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Trying to show that, where $(X,\mathscr{F})$ is a measurable space,

if $f^{-1}(A) \in \mathscr{F} $ whenever $A \in \mathscr{A}$,

then $f^{-1}(A) \in \mathscr{F} $ whenever $A \in \sigma(\mathscr{A})$

We should be assuming $f:X\rightarrow Y$ is a function where $X \in \mathscr{F}$ is nonempty. Also, $\mathscr{A}$ is a collection of subsets of Y.

aduh
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  • Doesn't $f$ need to be measurable? And if it is measurable, then doesn't this follow immediately from the definitions? (To be honest I am kind of spacing out, so I don't think I focused enough to understand it) – Chill2Macht Oct 14 '16 at 20:33
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    @William another way to state this question is "how do I prove: if $f$ is $\mathscr{A}$ measurable then it is $\sigma(\mathscr{A})$ measurable"? – Taylor Oct 14 '16 at 20:41
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    @Taylor That's not right. $\mathscr{A}$ is not assumed to be a $\sigma$-field. You may want to review the definition of a measurable function https://en.wikipedia.org/wiki/Measurable_function. – aduh Oct 14 '16 at 20:44
  • @aduh I should put "measurable" in quotes then. Good call – Taylor Oct 14 '16 at 20:47
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    @William The point is to show that to check the measurability of $f$, it's enough to check that the measurability condition holds on a collection of sets $\mathscr{A}$ that generates the $\sigma$-field associated with $Y$. For example, if $Y$ is $\mathbb{R}$ equipped with the Borel field, it's enough to check that $f^{-1}((-\infty, a]) \in \mathscr{F}$ to show that $f$ is measurable. – aduh Oct 14 '16 at 21:00
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    If $f$ is a map from $X$ to $Y$, what is the role of $\Omega$? –  Oct 14 '16 at 21:56

1 Answers1

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Let $\mathcal{B} = \{A \subset Y: f^{-1}(A) \in \mathscr{F} \}$ be the collection of subsets of $Y$ for which the statement holds. Then $\mathcal{B} \supset \mathscr{A}$. Moreover, $\mathcal{B}$ is a $\sigma$-field because

$$f^{-1}(\cup_iA_i) = \cup_if^{-1}(A_i)$$ $$f^{-1}(A^c) = (f^{-1}(A))^c.$$

(See this, for example.)

Therefore $\mathcal{B} \supset \sigma(\mathscr{A})$, so $f^{-1}(A) \in \mathscr{F}$ whenever $A \in \sigma(\mathscr{A})$.

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