Why is $\sigma(X) = \{ X^{-1}(B) \ |\text{$B$ borel set} \}$ when $X$ is measrable if $X^{-1}(U) |\text{U open set}$

Question

I am following a book on probability theory. it says $X$ is a random variable if $X^{-1}(U)$ is in the sigma algebra for all open sets $U$. But on the other hand $\sigma(X) = \{ X^{-1}(B) \ |\text{$B$ borel set} \}$ is the smallest sigma algebra to which $X$ is measurable. Why do we not define it $\sigma(X) = \{ X^{-1}(B) \ |\text{$U$ open set} \}$ instead?

Answer 1

In contrast with $\sigma(X)$ the collection $\{X^{-1}(U)\mid U\text{ open set }\}$ is not necessarily a $\sigma$-algebra, so it would be wrong to identify the first collection with the second.

It can be proved though that $\sigma( \{X^{-1}(U)\mid U\text{ open set }\})=\{X^{-1}(B)\mid B\text{ borel set }\}$.

Answer 2

For example, the identity map $\textrm{id} \colon \mathbf{R} \to \mathbf{R}$ is a random variable on $\mathbf{R}$. The two sets \begin{align} \sigma(\mathrm{id}) = \{ \mathrm{id}^{-1}(B) \mid B \colon \text{Borel} \} = \{ B \mid B \colon \text{Borel} \} \end{align} and \begin{align} \sigma'(\mathrm{id}) = \{ \mathrm{id}^{-1}(U) \mid U \colon \text{open} \} = \{ U \mid U \colon \text{open} \} \end{align} are different because $\{0\} \in \sigma(\mathrm{id})$ and $\{0\} \notin \sigma'(\mathrm{id})$. Furthermore, $\sigma'(\mathrm{id})$ is not a sigma algebra.