# Show that sets of maximum random variables is measureable

Given $$X_1,...,X_n$$ independent and identically distributed real-valued random variables on the probability space $$(\Omega,\mathbb{F},P)$$, I have to show that: $$(\max\{X_1,...,X_n\} \to \infty)\in\mathbb{F}$$ I have tried to give an answer below, however, I am unsure whether the argumentation is correct - especially regarding taking the maximum and using $$\cup$$ instead.

We know that the random variable can be written as: $$X:\ \Omega \to \mathbb{R}$$ and by definition it fullfills: $$X^{-1}((-\infty,c]) = \{\omega \in \Omega : X(\omega) \leq c\} \in \mathbb{F}$$ I will start by looking at the sets for a given sample of random variables with $$n$$ elements: $$\{ \omega : \max\{X_1(\omega),...,X_n(\omega)\}\leq c\ |\ \omega \in \Omega \}$$ So I can write $$\{X_1(\omega)\leq c\ |\ \omega \in \Omega\}\land...\land\{X_n(\omega)\leq c\ |\ \omega \in \Omega\}=\bigcap_{i=1}^n \{X_i(\omega)\leq c\ |\ \omega \in \Omega\},$$ and as $$c\in\mathbb{R}$$ can be arbitrary large, and that every random variable is measurable as it lives on the probability space, it must be true that $$(\max\{X_1,...,X_n\} \to \infty)\in\mathbb{F}$$

• I don’t see how the last part follows, at least not without doing a couple nontrivial extra steps. Commented Sep 14, 2019 at 19:36
• @spaceisdarkgreen any hints to what these nontrivial steps are? Commented Sep 15, 2019 at 4:22

I'm guessing you have a countable collection $$X_1, X_2, \ldots$$ of random variables; otherwise $$(\max\{X_1,\ldots,X_n\}\to \infty)$$ is hard to make sense of.
Note that $$\{\lim_{n\to \infty} \max\{X_1,\ldots, X_n\}= \infty\} = \Omega \setminus \{\lim_{n\to \infty} \max\{X_1,\ldots, X_n\} < \infty\}$$ and \begin{align} \{\lim_{n\to \infty} \max\{X_1,\ldots, X_n\} < \infty\} & = \cup_{N=1}^{\infty}\{\lim_{n\to \infty} \max\{X_1,\ldots, X_n\} < N\}. \\ & = \cup_{N=1}^{\infty} \{\max\{X_1, X_2\ldots\} < N\} \\ & = \cup_{N=1}^{\infty} \cap_{n=1}^{\infty}\{X_n < N\} \end{align}