Weak Law of Large Numbers Confusion If I have a sequence of IID random variables $\{X_i\}_{i=1}^{n}$ with $\mathbb{E}[X_i]=\mu$  then the WLLN implies that $\bar{X}_n$ converges to $\mu$. This is where my confusion lies. It is my understanding that RV are deterministic functions $X:\Omega\rightarrow \mathbb{R}$, where the random component comes from $\omega\in\Omega$. If we are defining $\bar{X}_n(\omega)=\frac{1}{n}\sum_{i=1}^{n}X_i({\omega})$, if $X_i$ are iid then it is not assumed that $X_i(\omega)=X_j(\omega)$ so have an average of different values. When we apply the weak law of large numbers in real life we take the average of a sequence of iid random variables, but doesn't the WLLN require that the RV's be evaluated in the same $\omega$, why are we able to assume that this holds. Am I understanding the WLLN wrong?
 A: To clarify a bit more about the role of the $\omega$ here: in this setup generally the space $\Omega$ is defined as a space of sequences and each $X_i$ is the $i$-th projection of each $\omega$.
A concrete way to define it: set a probability measure in $\mathbb{R}$, then this measure will represent the law of each $X_i$. It can be shown than then the infinite product space $\Omega:=\mathbb{R}^{\infty }:=\prod_{n\in \mathbb N }\mathbb{R}$ can be given also with a probability measure induced by the individual probability measures of each $\mathbb{R}$ probability space.
Then, as advised at the beginning, each $\omega\in \Omega$ is just an arbitrary real valued sequence. Then when you evaluates something like $\Pr[\bar X_n\in A ]$ you are measuring a set (of sequences) in $\Omega$. Then informally the weak law of large numbers say us that as $n$ increases then the amount of sequences in $\Omega$ such that the mean of it $n$-th first values are far away of $\mu$ becomes arbitrarily small in measure (i.e., in probability). Formally it assert that
$$
\forall \epsilon >0,\, \forall \delta >0,\,\exists N\in \mathbb N,\, \forall n\geqslant N: \Pr [|\bar X_n-\mu|\geqslant \delta ]<\epsilon 
$$
where $\mu:=\operatorname{E}[X_1]$ and $\Pr $ is the probability measure in $\Omega $.
In summary: each $\omega $ represent intuitively a "random sample" (a sequence of values), and remember that $\Pr [X\in A]:=\Pr (\{\omega \in \Omega : X(\omega )\in A\})=\Pr(X^{-1}(A))$, in this case $X=|\bar X_n-\mu|$ and $A=[\delta ,\infty )$, so we are measuring the probability that a random sample belongs to some subset of $\Omega $, in this case we are measuring the probability that, given a sequence, the mean of it $n$-th first values deviates from $\mu$ more than a given amount.
