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Let $(\Omega, {\cal F})$ and $(E, {\cal E})$ be measurable spaces and $X$ a r.v. from $\Omega$ into $E$ (i.e., $X$ is ${\cal F}/{\cal E}$-measurable). We assume that $\cal E$ is generated by ${\cal A}$, i.e., ${\cal E} = \sigma({\cal A})$. I wish to show the following important fact: $\sigma(X) = \sigma(X^{-1}{\cal A})$ where $\sigma(X) := \{X^{-1}A; A\in {\cal E}\}$ and $X^{-1}{\cal A} := \{X^{-1}A; A\in {\cal A}\}$.

Is the following proof correct? Or is there any other simpler proof of it?

Since $\sigma(X) \supset X^{-1}{\cal A}$, we have $\sigma(X) \supset \sigma(X^{-1}{\cal A})$.

To show the other inclusion, we note that for any $A\in {\cal A}$, $X^{-1}A \in X^{-1}{\cal A} \subset \sigma(X^{-1}{\cal A})$. Since $\cal A$ generates $\cal E$, this shows that the r.v. $X$ is $\sigma(X^{-1}{\cal A})$-measurable. By the minimal property of $\sigma(X)$ we can conclude that $\sigma(X) \subset \sigma(X^{-1}{\cal A})$. QED

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You might accept some of the previous answers you have received (the "ok"-sign on the left of the post) to encourage people to answer your question. – Michael Greinecker Sep 10 '12 at 15:06
@MichaelGreinecker: Thanks for the suggestion. I didn't know the system of acceptance. – Yamamoto Sep 10 '12 at 15:46
up vote 2 down vote accepted

Let $\mathcal F=\sigma(X^{-1}\mathcal A)$ and $\mathcal C=\{A\in \mathcal E\mid X^{-1}A\in\mathcal F\}$, then $\mathcal A\subset\mathcal C$ because $X^{-1}\mathcal A\subset\mathcal F$, and $\mathcal C\subset\mathcal E$ by definition, hence $\sigma(\mathcal A)\subset\sigma(\mathcal C)\subset\sigma(\mathcal E)$. Here are two key facts:

  1. $\sigma(\mathcal A)=\sigma(\mathcal E)$ by hypothesis, and $\sigma(\mathcal E)=\mathcal E$ because $\mathcal E$ is a $\sigma$-algebra, hence $\sigma(\mathcal C)=\mathcal E$.
  2. $\mathcal C$ is a $\sigma$-algebra, hence $\mathcal C=\sigma(\mathcal C)$.

Applying 1. and 2. yields $\mathcal C=\mathcal E$, in other words, for every $A$ in $\mathcal E$, $X^{-1}A$ is in $\mathcal F$. This means that $X^{-1}\mathcal E\subset\mathcal F$. On the other hand, since $\mathcal A\subset\mathcal E$, $\mathcal F\subset\sigma(X^{-1}\mathcal E)$. Since $\mathcal E$ is a $\sigma$-algebra, $X^{-1}\mathcal E$ is a $\sigma$-algebra, hence $\sigma(X^{-1}\mathcal E)=X^{-1}\mathcal E$. This yields $\mathcal F=X^{-1}\mathcal E=\sigma(X)$, the desired result.

The only nontrivial step in this proof is 2. $\mathcal C$ is a $\sigma$-algebra. If some problem arises when trying to prove it, just yell.

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I wonder. When I have an interest in one question, I see already an answer from you. Do you have any idea why? – Seyhmus Güngören Sep 10 '12 at 21:11
@SeyhmusGüngören Interest overlap + speed response:) – Sasha Sep 11 '12 at 0:11

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