# Explanation of $\lim\sup$ of a sequence of random variables in measure theory

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.

## 1 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.).

• I'm confused, can't we define each $X_n$ as being a measurable function from $\mathbb{R} \to \mathbb{R}$ with the Borel $\sigma$-algebra such that the preimage has the measure given by the C.D.F of the standard normal variable? Then we would have $X_n(\omega)$ is the same for every $n$? Also if I can't think of the $\limsup$ in that way what is the best way to think of the $\limsup$ of a sequence of random variables? – Abdul Hadi Khan Sep 16 '18 at 15:42
• Your $X_i$ would not be independent, contradicting the i.i.d. assumption. The way to think about them is the same as how you think about real number case, except now each $\omega$ can (and generally will) get a different sequence of real numbers, i.e. $\limsup Y_n$ is just another random variable. – user10354138 Sep 16 '18 at 15:50
• Usually iid sequence of random variables is to take a product probability space $\Omega:=(\Omega_1)^\mathbb{N}$, where $\Omega_1$ is a probability space modelling a single member of that iid random variable, and each member of that sequence corresponds to a different projection map. What you have described is just the diagonal of this product, which typically is null unless there are atoms. – user10354138 Sep 16 '18 at 15:54