The definition I have been given of the $\limsup\limits_{n \to \infty} Y_n$ where the $Y_n$ are random variables is that it is another random variable defined as $(\limsup\limits_{n \to \infty} Y_n)(\omega) = \limsup\limits_{n \to \infty} Y_n(\omega)$ for any $\omega \in \Omega$ where the $\lim\sup$ on the R.H.S is the standard $\lim\sup$ of a sequence of reals (as that is the definition of $\lim\sup$ for a sequence of measurable functions).
However I am then given the question to prove that if we have $X_i$ as IID standard $N(0,1)$ random variables then we get $\limsup\limits_{n \to \infty} \frac{X_n}{\sqrt{2\log(n)}} = 1$ almost surely. In this case if I use the definition given above we get that $X_n(\omega) = X_m(\omega)$ for any $n,m$ and $\omega$ as all the $X_i$ have the same distribution. So our sequence is then just decaying to $0$ so the $\lim\sup$ should be $0$ and not $1$ almost surely.
I'm sure that I'm the one who has a conceptual error somewhere but I can't see what. I know the question is not wrong as other people have asked it here. Please help me clear up this confusion.
Best Answer
Two random variables having the same distribution does not mean they take the same value on each $\omega$. Indeed, independence already kill that possibility unless the random variable is constant (a.s.).