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Suppose that a random variable $X_n \to X$ almost surely. Then, the probability statement form is:

$$ P\left(\{\omega \in \Omega: \lim_{n \to \infty} X_n(\omega) = X(\omega)\}\right) =1 $$

I am wondering what is the correct way to view this statement. Is it to say that the probability of events in $\Omega$ (the sample space) that cause the limit to converge all occur with probability one, meaning, they have a possibility of occuring, or is it viewed as, they must occur?

If my sample space were the roll of one die, then I have that $\Omega = \{1,2,3,4,5,6\}$. In this case, if I have some random variable $X_n$ where:

$$ P\left(\{\omega \in \{1,2,3,4,5,6\}: \lim_{n \to \infty} X_n(\omega) = X(\omega)\}\right) =1 $$

What is the correct way to interpret this statement?

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  • $\begingroup$ In your example (roll of one die), which $X_n$ and $X$ do you have in mind? $\endgroup$ – Did Oct 6 '16 at 6:55
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$\bbox[gainsboro,1pt]{\{\omega \in \Omega: Q(\omega)\}}$ is the event of outcomes in the space($\Omega$) where the condition $Q$ is true.

$\bbox[gainsboro,1pt]{\lim\limits_{n\to\infty} X_n(\omega)=X(\omega)}$ is the condition that the $n$ to $\infty$ limit of the sequence ${[X_n(\omega)]}_{n\in \Bbb N}$ is $X(\omega)$.

Often $\bbox[gainsboro,1pt]{\mathsf P(\{\omega\in\Omega: \lim\limits_{n\to\infty} X_n(\omega)=X(\omega)\})}$ is just written $\bbox[gainsboro,1pt]{\mathsf P(\lim\limits_{n\to\infty} X_n=X)}$ for short.

This probability being $1$ means that $\bbox[gainsboro,1pt]{\lim\limits_{n\to\infty} X_n=X}$ certainly occurs except perhaps for some zero measure set of outcomes.   That is, we have a point-wise convergence of the sequence.

That means that: $X_n\to X$ almost surely.

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