TITLE: $\sigma$-algebra generated by a random variable
QUESTION [5 upvotes]: How to show that $\sigma(\{X^{-1}(U):U \text{ open in }R\}) = \{X^{-1}(B): B \text{ Borel set in } R\}$.
I can show $\sigma(\{X^{-1}(U):U \text{ open in }R\}) \subseteq \{X^{-1}(B): B \text{ Borel set in } R\}$.
It remains to show the other direction. (Let's say the definiton of $\sigma$-algebra generated by a random variable is taken to be $\sigma(\{X^{-1}(U):U \text{ open in }R\})$. I know some may define it as the smallest $\sigma$-algebra $\mathcal{Y}$ such that X is $\mathcal{Y}$-measurable.)

REPLY [7 votes]: Let $\mathcal{U}=\{U\subseteq\mathbb{R}\mid U\text{ is open}\}$ denote the open subsets of $\mathbb{R}$ and $\mathcal{B}$ denote the Borel sets of $\mathbb{R}$. Then we want to show that $X^{-1}(\mathcal{B})\subseteq \sigma(X^{-1}(\mathcal{U}))$.
To that end show that $$\mathcal{A}:=\{B\subseteq\mathbb{R}\mid X^{-1}(B)\in \sigma(X^{-1}(\mathcal{U}))\}$$ is a sigma-algebra. Then use the fact that $\mathcal{U}\subseteq\mathcal{A}$ to conlude that $\sigma(\mathcal{U})\subseteq\mathcal{A}$. Now this means that
$$
X^{-1}(B)\in\sigma(X^{-1}(\mathcal{U})),\quad \text{for all}\,\, B\in\sigma(\mathcal{U}),
$$
or in other words
$$
X^{-1}(\mathcal{B})\subseteq\sigma(X^{-1}(\mathcal{U})).
$$