What is this theorem about measurable functions saying?

[kalish1]

**Theorem:**

Let $(\Omega,\mathcal{F})$ be a measurable space and let $f:\Omega \rightarrow Y$ be a given function. Let $\mathcal{A}$ be a collection of subsets of $Y$.

If $f^{-1}(A) \in \mathcal{F}$ for every $A \in \mathcal{A}$, then $f^{-1}(A) \in \mathcal{F}$ for every $A \in \sigma(\mathcal{A})$.

**Questions**:
I understand that $\Omega$ is the underlying set and $\mathcal{F}$ is a class of subsets of $\Omega$ that is closed under finite unions and complementation (i.e., $\mathcal{F}$ is a field). But what is $Y$? And what is the purpose of defining $\mathcal{A}$ as a *collection of subsets* of $Y$, with $A$ an element in that collection? Moreover, what is the message of the theorem?

The notation is throwing me off.

This question has been crossposted on real analysis - What is this theorem about measurable functions saying? - Mathematics Stack Exchange

 

[Euge]

**Theorem:**

Let $(\Omega,\mathcal{F})$ be a measurable space and let $f:\Omega \rightarrow Y$ be a given function. Let $\mathcal{A}$ be a collection of subsets of $Y$.

If $f^{-1}(A) \in \mathcal{F}$ for every $A \in \mathcal{A}$, then $f^{-1}(A) \in \mathcal{F}$ for every $A \in \sigma(\mathcal{A})$.

**Questions**:
I understand that $\Omega$ is the underlying set and $\mathcal{F}$ is a class of subsets of $\Omega$ that is closed under finite unions and complementation (i.e., $\mathcal{F}$ is a field). But what is $Y$? And what is the purpose of defining $\mathcal{A}$ as a *collection of subsets* of $Y$, with $A$ an element in that collection? Moreover, what is the message of the theorem?

The notation is throwing me off.

This question has been crossposted on real analysis - What is this theorem about measurable functions saying? - Mathematics Stack Exchange

Hi kalish,

There is nothing particular about $Y$, except that it's the codomain of $f$. The symbol $\sigma(\mathcal{A})$ denotes the sigma-algebra generated by $\mathcal{A}$. It's important to have $\mathcal{A}$ to be a collection of subsets of $Y$, or else $\sigma(\mathcal{A})$ is undefined.

Note that $(Y, \sigma(\mathcal{A}))$ is a measurable space. So the theorem is saying that the function $f$ is measurable as a map from $(\Omega, \mathcal{F})$ to $(Y, \sigma(\mathcal{A}))$ if $\mathcal{F}$ contains all the preimages $f^{-1}(A)$, $A\in \mathcal{A}$. In particular, if $f$ is a function from $(X, \mathcal{B}_X)$ to $(Y, \mathcal{B}_Y)$ (here $\mathcal{B}_X$ and $\mathcal{B}_Y$ denote the Borel algebras on topological spaces $X$ and $Y$, respectively), then $f$ is a Borel function if the preimage of every open set in $Y$ is a Borel set in $X$.