# Is $\{w:\lim_{n \rightarrow \infty}S_n/n \in A\}$ in tail $\sigma$-field?

Suppose $\{X_n\}$ are independent random variables on $(\Omega, \mathcal{F},P)$. Set $S_n:=\sum_{i=1}^{n}X_i$.

Suppose $\frac{1}{n}S_n \rightarrow Y \;a.s.$ for some real-valued random variable $Y$. How to show that for any set $A \subset R$ $$E:=\{w:Y \in A\} \in \mathcal{T}?$$ Here $\mathcal{T}$ represents the tail $\sigma$-field.

The trouble I meet:

If for any $t \in N^{+}$, $$\{\lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i}{n} \in A\} = \{\lim_{n \rightarrow \infty}\frac{\sum_{i=t}^{n}X_i}{n} \in A\},$$ we actually finished the proof.

My puzzle about above equation: It's clear that the left hand side implies the right hand side. But I think the right hand side does not necessarily implies the left hand side. What if $X_j(w) = \infty$ for some $j < t$?

I didn't use the fact that $Y$ is a real-valued random variable. I think it may be the key. But How to use it?

• Hint: If $X_k(w) = \pm\infty$ for some $k$, then, for all $n>k$, if $\frac{\sum_{i=1}^{n}X_i(w)}{n}$ is defined, we have $\frac{\sum_{i=1}^{n}X_i}{n}=\pm\infty$. But $Y= \lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i}{n}$ a.s., and $Y$ is a real-valued random variable, so, for all $k\in \mathbb{N}$, $X_k$ is finite a.s.. So, since the countable union of zero measure set is a zero measure set, there a set $D\subset \Omega$ such that $P(D)=0$ and for all $w\in \Omega - D$ and all $k\in \mathbb{N}$, we have $X_k(w)\neq \pm\infty$. – Ramiro Dec 4 '15 at 11:45
• What is $R$ in "$A\in R$"? – Jack Dec 6 '15 at 22:02
• @Jack I guess "$A \in R$" is meant to be $A\subseteq \mathbb{R}$. – Ramiro Dec 10 '15 at 10:26

Case 1: For all $n\in \mathbb{N}^+$, $X_n$ are real-valued random variables.

In this case, for all $n\in \mathbb{N}^+$ and for all $w\in \Omega$, $X_n(w)\neq \pm\infty$.

So, we have that, for any $A\subseteq \mathbb{R}$ and for any $t\in \mathbb{N}^+$

$$\left\{w : \lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i(w)}{n} \in A\right\}=\left\{w : \lim_{n \rightarrow \infty}\frac{\sum_{i=t}^{n}X_i(w)}{n} \in A\right\}$$ and we can easily conclude that $$\left\{w : \lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i(w)}{n} \in A\right\}\in \mathcal{T} \:\:\:\:\:\:\:\: (*)$$ where $\mathcal{T}$ is the tail $\sigma$-field.

Remark for case 1: On the other hand, since $\lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i}{n} \rightarrow Y \;a.s.$, we have that $$P\left(\left\{w : \lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i(w)}{n} \in A\right\}\:\Delta\:\{w:Y(w) \in A\}\right)=0$$ But, this information, even combined with $(*)$, is NOT enough to prove that that $\{w:Y(w) \in A\} \in \mathcal{T}$.

Case 2: The random variables $X_n$ may have infinity values ($\pm\infty$).

In this case, the result is false, as shown in the following counterexemple.

Let, for each $n\in \mathbb{N}^+$, let $\Omega_n=\{0,1\}$, $\mathcal{F}_n=2^{\Omega_n}$ and $P_n$ the probability defined on $\mathcal{F}_n$ by $P_n({0})=1$, $P_n({1})=0$.

Let $\Omega=\prod_{n=1}^{+\infty}\Omega_n$, $\mathcal{F}=\prod_{n=1}^{+\infty} \mathcal{F}_n$ and $P=\prod_{n=1}^{+\infty} P_n$. For each $n\in \mathbb{N}^+$, $\pi_n: \Omega \to \Omega_n$ is the projection on the $n$th factor.

Let us, for each $n\in \mathbb{N}^+$, define the random variable $X_n$ by: X_n(w)= \left\{\begin{aligned} &0 &\textrm{ if } \pi_n(w)=0; \\ &+\infty &\textrm{ if } \pi_n(w)=1 \end{aligned} \right. It is easy to see that $\{X_n\}$ are independent random variables on $(\Omega, \mathcal{F},P)$. Let $Y$ be a random variable defined on $(\Omega, \mathcal{F},P)$ by Y(w)= \left\{\begin{aligned} &0 &\textrm{ if, for all } n\in \mathbb{N}^+, \pi_n(w)=0; \\ &1 &\textrm{ if there is } n\in \mathbb{N}^+ \textrm{ s.t. } \pi_n(w)=1 \end{aligned} \right. Now, let $$E= \left\{w : \textrm{ for all } n\in \mathbb{N}^+, \pi_n(w)=0 \right\}$$ It is easy to see that $$\left\{w :\lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i(w)}{n} = Y(w)\right\}=E$$
and $$P(E)=1$$ So $\frac{\sum_{i=1}^{n}X_i}{n}$ converges $Y$ a.s.

Now take $A=\{0\}$, we have $$\left\{w :\lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i(w)}{n} \in A\right\}=\left\{w : Y(w) \in A \right\}=E$$

And, it is easy to see that $E\notin \mathcal{T}$.

If the random variables take values in $(\mathbb R, \mathscr B(\mathbb R))$ rather than $(\overline{\mathbb R}, \mathscr B(\overline{\mathbb R}))$, then none of the random variables or $Y$ can ever be infinite. I ran into a similar problem.

If the random variables indeed take values in $(\overline{\mathbb R}, \mathscr B(\overline{\mathbb R}))$...

Say for example $t=2$ and $X_1(\omega) = \infty$, then...I still don't see what's the problem. Actually, we know that if indeed

$$Y = \lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i}{n} \ \text{a.s.}$$

then $\forall t \in \mathbb N$,

$$\lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i}{n} = \lim_{n \rightarrow \infty}\frac{\sum_{i=t}^{n}X_i}{n} \ \text{a.s.}$$

This is because the limit of a sequence of numbers or functions doesn't depend on any finite subset of terms. Hence, the equality in question follows from the aforementioned.

So yeah, assuming the limit indeed exists, I don't think it should matter w/c of the random variables are infinite. Are you thinking something like:

$$Y = \lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i}{n} = \infty \ \text{a.s.}$$

$$\lim_{n \rightarrow \infty}\frac{\sum_{i=2}^{n}X_i}{n} < \infty \ \text{a.s.}$$

?

That would contradict the above fact. If for example $Y = X_1 = \infty \ \text{a.s.}$, then it indeed follows that

$$\lim_{n \rightarrow \infty}\frac{\sum_{i=2}^{n}X_i}{n} = \infty \ \text{a.s.}$$

Another thing to note is almost surely. Perhaps the equality in question actually holds only almost surely.

• Does the anonymous downvoter (Did on a second account?) care to explain the downvote? – BCLC Dec 4 '15 at 6:42
• I don't understand your whole answer - why are you talking about $\overline{\mathbb{R}}$? The assumption is that Y is real-valued, so all your whole debate is hypothetical. – Olorun Dec 4 '15 at 11:51
• @Olorun Depends on the textbook I think? – BCLC Dec 4 '15 at 12:09