# equality between two sigma algebras

Take two sets $E_1$ and $E_2$, and assume $f$ is a function $E_1 \to E_2$. Take now a family of subsets of $E_2$ and call it $(O_i)_{i\in I}$, and consider the family $\left(f^{-1}(O_i)\right)_{i\in I}$ of subsets of $E_1$. Call $\mathcal B_1$ the $\sigma$-algebra on $E_1$ generated by $\left(f^{-1}(O_i)\right)_{i\in I}$ and $\mathcal B_2$ the $\sigma$-algebra on $E_2$ generated by $(O_i)_{i\in I}$

Is it true that $\mathcal B_1 = f^{-1}(\mathcal B_2)$ ? What I see is that at least $\mathcal B_1 \subset f^{-1}(\mathcal B_2)$.

If this is not true how would one prove that for a measurable function $f: E \to \overline{\mathbb{R}}$ the preimage of any borel set of $\overline{\mathbb{R}}$ is a measurable set of $E$, when measurability has been defined by: $f^{-1}(]c;+ \infty[)$ is a measurable set for any $c\in\mathbb{R}$

For any subcollection $\mathcal{C}$ of $\wp\left(E_{1}\right)$ or $\wp\left(E_{2}\right)$ let $\sigma\left(\mathcal{C}\right)$ denote the $\sigma$-algebra generated by $\mathcal{C}$.

Instead of $\left(O_{i}\right)_{i\in I}$ I will write $\mathcal{V}$ so that $\sigma\left(\mathcal{V}\right)$ denotes the $\sigma$-algebra generated by $\mathcal{V}$.

Defining $f^{-1}\left(\mathcal{C}\right)=\left\{ f^{-1}\left(C\right)\mid C\in\mathcal{C}\right\}$ for any subcollection $\mathcal{C}$ of $\wp\left(E_{2}\right)$ you are actually asking whether the following statement is true: $$f^{-1}\left(\sigma\left(\mathcal{V}\right)\right)=\sigma\left(f^{-1}\left(\mathcal{V}\right)\right)$$ The answer to that question is: "yes".

Start by proving the following statements:

• $f^{-1}\left(\mathcal{A}\right)$ is a $\sigma$-algebra whenever $\mathcal{A}\subseteq\wp\left(E_{1}\right)$ is a $\sigma$-algebra.
• $\left\{ A\in\wp\left(E_{2}\right)\mid f^{-1}\left(A\right)\in\mathcal{B}\right\}$ is a $\sigma$-algebra whenever $\mathcal{B}\subseteq\wp\left(E_{1}\right)$ is a $\sigma$-algebra.

They are not too difficult to prove, especially because preimages are quite nice to work with.

The first statement tells us immediately that $f^{-1}\left(\sigma\left(\mathcal{V}\right)\right)$ is a $\sigma$-algebra, and this of course with $f^{-1}\left(\mathcal{V}\right)\subseteq f^{-1}\left(\sigma\left(\mathcal{V}\right)\right)$.

This allows the conclusion that $\sigma\left(f^{-1}\left(\mathcal{V}\right)\right)\subseteq f^{-1}\left(\sigma\left(\mathcal{V}\right)\right)$.

The second statement tells us that $\left\{ A\in\wp\left(E_{2}\right)\mid f^{-1}\left(A\right)\in\sigma\left(f^{-1}\left(\mathcal{V}\right)\right)\right\}$ is a $\sigma$-algebra and this with $\mathcal{V}\subseteq\left\{ A\in\wp\left(E_{2}\right)\mid f^{-1}\left(A\right)\in f^{-1}\left(\sigma\left(\mathcal{V}\right)\right)\right\}$.

This allows the conclusion that $\sigma\left(\mathcal{V}\right)\subseteq\left\{ A\in\wp\left(E_{2}\right)\mid f^{-1}\left(A\right)\in\sigma\left(f^{-1}\left(\mathcal{V}\right)\right)\right\}$ or equivalently $f^{-1}\left(\sigma\left(\mathcal{V}\right)\right)\subseteq\sigma\left(f^{-1}\left(\mathcal{V}\right)\right)$.

• Thank you very much, this seems completely correct to me ! – incas Aug 29 '16 at 12:54
• You are very welcome. – drhab Aug 29 '16 at 15:15