Suppose we have sequence of random variables $\{X_n\mid n\in\mathbb{N}\}$, defined on a probablity space $(\Omega,\mathcal{F},\mathbb{P})$. Then we define $(\inf_{n\in\mathbb{N}}X_n)(\omega)=\inf_{n\in\mathbb{N}}X_n(\omega),\quad\forall{\omega\in\Omega}$. I understand the function $(\inf_{n\in\mathbb{N}}X_n)(\cdot)\equiv Z(\cdot)$ as "no bigger" than any function $X_n$ ($\forall\omega\in\Omega\quad Z(\omega)\leq X_n(\omega),\forall{n\in\mathbb{N}}$), and also, "no smaller" than any other function with this property ($\forall\omega\in\Omega\quad Z(\omega)\geq Z^{'}(\omega)\mid Z^{'}(\omega)\leq X_n(\omega),\forall{n\in\mathbb{N}}$). I would like to prove a statement below: $$ \{\omega\in\Omega\mid Z(\omega)>a\}=\bigcap_{n=1}^{\infty}\{\omega\in\Omega\mid X_n(\omega)>a\},a\in\mathbb{R}\quad? $$ I know, that $Z(\omega)>a\Rightarrow\bigwedge_{n\geq1}X_n(\omega)>a$, because $Z$ is "no bigger" than a sequence. Since, left side is a subset of a right side. How can i prove an inverse implication?


1 Answer 1


I don't think the other inclusion is true.

Let $x_0 \in \mathbb R$ and define,

\begin{align*} X_n \ \colon \ &\Omega \longrightarrow \mathbb R\\ & \omega \longmapsto x_0+n^{-1} \end{align*}

Then $Z(\omega) = x_0$.

Now, $\forall n$, $\left \{ \omega : X_n(\omega) > x_0 \right \}= \Omega$ thus, $$ \bigcap_{n\geq 1}\left \{ \omega : X_n(\omega) > x_0 \right \} = \Omega $$ While $\left \{ \omega : Z(\omega) > x_0 \right \} = \emptyset.$

  • $\begingroup$ Thank you for an answer. It's interesting, because i met in some sources this statement, as a proof of the claim that $Z(\cdot)$ is measurable. I'm confused now. $\endgroup$ Jul 26, 2020 at 14:18
  • 1
    $\begingroup$ @Mentossinho When you work with $\inf$ or $\sup$ strict inequalities have no reason to be preserved, for example if we have $\forall n, X_n > t$ the only thing we can say about the $\inf$ is that $\inf X_n \geq t$. Thus measurability of $Z$ is better studied by looking at the class of sets $\{\omega : Z(\omega) \geq t \}$ and show that such sets are equal to $\bigcap \{ \omega : X_n \geq t \}$. $\endgroup$
    – periwinkle
    Jul 26, 2020 at 16:52
  • $\begingroup$ @​winperikle: very hepful information! Could you clarify, why is it true that $\forall n, X_n(\omega) > t\Rightarrow\inf X_n(\omega) \geq t$ and how i should start to prove, that $\bigcap\{\omega\mid X_n(\omega) \geq t \}\subseteq\{\omega\mid Z(\omega) \geq t \}$? $\endgroup$ Jul 26, 2020 at 17:28
  • $\begingroup$ $\inf X_n(\omega)$ is the largest quantity that is $\leq$ than all $X_n(\omega)$. Since $t$ is $<$ than all $X_n(\omega)$, we have $t \leq \inf X_n(\omega)$. The set $\{ \omega : \forall n, X_n(\omega) \geq t \}$ is an intersection because of the $\forall n$. Finally, the implication $\Rightarrow$ translates into a set inclusion. $\endgroup$
    – periwinkle
    Jul 26, 2020 at 17:59
  • $\begingroup$ @​winperikle i think i've got it! Thank your for your time. $\endgroup$ Jul 26, 2020 at 21:11

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