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seen May 18 '13 at 10:40

On extended leave from the site. I visit occasionally, but irregularly, mostly without logging in. For the time being it is impossible to predict if and when I will find the time to contribute again.


Dec
23
revised banach-spaces wiki excerpt
I removed the specification of the ground field since there is absolutely no reason to exclude p-adic Banach spaces. There's also no need to define Cauchy sequences here.
Dec
20
reviewed Close multiple choice question on holomorphic functions
Dec
20
reviewed Reject suggested edit on Mathematical difference between white and black notes in a piano
Dec
20
comment Inner regularity property of Radon measures in metric spaces
You understand Michael's answer correctly. You can find proofs in many places, for example Theorem 3.2 in Parthasarathy's Probability Measures on Metric Spaces or Theorem 17.11 of Kechris's Classical Descriptive Set Theory.
Dec
20
comment Inner regularity property of Radon measures in metric spaces
I see. Here's the idea: The example of a non-tight Radon measure on an uncountable disjoint union of $\mathbb{R}$ I linked you to can be modified to use the Cantor set with the "coin-flipping measure" instead of $\mathbb{R}$. Thus the task is essentially reduced to find a closed subspace of a metric space which is homeomorphic to an uncountable disjoint union of Cantor sets. This can be done using this construction and a locally finite disjoint family of open sets (and some effort). I'm not so sure if the actual construction is all that enlightening.
Dec
19
awarded  Enlightened
Dec
19
awarded  Nice Answer
Dec
19
comment Inner regularity property of Radon measures in metric spaces
I didn't work it out in full detail, but I think that one can use the example I alluded to above to show that on every non-separable complete metric space without isolated points there is an outer measure as in Evans-Gariepy, but that the associated measure (from Carathéodory) will not be tight while satisfying all your conditions. On the other hand, one can always construct a tight version of it which will fail to be outer regular.
Dec
19
reviewed Reject suggested edit on Inequality involving expectation
Dec
19
comment Weak-to-weak continuous operator which is not norm-continuous
@Mike: that's my favorite example, too :-) It may be worth adding that the summation functional is adjoint to the limit functional on $c$ (convergent sequences). That is, if you view $\ell^1$ as the dual space of $c$ via the pairing $\langle x,y \rangle_{\ell^1,c} = \left(\sum x_n\right) \lim y_n + \sum x_n y_n$ then $T$ is adjoint to $S(y) = (\lim y_n) \cdot e_1$.
Dec
19
comment Inner regularity property of Radon measures in metric spaces
What kind of spaces are you interested in? Do you care about large spaces (not $\sigma$-compact)? Would you mind adding completeness? Have you considered the example of Lebesgue measure times counting measure on the reals times the discrete reals (it has two incarnations: one is inner regular but not outer regular and the other is outer regular but not inner regular)?
Dec
19
reviewed Approve suggested edit on Do there exist any odd prime powers that can be represented as $n^4+4^n$?
Dec
19
comment Why for a compact metric probability space, any Borel subset can be approximated by compact set?
@Jochen: The passage on Wikipedia is correct in that they are talking about inner regularity with respect to closed sets as opposed to inner regularity with respect to compact sets. The former holds for every semi-finite Borel measure on a metric space while the latter needs additional assumptions such as the ones you note.
Dec
16
reviewed Reject suggested edit on What is the most elegant proof of the Pythagorean theorem?
Dec
16
comment Abstract characterization of Borel $\sigma$-algebras
If you are willing to enter the realm of measure algebras then there's no need of invoking Maharam's classification theorem. The much more elementary Stone representation theorem gives you an honest topological measure space whose measure algebra is (canonically isomorphic to) the one you started with. Since Jon is an operator algebraist it might be worth pointing out that the resulting space is the same as the Gelfand spectrum of $L^\infty(\mu)$. In this context variants of this result also go by the name of "Mackey's point realization theorem".
Dec
15
reviewed Approve suggested edit on Simple Fourier Series
Dec
11
reviewed Leave Open How do mathematicians think about the existence of numbers?
Dec
10
comment Does the open mapping theorem imply the Baire category theorem?
@MattN. Thank you very much, but I agree with Asaf's assessment. A bounty would not be of use in this case. I doubt the answer is known at all, that's why I didn't take the question to MO either.
Dec
10
comment Does the open mapping theorem imply the Baire category theorem?
Thank you for your answer and I apologize for the confusion. As Asaf pointed out, this does indeed not address my question. I asked: "can one deduce the Baire category theory theorem for complete metric spaces (or the axiom of dependent choice) from the open mapping theorem (in whatever form)"? Indeed I linked to Schechter's book giving seven equivalent conditions on topological vector spaces (among which the open mapping theorem, the uniform boundedness, and so on). I know Jarchow's book, but my question is not "for which topological vector spaces does the OMT hold"?
Dec
2
awarded  Enlightened