X_{n+2}, . . .)$ is the smallest sigma algebra with respect to which $X_i$ is measurable for $i=n+1,n+2,....$. Finite intersections of sets of the type $X_i^{-1}(E)$ where $E$ is a Borel set in $\mathbb R$ and $i>n$ form an algebra and the sigma algebra generated by this algebra equals $ \sigma(X_{n+1}, X_{n+2}, . . .)$.