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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: $$\{\bar X_n\}_{n=1}^\infty$$
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    $\begingroup$ 1. Yes. 2. All the random variables $X_n$ and $\bar X_n$ are defined on $(\Omega,\mathcal F,P)$. $\endgroup$ – Did May 12 '16 at 6:10
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  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).

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  • $\begingroup$ 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? $\endgroup$ – julian.marr May 12 '16 at 19:11
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    $\begingroup$ Yes, exactly, that is correct. $\endgroup$ – Lost in a Maze May 13 '16 at 5:22

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