# Strong Law of Large Numbers and Convergence a.e.

Let $(\Omega,\mathcal F,P)$ be a probability space. A sequence of r.v.'s $X_n$ converges a.e. to $X$ if and only if there exists a null set $N$, such that:

$\forall \omega\in\Omega\setminus N:\lim_{n\rightarrow\infty}X_n(\omega)=X(\omega)$ finite.

As I understans this, implicit in this definition is that both $\{X_n\}_{n=1}^\infty$ and $X$ are defined on the same probability space $(\Omega,\mathcal F,P)$.

How does this apply to the strong law of large numbers: $\bar X_n\rightarrow_{a.e.}\mathbb E(X)$.

My doubts are the following:

1. Should I treat $\bar X_n$ as converging almost surely to a constant $\mathbb E(X)$?
2. If $X_n$ is defined on $(\Omega,\mathcal F,P)$, then is $\bar X_n$ also defined on $(\Omega,\mathcal F,P)$? If not, how do we construct a probability space for the sequence: $$\{\bar X_n\}_{n=1}^\infty$$
• 1. Yes. 2. All the random variables $X_n$ and $\bar X_n$ are defined on $(\Omega,\mathcal F,P)$.
– Did
May 12 '16 at 6:10

1. Yes, $$E(X)$$ is constant, so $$\bar{X}_n \to E(X)$$ a.e. just means $$P(\omega\,:\, \lim_{n \to \infty}\bar{X}_n(\omega) = E(X)) = 1.$$
If you like, just consider the (constant) random variable $$Y$$ on $$\Omega$$ defined by $$Y(\omega) = E(X)$$ for all $$\omega$$, then $$\bar{X}_n \to Y$$ a.e.
2. If $$X_1,\ldots, X_n$$ are $$\mathcal{F}$$-measurable, then so is $$\bar{X}_n$$ (this is just using the fact that measurability is preserved under addition and multiplication).
• One final doubt: for a given $\omega_0\in\Omega\setminus N$, this determines a sequence $X_1(\omega_0),...,X_n(\omega_0),...$ which in turn determines the sequence: $\bar X_1(\omega_0)=X_1(\omega_0), \bar X_2(\omega_0)=0.5*X_1(\omega_0)+0.5*X_2(\omega_0)$ and so on. So this sequence of means will converge to the number $\mathbb E(X)$. Is that correct? May 12 '16 at 19:11