# $\inf$ of a sequcence of random variables bigger than some $a\in\mathbb{R}$

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?

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 $$\left \{ \omega : Z(\omega) > x_0 \right \} = \emptyset.$$

• 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. – Mentossinho Jul 26 '20 at 14:18
• @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 $\bigcap \{ \omega : X_n \geq t \}$. – winperikle Jul 26 '20 at 16:52
• @​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 $\bigcap\{\omega\mid X_n(\omega) \geq t \}\subseteq\{\omega\mid Z(\omega) \geq t \}$? – Mentossinho Jul 26 '20 at 17:28
• $\inf X_n(\omega)$ 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. – winperikle Jul 26 '20 at 17:59
• @​winperikle i think i've got it! Thank your for your time. – Mentossinho Jul 26 '20 at 21:11