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If the integrals of $f$ over all sets in a generator are strictly smaller than the integrals of $g$, do we have $f<g$?
Yes for $=$, no for $\leq, \geq, <$ or $>$.
Here is a counter-example for $>$. Let $(\Omega, \mathcal A) = (\mathbb R, \mathcal B)$ with $\mathcal B$ the Borel sigma-algebra, and $E = \{(-\...
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Accepted
When is $\mathbb E[F(S)\mid S=s]= \mathbb E[F(s)]$ true?
I am looking for a sufficient condition under which it is true that $$\Bbb E[F(S)\mid S=s] = \Bbb E[F(s)]$$
$S$ can be seen as a "selector" r.v., each value of which just picks out one r.v. ...
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If the integrals of $f$ over all sets in a generator are strictly smaller than the integrals of $g$, do we have $f<g$?
To give some positive (though possibly trivial) examples:
If $\Omega$ is countable and we choose its one-element subsets as generator $\mathcal E$, then the statement holds for $\sim\in\{<,\le,=,\...
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