Generate $\sigma(X)$ for RV $X$ where $X(\omega):=\max\{a,\omega\}$. I am not sure how to generate $\sigma(X)$ for RV $X$ when $X(\omega):=\max\{a,\omega\}$ for a probability $(\Omega,\mathcal{F},\textbf{P})$ with a continuous density $f(\omega)$ and $\Omega=\mathbb{R}$.
My intuition tell me, $\sigma(X)$ is just a Borel subset of $(a,\infty)$ because the pre-image of $X$ is just the set $(a,\infty)$.
However, I am not sure if such an intuition is correct. Any guidance or hint would be appreciated.
 A: There are a few elementary results needed. Suppose $(\mathrm{Y}, \mathscr{Y})$ is a measurable space and $\mathrm{X}$ is a set. For any subset $\mathscr{H}$ of $\mathscr{Y}$ and any function $f:\mathrm{X} \to \mathrm{Y}$ write
$$
f^{-1}(\mathscr{H}) = \{ f^{-1}(\mathrm{H}) \mid \mathrm{H} \in \mathscr{H} \}.
$$

*

*If $f:\mathrm{X} \to \mathrm{Y}$ is any function, then $f^{-1}(\mathscr{Y})$ is a sigma field in $\mathrm{X}.$


*If $\mathscr{S}$ is any subset of $\mathscr{Y},$ then $f^{-1}(\sigma(\mathscr{S})) = \sigma(f^{-1}(\mathscr{S})).$
Thus, $X^{-1}(\mathscr{B}_\mathbf{R}) = \sigma(X^{-1}(\{(-\infty, t] \mid t \in \mathbf{R}\})) = \sigma(\{(-\infty, t] \mid t \geq a\}) =: \mathscr{K}.$ Denote by $\mathscr{F}$ the set of all unions $z \cup \mathrm{A}$ such that $z \in \{\varnothing, (-\infty, a]\}$ and $\mathrm{A}$ is a Borel set in $(a, \infty).$ The aim is to show that $\mathscr{K} = \mathscr{F}.$

*

*It easily follows that $\mathscr{F}$ is a sigma algebra.

*The inclusion $\subset$ follows since $\{(-\infty, t] \mid t \geq a\} \subset \mathscr{F}$ (and since $\mathscr{F}$ is a sigma algebra).

*The inclusion $\supset$ follows since $\mathscr{K}$ contains $\varnothing,$ $(-\infty, a]$ and $(c, d]$ for $a < c < d.$
