RG.
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Report Thread starter 6 years ago
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 \mathrm{Given\ a\ sequence\ \{X_n\}^\infty_{n=1}\ of\ random\ variables\ on\ a\ measurable\ space\ \left(\Omega,\mathcal{F}\right),  \ prove\ that:}

 \mathrm{\underset{n} sup\ X_n\ and\ \underset{n}inf\ X_n\ are\ random\ variables}

 \mathrm{Hint:\ Show\ that:}

 \{\omega \in \Omega: \underset{n}sup\ X_n(\omega) \geq t \} = \underset{n} \bigcup\ \{\omega \in \Omega: X_n(\omega) \geq t \}

 \mathrm{and:}

 \{\omega \in \Omega: \underset{n}inf\ X_n(\omega) \leq t \} = \underset{n} \bigcup\ \{\omega \in \Omega: X_n(\omega) \leq t \}


Now, here's where my problem comes in:

I know that the union of measurable sets is measurable, and if I could just state the above, I'd be fine, but it's asking me to show it, and I have no idea how to start

Does anyone have any idea where I can start to solve this?
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DFranklin
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Maybe (probably!) I'm being stupid, but is the first identity actually true?

Take the measurable space as [0,1] with standard measure (though it hardly matters what we choose), and define X_n(x) = 1 - 1/n (so a constant function).
Then \{\omega \in \Omega : X_n(\omega) \ge 1\} = \emptyset, but \sup_n X_n(\omega) = 1 \forall \omega so \{\omega \in \Omega : \sup_n X_n(\omega) \ge 1\} = \Omega.
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