From Probability of a supremum of a sequence of independent random variables
Person asks the question
" Suppose $\{X_n\}$ is an independent sequence of random variables. Show that $$P(\sup X_n<\infty)=1$$ if and only if $$\sum_n P(X_n>M)<\infty$$ for some $M.$"
My question is what is the meaning of $\sup X_n\ $.
It seems like it's either
$\sup X_n=\sup\ \{\omega:X_n(\omega)\} $ for all n
or
$\sup X_n=\sup\ \{\omega:X_n(\omega)\ \ \text{for all n}\}.$
It would be great if anyone can clarify this.
To make sense of the first statement, we have to make sense of the statement:
$$ \sup X_n < \infty $$ More formally, this is an event, as in we must consider the set: $$ \left\{\omega \in \Omega : \sup_{n \in \mathbb{N}} X_n(\omega) < \infty\right\} $$ Thus, we are considering all sample points $\omega$ for which the supremum of the sequence of real numbers, $X_n(\omega)$ (Remember the sample point $\omega$ is fixed) is less than $\infty$.
This in mind, the statement: $$ \mathbb{P}\left(\sup_n X_n < \infty\right) = 1 $$ is shorthand for the statement: $$ \mathbb{P}\left(\left\{\omega \in \Omega : \sup_{n \in \mathbb{N}} X_n(\omega) < \infty\right\} \right) = 1 $$