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 $$ (\max\{X_1,...,X_n\} \to \infty)\in\mathbb{F} $$

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

1 Answer 1


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}


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