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!


1 Answer 1


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.


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