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Let $(\Omega,\mathcal{F})$ be a measure space and $X$ mapping from $\Omega$ to $\mathbb{R}$. Assume that $X^{-1}((a,b])\in \mathcal{F}$ for all intervals. Prove that $X$ is a random variable.

First, I have shown that $\mathcal{B}(\mathcal{R})$ is the $\sigma$-algebra generated by all intervals of the form $(a,b]$. Using this, I want to show that $X^{-1}(B)\in\mathcal{F}$ for all $B \in \mathcal{B}(\mathcal{R})$ (Borel) and thus $X$ is random variable. But, I don't know how to do the last step. Can you please help me?

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Hint: Show that

$$\mathcal{D} := \{B \in \mathcal{B}(\mathbb{R}); X^{-1}(B) \in \mathcal{F}\}$$

is a Dynkin system. Conclude from the fact that

$$\mathcal{G} := \{(a,b]; a<b\}$$

is contained in $\mathcal{D}$ and that $\mathcal{G}$ is a $\cap$-stable generator of $\mathcal{B}(\mathbb{R})$ that $$\mathcal{D} = \sigma(\mathcal{G}) = \mathcal{B}(\mathbb{R}).$$

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  • $\begingroup$ Thanks so much for the hint. Can you please provide me with more details, or a simpler solution. I have not ever heard Dynkin system and stable generator. Learning that takes me a lot of time. $\endgroup$ Feb 8 '16 at 20:47
  • $\begingroup$ @Susan I'm not aware of an easier proof. Have you heard of $\lambda$-systems? $\endgroup$
    – saz
    Feb 9 '16 at 6:30
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    $\begingroup$ @Susan How about asking more precise questions? Where exactly are you stuck? To show that it is a Dynkin system you have to verify certain properties... so where do you have trouble? (And yes, I know that they are the same... ) $\endgroup$
    – saz
    Feb 9 '16 at 15:47
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    $\begingroup$ @BCLC In which way is this any different from what I suggested in my answer? We need that $\mathcal{G}$ is $\cap$-stable (aka a $\pi$-system) to conclude that the Dynkin system generated by $\mathcal{G}$ equals the $\sigma$-algebra generated by $\mathcal{G}$. $\endgroup$
    – saz
    Feb 10 '16 at 13:12
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    $\begingroup$ @BCLC It's pretty clear what "$\cap$-stable" means, isn't it? It is stable under intersections. In contrast, "$\pi$-system" is just a fancy name which doesn't reveal anything about its meaning. $\endgroup$
    – saz
    Feb 10 '16 at 15:11

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