How to prove part (3)?

[runic pawn]

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[proud plank]

Use the proposition from before here $\sigma(\mathcal{G}_{0})=\mathcal{G}$

[sinful perch]

😑 rotoR

[proud plank]

The idea here is that you want to prove that $X^{-1}(\mathcal{G}) \subset \mathcal{F}$

[sinful perch]

😑 rotoR

[proud plank]

Using the beginning of the proposition and the definition of what a generated sigma algebra is you can deduce the answer

[proud plank]

@thorn anvil

[thorn anvil]

ummm

I'm still stuck

@proud plank ummm how do I show it is a subset

and why am I showing that

[proud plank]

Okay let’s start from the beginning given the proposition you want to prove that $\forall B \in \mathcal{G}, X^{-1}(B) \in \mathcal{F}$

[sinful perch]

😑 rotoR

[proud plank]

Which is basically asking you to prove that $X^{-1}(\mathcal{G}) \subset \mathcal{F}$ do you agree ?

[sinful perch]

😑 rotoR

[proud plank]

As defined in proposition (1)

Now with proposition (2) we have $X^{-1}(\mathcal{G})=X^{-1}(\sigma(\mathcal{G}_0))=\sigma(X^{-1}(\mathcal{G}_0))$

[sinful perch]

😑 rotoR

[proud plank]

with $X^{-1}(\mathcal{G}_0)={X^{-1}(B), B \in \mathcal{G}_0 }$

[sinful perch]

😑 rotoR

[proud plank]

Well what do you know as we’ve assumed, $X^{-1}(\mathcal{G}_0}) \subset \mathcal{F}$

[sinful perch]

😑 rotoR
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[proud plank]

You can conclude from there, use the definition of what the generated sigma algebra is