Timeline for A characterization of $L_1(\mu)$ in $L_\infty(\mu)^*$
Current License: CC BY-SA 4.0
14 events
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| Jun 19, 2018 at 14:40 | comment | added | Robert Furber | @SergeiAkbarov I wrote $\mu$ instead of $\kappa$ by force of habit. In the first example, $\kappa$ is defined on the countable-cocountable $\sigma$-algebra, and is not localizable. In the second example, it is defined on the powerset $\sigma$-algebra, and so is localizable. A measure is localizable iff it is semi-finite and the algebra of measurable sets modulo nullsets is a complete Boolean algebra. As the only nullset of the counting measure is the empty set, this comes down to the fact that the countable-cocountable $\sigma$-algebra is not a complete Boolean algebra, but the powerset is. | |
| Jun 19, 2018 at 14:36 | history | edited | Robert Furber | CC BY-SA 4.0 |
Changed $\mu$ to $\kappa$ where it should be.
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| Jun 5, 2018 at 16:28 | comment | added | Sergei Akbarov | Robert, excuse me again, as far as I understand in your first example $\kappa$ and $\mu$ are the same. And if so, $\mu$ is localizable in this case, isn't it? | |
| May 31, 2018 at 19:57 | history | edited | Robert Furber | CC BY-SA 4.0 |
Fixed another missing *.
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| May 31, 2018 at 18:13 | history | edited | Robert Furber | CC BY-SA 4.0 |
Fixed failure to indicate that the dual space of $L^\infty$ was meant.
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| May 31, 2018 at 18:12 | comment | added | Robert Furber | @SergeiAkbarov Quite right, I'll just fix it. | |
| May 31, 2018 at 17:53 | comment | added | Sergei Akbarov | Ah, that's why the integral $\int f d\mu$ exists for each $f\in L^\infty(X,{\mathcal P},\kappa)$... Excuse me. By the way you should write $L^\infty(X,{\mathcal P},\kappa)^*$ (with the star *) when you say that the functional $f\mapsto \int f d\mu$ belongs to $L^\infty(X,{\mathcal P},\kappa)^*$. | |
| May 31, 2018 at 17:22 | comment | added | Robert Furber | @SergeiAkbarov It's a probability measure, and therefore finite in the sense of measure theory, unless you mean something else. | |
| May 31, 2018 at 15:33 | comment | added | Sergei Akbarov | Robert, I don't understand something in your second example. This measure $\mu:{\mathcal P}(X)\to [0,1]$, it must be finite, isn't it? | |
| May 28, 2018 at 11:05 | comment | added | Robert Furber | @PietroMajer It's also relatively easy to prove that $\ell^\infty(X)$ is the dual of $\ell^1(X)$ for all sets $X$, once you know it's true, even without the more general theorem. | |
| May 28, 2018 at 10:58 | comment | added | Robert Furber | @PietroMajer The first one is, the second one isn't (which is why I inncluded a second one). $L^\infty(X)$ is the dual of $L^1(X)$ iff $X$ is localizable as a measure space. See, for instance: jstor.org/stable/2372178 | |
| May 28, 2018 at 7:46 | comment | added | Pietro Majer | These non $\sigma$-finite examples are also counterexamples to $L^\infty(X)$ being the dual of $L^1(X)$, aren't they? | |
| May 28, 2018 at 2:35 | history | edited | Robert Furber | CC BY-SA 4.0 |
changed slip-up "countable measure" -> "counting measure"
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| May 28, 2018 at 2:18 | history | answered | Robert Furber | CC BY-SA 4.0 |