I have a problem with tail events. At the top of page 19 of Stefan Grosskinsky's lecture notes, it is pointed out that $A := \{\omega: \lim_{n\rightarrow\infty}X_n(\omega) \text{ exists}\}$ is a tail event, where $(X_n)_{n \in \mathbb{N}}$ are random variables defined on probably space $(\Omega, \mathcal{A}, \mathbb{P})$, taking values in $(\mathbb{R}, \mathcal{B})$, where $\mathcal{B}$ is the Borel $\sigma$-algebra on $\mathbb{R}$. Refer to bottom of page 18 for the definition of tail events. While this certainly sounds intuitive, I wanted to see if I can show more rigorously how this is true. I need to show that $$\forall M \in \mathbb{N}, \quad A \in \mathcal{T}_M,$$ where $\mathcal{T}_M := \sigma(X_{M+1}^{-1}(\mathcal{B}), X_{M+2}^{-1}(\mathcal{B}), \ldots)$.

I realise the following. \begin{equation*} \begin{split} A &= \{\omega: \exists c \in \mathbb{R}, \forall \epsilon > 0, \exists N \in \mathbb{N}, \forall n > N, X_n(\omega) \in (c-\epsilon, c+\epsilon)\} \\ &= \bigcup_{c \in \mathbb{R}} \bigcap_{\epsilon > 0} \bigcup_{N \in \mathbb{N}} \bigcap_{n > N} X_n^{-1}(c-\epsilon, c+\epsilon). \end{split} \end{equation*} After a long night's struggle I managed to convince myself of the following. I will skim the details for now but please do ask me for more detail if you're interested. Certainly $X_n^{-1} (c-\epsilon, c+\epsilon) \in X_n^{-1}(\mathcal{B})$ for each particular $n \in \mathbb{N}$. So for a particular $N \in \mathbb{N}$, we have $\bigcap_{n>N}X_n^{-1}(c-\epsilon, c+\epsilon) \in \sigma(X_{N+1}^{-1}(\mathcal{B}), X_{N+2}^{-1}(\mathcal{B}), \ldots) := \mathcal{T}_N$, by definition of $\sigma$-algebra (in particular the closure of $\sigma$-algebrae under countable intersections). Notice that $\bigcap_{n>N}X_n^{-1}(c-\epsilon, c+\epsilon)$ tends \emph{up} as $N \rightarrow \infty$. So $$\bigcup_{N\in\mathbb{N}} \bigcap_{n>N}X_n^{-1}(c-\epsilon, c+\epsilon) = \lim_{N \rightarrow \infty} \bigcap_{n>N}X_n^{-1}(c-\epsilon, c+\epsilon),$$ and so it is certainly a member of $\mathcal{T}_M$ for any finite $M \in \mathbb{N}$. So far so good, we have shown that $\forall M \in \mathbb{N}$, $\bigcup_{N\in\mathbb{N}} \bigcap_{n>N}X_n^{-1}(c-\epsilon, c+\epsilon) \in \mathcal{T}_M$. Now we only need to deal with the $\bigcap_{\epsilon>0}$ and the $\bigcup_{c \in \mathbb{R}}$, bearing in mind however that $\sigma$-algebrae are closed under \emph{countable} intersections and unions, and that $\bigcap_{\epsilon>0}$ and $\bigcup_{c \in \mathbb{R}}$ are uncountable union and intersection. I can see that the intersection $\bigcap_{\epsilon > 0}$ can be changed to a countable intersection $\bigcap_{\epsilon \in \mathbb{Q}, \epsilon > 0}$ without changing the set. That is, for a fixed $c \in \mathbb{R}$, we have $$\bigcap_{\epsilon > 0} \bigcup_{N \in \mathbb{N}} \bigcap_{n > N} X_n^{-1}(c-\epsilon, c+\epsilon) = \bigcap_{\epsilon \in \mathbb{Q}, \epsilon > 0} \bigcup_{N \in \mathbb{N}} \bigcap_{n > N} X_n^{-1}(c-\epsilon, c+\epsilon).$$ Again, please ask me for details if you're interested but I won't go into it here. Now comes the key point. First (for short-hand notation) define $$A_c := \bigcap_{\epsilon > 0} \bigcup_{N \in \mathbb{N}} \bigcap_{n > N} X_n^{-1}(c-\epsilon, c+\epsilon)$$ for each $c \in \mathbb{R}$. I can't turn the \emph{uncountable} union $\bigcup_{c \in \mathbb{R}}$ into a \emph{countable} one. That is, it seems to me that $$\bigcup_{c \in \mathbb{R}} A_c \neq \bigcup_{c \in \mathbb{Q}} A_c.$$ For example, think about the event that $X_n$ tend to an irrational number. So it seems to me that, at least from this analysis, we cannot say that $A := \{\omega: \lim_{n\rightarrow\infty}X_n(\omega) \text{ exists}\}$ is a member of $\mathcal{T}_M$ (or indeed the tail $\sigma$-algebra $\mathcal{T}$). At most we can say that $\{\omega: \lim_{n\rightarrow\infty}X_n(\omega) \text{ exists in }\mathbb{Q}\}$ is a member of $\mathcal{T}_M$ for every $M \in \mathbb{N}$ (and so is in $\mathcal{T}$).

Am I wrong? Is my analysis wrong? Please feel free to ask me to expand on any part of this. This is a long and complicated (tedious rather) question I know. I'd really really appreciate any help. Thank you very much for your time!


1 Answer 1


Since $\mathbb{R}$ is complete, the set of convergent sequences is exactly the set auf Cauchy sequences and one can formalize what it means that $(X_n(\omega))$ is a Cauchy sequence with countably many operations. You can find the details here. Now whether a sequence converges or not does not depend on any finite terms, so the set is measurable for all $\mathcal{T}_M$ and hence for $\mathcal{T}$.

Given that, the set of sequences that converge to an irrational number is a tail event too. You can let for each $q\in\mathbb{Q}$ take the event $B_q$ that the sample sequence converges to $q$. You have already shown this to be measurable. Then $A\backslash\big(\bigcup_{q\in\mathbb{Q}}B_q\big)$ is the event that the sample realization converges to an irrational number.

  • $\begingroup$ Ah yes of course!!!!! Cauchy!!!! I wanted to give you a vote up but I don't have enough reputation to do that, sorry!!! I'll do that later when I have more reputation. I hope I'll have the privilege of your answers to my future questions. Thank you very much!!! $\endgroup$
    – Ray
    Feb 13, 2014 at 22:25

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