Let $(\Omega, \mathcal F)$ be a measurable space (especially a probability space), and let $Y : \Omega \to \mathbb R$ be measurable. Then we define the sigma algebra generated by $Y$ to be
$$
\sigma(Y) = \{Y^{-1}(B) \mid B \in \mathcal B(\mathbb R)\},
$$
where $\mathcal B(\mathbb R)$ are the borel sets of $\mathbb R$.
Intuitively it seems to me that the goal of the definition of $\sigma(Y)$ is that it "collapses the level sets" — that is, an alternative definition might be
$$
\sigma'(Y) = \{A \in \mathcal F \mid Y^{-1}(Y(A)) = A\}.
$$
Are these definitions equivalent? I can easily prove that $\sigma(Y) \subseteq \sigma'(Y)$, but not the other way around — nor can I easily think of a counterexample.
Best Answer
They are not equivalent. Is is not true that $\sigma'(Y)\subset \sigma (Y)$. Given that $Y^{-1}(Y(A))=A$ how do you find the Borel set $B$?. The reverse inclusion is always true. Use a simple set theoretic argument for this.
For an explicit counterexample consider $\mathbb R$ with the sigma algebra of all subsets and let $Y$ be the identity map on $\mathbb R$. Then $\sigma'(Y)$ contains all subsets of $\mathbb R$ whereas $\sigma(Y)$ is the Borel sigma algebra of $\mathbb R$.