# Generating a $\sigma$-algebra [duplicate]

Possible Duplicate:
Preimage of generated $\sigma$-algebra

I wish to prove the following:

"Let $X$ be a set and $\mathcal{A}$ a family of subsets of $X$, and $\Sigma_{\mathcal{A}}$ the $\sigma$-algebra of subsets of $X$ generated by $\mathcal{A}$. Suppose that $Y$ is another set and $f : Y \rightarrow X$ a function. Then $\left\{f^{-1} \left[E \right] : E \in \Sigma_{\mathcal{A}} \right\}$ is the $\sigma$-algebra of subsets of $Y$ generated by $\left\{ f^{-1} \left[A\right] : A \in \mathcal{A} \right\}$."

I understand that the $\sigma$-algebra of subsets of $Y$ generated by $\left\{ f^{-1} \left[A\right] : A \in \mathcal{A} \right\}$ is defined to be $$\bigcap \left\{ \Sigma : \Sigma \text{ is a } \sigma \text{-algebra of subsets of }Y, \left\{ f^{-1} \left[A\right] : A \in \mathcal{A} \right\} \subseteq \Sigma\right\}.$$

I'm not sure how this leads to the desired result, though. Any help much appreciated.

## marked as duplicate by Nate Eldredge, t.b., Willie WongMar 26 '12 at 12:36

• Something to consider: What do you mean by "$f$ is a measurable function"? (Indeed, in a sense, that's precisely what you're trying to create here.) :) – cardinal Jan 24 '12 at 2:23
• I think what you are saying is that from the way we define the domain of $\quad f \quad$ it will necessarily be a measurable function right ? Bit new to this material, so just wondering whether I understand you correctly. (post amended accordingly...) – Beltrame Jan 24 '12 at 2:32
• I understand $f:D \rightarrow \mathbb{R}$ to be a measurable function if any, or equivalently all, of the following are true: (i) ${x : f(x) < a} \in \Sigma_D$ for every $a \in \mathbb{R}$; (ii) ${x : f(x) \leq a} \in \Sigma_D$ for every $a \in \mathbb{R}$; (iii) (i) ${x : f(x) > a} \in \Sigma_D$ for every $a \in \mathbb{R}$; (iv) (i) ${x : f(x) \geq a} \in \Sigma_D$ for every $a \in \mathbb{R}$, where $D \subseteq X$ and $\Sigma_D$ is the subspace $\sigma$-algebra of subsets of $D$. – Harry Williams Feb 6 '12 at 22:12