# To prove a map from a measurable space to a set with a $\sigma$-algebra generated by a family of subsets.

Suppose $$(X,m)$$ is a measurable space and $$Y$$ is a set. Given $$\mathscr{S}\subset P(X)$$, a family of subsets of $$Y$$, we can construct a smallest $$\sigma$$-algebra $$\sigma (\mathscr{S})$$ containing $$\mathscr{S}$$, which is the intersection of all the $$\sigma$$-algebra containing $$\mathscr{S}$$.

Claim: $$f:(X,m)\to (Y,\sigma (\mathscr{S}))$$ is measurable iff $$f^{-1}(S)\in m$$ for any $$S\in \mathscr{S}$$.

Whether the claim is correct?

Note that in a similar case in general topology, it only suffices to prove the preimage of all the set in a basis is open, if we want to prove a map form a topogoly space to a set with a topology generated by a basis.

In the case of measurable space, the definition of $$\sigma (\mathscr{S})$$ is rather abstract and its members don't have specific relation with $$\mathscr{S}$$. However, as we all know, when it comes to a topology generated by a family ofsubsets, its members are clear, which is union of the subsets.

I don't know how to use the definition of $$\sigma (\mathscr{S})$$ to prove the claim. Any help would be appreciated.

• The analogy with topological spaces goes even further: it suffices to prove that the preimages of sets in a subbasis are open. On a subbasis no further restrictions are imposed (this in contrast with a basis), just like no further restrictions are imposed on the collection $\mathscr S$ in your question. – drhab Feb 14 at 8:16

## 3 Answers

If $$f$$ is measurable and $$S \in \mathcal S$$ then it is obvious that $$f^{-1}(S) \in m$$. For the converse let $$\mathcal G=\{B \in \sigma(\mathcal S): f^{-1}(B) \in m\}$$. A simple verification shows that this is a sigma algbera. If $$f^{-1}(S) \in m$$ for all $$S \in \mathcal S$$ then $$\mathcal S \subset \mathcal G$$ and hence $$\sigma(\mathcal S) \subset \mathcal G$$. This proves that $$f$$ is measurable.

Let $$\mathscr{F}$$ be the collection $$V\in \sigma(\mathscr{S})$$ such that $$f^{-1} (V) \in m$$. Then it's clear that $$\mathscr{F}$$ is a $$\sigma$$-algebra that contains $$\mathscr{S}$$ and thus $$\sigma(\mathscr{S}) \subseteq \mathscr{F}$$. Hence, $$f$$ is measurable

Let $$X,Y$$ be sets, let $$f:X\to Y$$ be a function and let $$\mathscr S$$ be a subcollection of powerset $$\wp(Y)$$.

A very nice general truth (put it in your math-luggage!) in this context is the following:$$f^{-1}(\sigma(\mathscr S))=\sigma(f^{-1}(\mathscr S))\tag1$$

This can be applied immediately: if $$f^{-1}(\mathscr S)\subseteq m$$ where $$m\subseteq\wp(X)$$ is a $$\sigma$$-algebra then we are allowed to conclude that: $$\sigma(f^{-1}(\mathscr S))\subseteq m$$

which according to $$(1)$$ is exactly the same as: $$f^{-1}(\sigma(\mathscr S))\subseteq m$$

For a proof of $$(1)$$ see this answer.