Timeline for Why is $\mathcal{E}(X)=\mathcal{E}(X,X^*)$?
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5 events
| when toggle format | what | by | license | comment | |
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| Nov 18, 2015 at 15:25 | comment | added | Gerald Edgar | houda's question is OK even when $X$ is non-separable, but Nate's isn't. | |
| Nov 18, 2015 at 14:47 | comment | added | Nate Eldredge | A somewhat less obvious fact (though still not too hard to prove) is that both $\sigma$-algebras equal the Borel $\sigma$-algebra on $X$. In particular, the norm and weak topologies have the same Borel sets. | |
| Nov 18, 2015 at 12:52 | comment | added | Andreas Blass | I agree with Simon that this is rather obvious, but let me give a hint for the part that might be slightly less obvious. Consider the family of subsets $C_0\subseteq\mathbb R^n$ such that the corresponding $C$ (as in your definition of $\mathcal E(X)$) is in $\mathcal E(X,X^*)$. Check that this family is a $\sigma$-algebra containing the basic open sets in $\mathbb R^n$ (rectangular boxes) and therefore containing all the Borel sets in $\mathbb R^n$. | |
| Nov 18, 2015 at 12:44 | comment | added | Simon Henry | Unless I'm missing something, that actually seem obvious: the $\sigma$-algebra you defined make all the linear form measurable, and conversely, any $\sigma$-algebra that makes the linear form measurable have to contain those cylindrical sets... | |
| Nov 18, 2015 at 12:34 | history | asked | Heidy | CC BY-SA 3.0 |