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Strong law of large numbers (SLLN) says if $X_1, X_2, \dots$ are iid random variables with expectation $\mu$, then $\bar{X}_n \to \mu$ almost surely, or $$P\left(\lim_{n\to \infty} \bar{X}_n = \mu\right)=1.$$

If we use the probability space $(\Omega, \mathcal{F}, P)$ and the concept of the random variable as a function from $\Omega$ to $\mathbb{R}$, we can write $$P\Big( \omega \in \Omega : \lim_{n \to \infty} \bar{X}_n(\omega) = \mu \Big) = 1.$$

Now suppose we have a sample space from tossing a fair six-sided die, the sample space $\Omega=\{1,2,3,4,5,6\}$ and the probability is $P(\{i\})=1/6$. $X_i(\omega)=\omega$ are discrete uniform on $\{1,2,3,4,5,6\}$.

How do I understand $\bar{X}_n(\omega)$? For example, if I get a number $1$ from the die, what is $\bar{X}_n(\{1\})$? Since there is pointwise convergence, $\bar{X}_n(\{1\}) \to 3.5$?

Thank you for the help.

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    $\begingroup$ You have a different probability space for the distribution of multiple rolls than for a single roll. Indeed the probability space on which $\overline{X}_n$ is defined must have at least $6^n$ elements. Typically we don't worry about this and instead just fix a single huge probability space to start with, which has enough structure to do whatever we want, and then go from there. $\endgroup$ – Ian May 12 '16 at 11:40

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