# Sigma algebra generated by random variable on a set with generators

I'm having trouble proving an intuitive result I found in these lecture notes I'm using for self-study (1.2.14 there).

Suppose $$X$$ is a $$(\mathbb{S}, \mathcal{S})$$-valued random variable (from $$(\Omega, \mathcal{F})$$), and furthermore $$\mathcal{S} = \sigma(\mathcal{A})$$. If $$\mathcal{F}^X$$ is the $$\sigma$$-algebra generated by $$X$$ in $$\Omega$$, we want to show that $$\mathcal{F}^X = \sigma(\{X^{-1}(A) : A \in \mathcal{A}\})$$.

It's easy to prove that $$\mathcal{F}^X \supset \sigma(\{X^{-1}(A) : A \in \mathcal{A}\})$$, by noticing that (i) $$\mathcal{F}^X$$ is a $$\sigma$$-algebra, and that (ii) it contains $$\{X^{-1}(A) : A \in \mathcal{A}\}$$. But I believe I'm missing the right proof strategy for the other direction. Just appealing to the definitions and the tools developed so far (e.g. the $$\pi-\lambda$$ theorem) didn't take me very far.

I think I get the spirit of the claim. Basically, it says that if you have a set of generators $$\mathcal{A}$$ of $$\mathcal{S}$$, to obtain $$\mathcal{F}^X$$ you can either take the inverse images of all sets generated by $$\mathcal{A}$$, or you can take the inverse images of just the sets in $$\mathcal{A}$$ and then use those to generate a $$\sigma$$-algebra. So, the order of the operations of "taking inverse images" and "generating a $$\sigma$$-algebra" doesn't matter. Is this understanding correct?

Any hint on a direction that might work for the proof would be extremely appreciated!

So to start, it looks like in the notes that it was already shown that:

$$\mathcal{F}^X = \{X^{-1}(B): B \in \sigma(\mathcal{A})\}$$ and that this is in fact a $$\sigma-alg$$ so I will be starting from there.

Now the goal is to show that $$\sigma(\{X^{-1}(A): A \in \mathcal{A}\}) = \mathcal{F}^X$$

As stated $$LHS \subseteq RHS$$ by:

\begin{align} \{X^{-1}(A): A \in \mathcal{A}\} &\subseteq \{X^{-1}(B): B \in \sigma(\mathcal{A})\}\\ \implies \sigma(\{X^{-1}(A): A \in \mathcal{A}\}) &\subseteq \sigma(\{X^{-1}(B): B \in \sigma(\mathcal{A})\}) = \{X^{-1}(B): B \in \sigma(\mathcal{A})\} \end{align}

Now for $$RHS \subseteq LHS$$ we need to exploit the properties of measurability, which ensures that the map $$X^{-1}: \mathcal{B} \to \sigma(\mathcal{A})$$ preserves all set properties.

Now define $$\Sigma^{'} = \{B \in \sigma(\mathcal{A}): X^{-1}(B) \in \sigma(\{X^{-1}(A): A \in \mathcal{A}\})\}$$. Now we will proceed to show that this is in fact a $$\sigma-alg$$.

a) Since $$\sigma(\{X^{-1}(A): A \in \mathcal{A}\})$$ is a $$\sigma-alg$$ on $$\mathbb{S}$$, $$\mathbb{S} \in \Sigma^{'}$$

b) For $$A \in \Sigma^{'}$$, it must be that $$A^c \in \Sigma^{'}$$.

By set properties of the map $$X^{-1}$$, $$X^{-1}(A^c) = (X^{-1}(A))^c$$ and it must be that $$(X^{-1}(A))^c \in \sigma(\{X^{-1}(A): A \in \mathcal{A}\})$$ by $$\sigma-alg$$ properties since $$X^{-1}(A) \in \sigma(\{X^{-1}(A): A \in \mathcal{A}\})$$ by definition.

c) For $$A_1,A_2, \dots$$, $$A_i \in \Sigma^{'}$$ the countable union $$\cup_{i}A_i \in \Sigma^{'}$$

Similarly this follows since $$X^{-1}(\cup_{i}A_i) = \cup_i X^{-1}(A_i)$$

Thus by a),b), c) $$\Sigma^{'}$$ is a $$\sigma-alg$$ on $$\mathbb{S}$$ for which $$X$$ is measurable. Since $$\mathcal{F}^X$$ must be the smallest such $$\sigma-alg$$ it must be that $$RHS \subseteq LHS$$ and thus $$RHS = LHS$$

Re: intuition, I think that's the basic idea. From my limited understanding, measurability has deep connections with generating sets. Williams (Probability with Martingales section 3.13) has a good discussion about the intuitive significance of generated $$\sigma-alg$$s.

The way I understand it is the generated $$\sigma-alg$$ is the set of events $$F$$ for which for each and every $$s\in\mathbb{S}$$ we can decide whether $$F$$ has occured or not on the basis of the information observed through the random variable $$X(s), s \in \mathbb{S}$$. I think this gives some insight into why "the operations of "taking inverse images" and "generating a σ-algebra" doesn't matter.