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:
- Should I treat $\bar X_n$ as converging almost surely to a constant $\mathbb E(X)$?
- 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: $$\{\bar X_n\}_{n=1}^\infty$$
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.
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).