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Oct 6, 2015 at 17:44 history edited Johannes Hahn CC BY-SA 3.0
Fixed notation errors. Hopefully everyone understands now who is integrated over whom.
Oct 6, 2015 at 15:14 review Close votes
Oct 10, 2015 at 14:15
Oct 6, 2015 at 14:12 comment added Giovanni De Gaetano Thanks, now I get it. But I still think that your defining formula should read $$(\forall x^* \in E^*)\quad x^*(f) = \int_\Omega x^*(f(\omega)) d\mu(\omega).$$ Then $\int_E f d \mu \in E^{**} \simeq E$ is the mapping $(\int_E f d \mu)(x^*) = x^*(f)$.
Oct 6, 2015 at 14:07 comment added AB_IM It's okay, I clarified the notation further.
Oct 6, 2015 at 14:05 history edited AB_IM CC BY-SA 3.0
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Oct 6, 2015 at 14:04 answer added AB_IM timeline score: 1
Oct 6, 2015 at 14:02 comment added Giovanni De Gaetano Do you mean $x^*(f)$ instead of $x^*(e)$? And I still don't get what is denoted by $\int_E f d \mu$, because the expression above seems to be dependent on the choice of $x^* \in E$. Moreover, $x$ is an element of $\Omega$ and $x^*$ of $E$? I apologize for the annoying intrusion...
Oct 6, 2015 at 13:53 comment added AB_IM I edited my post to clarify my notation.
Oct 6, 2015 at 13:52 history edited AB_IM CC BY-SA 3.0
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Oct 6, 2015 at 13:46 history edited AB_IM CC BY-SA 3.0
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Oct 6, 2015 at 13:40 comment added Liviu Nicolaescu I meant, what do you mean by the integral of $f$ over $E$ when $f$ is defined on $\Omega$?
Oct 6, 2015 at 13:11 comment added AB_IM Its the unique $e\in E$ solving: $(x^* \in E^*)\, x^*(e) = \int x^*(f) $ (where the right is the Lebesgue integral).
Oct 6, 2015 at 13:01 comment added Liviu Nicolaescu Can you define $\int_E f$ where $f:\Omega\to E$?
Oct 6, 2015 at 12:48 history asked AB_IM CC BY-SA 3.0