Denote the Borel sigma-algebra on $\mathbb{R}$ by $\mathscr{B}$, and let $X$ be a real-valued random variable on a probability space $(\Omega, F, P)$, and define $C = \{X^{-1}[(-\infty, a]]\mid a\in \mathbb{R}\}$. It is known that $C$ is a $\pi$-system. We would like to show that $\sigma(C) = \sigma(X)$, i.e. the $\sigma$-algebra generated from the $\pi$-system is equalt to the algebra generated from the random variable. However, as trivial as it might seem I am currently struggling to show that $X$ is even $\sigma(C)$-measurable. I mean that by definition, $X$ is $\sigma(C)/\mathscr{B}$-measurable if $\forall A \in \mathscr{B}: X^{-1}(A) \in \sigma(C)$. Let $A \in \mathscr{B}$ be arbitrary. Then $X^{-1}(A) = \{\omega \in \Omega\mid X(\omega) \in A\}$, and we know that $A \in \sigma\left(\{(-\infty, x]\mid x \in \mathbb{R}\}\right)$, but what then? How do I progress from here, and how can I show that $\sigma(X) \subseteq \sigma(C)$?
Edit: Can I state something like, as $\{(-\infty, x]\mid x \in \mathbb{R}\}$ generates $\mathscr{B}$, then for $A \in \mathscr{B}$ it follows that $A = \bigcup_{n=1}^\infty \underbrace{(-\infty, a_n]}_{=I_n}$ for some sequence $(a_n)_{n\in \mathbb{N}}$, so that
$X^{-1}(A) = \{\omega \in \Omega\mid X(\omega) \in A\} = \{\omega \in \Omega\mid X(\omega) \in \bigcup_{n=1}^\infty I_n\} = \bigcup_{n=1}^\infty \{\omega \in \Omega\mid X(\omega) \in I_n\}$ so that $X^{-1}(A) = \bigcup_{n=1}^\infty X^{-1}(I_n) \in \sigma(C)$
for showing that $X$ is $\sigma(C)/\mathscr{B}$-measurable?
Best Answer
Lemma 1. The Borel $\sigma$-algebra $\mathscr{B}$ of $\mathbb{R}$ is generated by intervals of the form $(-\infty,a]$ for $a\in\mathbb{R}$.
Proof. This is Proposition 1.2 in Gerald Folland's Real Analysis: Modern Techniques and Applications.
Lemma 2. For a family of subsets (of $\Omega$ or $\mathbb{R}$) $\mathcal{B}$, write $\sigma(\mathcal{B})$ for the $\sigma$-algebra generated by $\mathcal{B}$. Then for any $\mathcal{A}\subseteq 2^{\mathbb{R}}$, and for any random variable (measurable function) $X:\Omega\to\mathbb{R}$, we have $$\sigma(X^{-1}(\mathcal{A}))=X^{-1}(\sigma(\mathcal{A})).$$ Proof. It is routinely verified that $X^{-1}(\sigma(\mathcal{A}))$ is a $\sigma$-algebra that contains $X^{-1}(\mathcal{A})$. Since $\sigma(X^{-1}(\mathcal{A}))$ is by definition the smallest such $\sigma$-algebra, it follows that $\sigma(X^{-1}(\mathcal{A}))\subseteq X^{-1}(\sigma(\mathcal{A}))$. For the other inclusion, let us first define $$\mathcal{E}:=\{B\subseteq\mathbb{R}:X^{-1}(B)\in\sigma(X^{-1}(\mathcal{A}))\},$$ and note that $\mathcal{E}$ is a $\sigma$-algebra on $\mathbb{R}$. Since $\mathcal{A}\subseteq\mathcal{E}$, it then follows that $\sigma(\mathcal{A})\subseteq \mathcal{E}$. Therefore, $X^{-1}(\sigma(\mathcal{A}))\subseteq \sigma(X^{-1}(\mathcal{A})).$ This finishes the proof.
Claim. $X$ is $\sigma(C)/\mathscr{B}$-measurable.
Proof. Let $\mathcal{A}:=\{(-\infty,a]:a\in\mathbb{R}\}$. By definition $X^{-1}(\mathcal{A})\subseteq \sigma(C)$. Since $\sigma(C)$ is a $\sigma$-field, using Lemmas 1 and 2, we get $$\sigma(X^{-1}(\mathcal{A}))\subseteq \sigma(C)\implies X^{-1}(\sigma(\mathcal{A}))\subseteq \sigma(C)\implies X^{-1}(\mathscr{B})\subseteq \sigma(C),$$ and thus $X$ is $\sigma(C)/\mathscr{B}$-measurable, as was to be shown.