[proofplan] The proof is a direct use of measurability. Each threshold set in $\Omega$ is the preimage under $X$ of a standard open or closed interval in $\mathbb R$. Since open and closed intervals are Borel sets, measurability of $X$ forces all four preimages to belong to $\mathcal F$. [/proofplan] [step:Pull back the four Borel threshold intervals under the measurable random variable] Let $(\Omega,\mathcal F,\mathbb P)$ be the probability space. Let $\mathcal B(\mathbb R)$ denote the [Borel $\sigma$-algebra](/page/Borel%20Sigma%20Algebra) on $\mathbb R$, meaning the $\sigma$-algebra generated by the open subsets of $\mathbb R$. Let \begin{align*} X:(\Omega,\mathcal F)&\to(\mathbb R,\mathcal B(\mathbb R)) \end{align*} be the given real-valued random variable, so $X$ is a [measurable map](/page/Measurable%20Map). Fix $a\in\mathbb R$. Define the four subsets of $\Omega$ by \begin{align*} E_{\le a} &:= \{\omega\in\Omega:X(\omega)\le a\},& E_{<a} &:= \{\omega\in\Omega:X(\omega)<a\},\\ E_{\ge a} &:= \{\omega\in\Omega:X(\omega)\ge a\},& E_{>a} &:= \{\omega\in\Omega:X(\omega)>a\}. \end{align*} The intervals $(-\infty,a)$ and $(a,\infty)$ are open subsets of $\mathbb R$, hence belong to $\mathcal B(\mathbb R)$. The intervals $(-\infty,a]$ and $[a,\infty)$ are closed subsets of $\mathbb R$, hence also belong to $\mathcal B(\mathbb R)$ because $\mathcal B(\mathbb R)$ is a $\sigma$-algebra containing all open sets and therefore containing their complements. By measurability of $X$, the preimage under $X$ of every set in $\mathcal B(\mathbb R)$ belongs to $\mathcal F$. Therefore \begin{align*} E_{\le a} &= X^{-1}((-\infty,a])\in\mathcal F,\\ E_{<a} &= X^{-1}((-\infty,a))\in\mathcal F,\\ E_{\ge a} &= X^{-1}([a,\infty))\in\mathcal F,\\ E_{>a} &= X^{-1}((a,\infty))\in\mathcal F. \end{align*} Since $a\in\mathbb R$ was arbitrary, all four threshold sets are events for every real threshold $a$. [guided] Let $(\Omega,\mathcal F,\mathbb P)$ be the probability space. Let $\mathcal B(\mathbb R)$ denote the [Borel $\sigma$-algebra](/page/Borel%20Sigma%20Algebra) on $\mathbb R$, that is, the $\sigma$-algebra generated by the open subsets of $\mathbb R$. Let \begin{align*} X:(\Omega,\mathcal F)&\to(\mathbb R,\mathcal B(\mathbb R)) \end{align*} be the given real-valued random variable. By definition, this means $X$ is a [measurable map](/page/Measurable%20Map): whenever $B\in\mathcal B(\mathbb R)$, its preimage $X^{-1}(B)$ belongs to $\mathcal F$. Fix a real number $a\in\mathbb R$. We want to prove that the four threshold sets are events, meaning that they are elements of $\mathcal F$. Define these four subsets of $\Omega$ by \begin{align*} E_{\le a} &:= \{\omega\in\Omega:X(\omega)\le a\},& E_{<a} &:= \{\omega\in\Omega:X(\omega)<a\},\\ E_{\ge a} &:= \{\omega\in\Omega:X(\omega)\ge a\},& E_{>a} &:= \{\omega\in\Omega:X(\omega)>a\}. \end{align*} The key point is that each of these sets is not being constructed directly in $\Omega$; it is the preimage of a simple interval in $\mathbb R$. The intervals $(-\infty,a)$ and $(a,\infty)$ are open subsets of $\mathbb R$, so they lie in $\mathcal B(\mathbb R)$ by the defining property of the Borel $\sigma$-algebra. The intervals $(-\infty,a]$ and $[a,\infty)$ are closed subsets of $\mathbb R$. Equivalently, each is the complement in $\mathbb R$ of an open interval: \begin{align*} (-\infty,a] &= \mathbb R\setminus(a,\infty),\\ [a,\infty) &= \mathbb R\setminus(-\infty,a). \end{align*} Since $\mathcal B(\mathbb R)$ is a $\sigma$-algebra, it is closed under complements. Hence $(-\infty,a]$ and $[a,\infty)$ also belong to $\mathcal B(\mathbb R)$. Now apply the measurability of $X$ to each of these four Borel intervals. Because the preimage of every Borel set under $X$ belongs to $\mathcal F$, we obtain \begin{align*} \{\omega\in\Omega:X(\omega)\le a\} &= X^{-1}((-\infty,a])\in\mathcal F,\\ \{\omega\in\Omega:X(\omega)<a\} &= X^{-1}((-\infty,a))\in\mathcal F,\\ \{\omega\in\Omega:X(\omega)\ge a\} &= X^{-1}([a,\infty))\in\mathcal F,\\ \{\omega\in\Omega:X(\omega)>a\} &= X^{-1}((a,\infty))\in\mathcal F. \end{align*} Thus all four threshold sets are elements of $\mathcal F$. Since the real number $a$ was arbitrary, the conclusion holds for every $a\in\mathbb R$. [/guided] [/step]