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Denote by $\ell^{\infty}$ the Banach space of bounded sequences (real or complex). By an elementary argument, the dual space of $\ell^{\infty}$ consists of all bounded finitely additive signed measures on $\mathcal{P}(\mathbb{N})$; see here or here. On the other hand, since $\ell^{\infty}$ is the same as $C(\beta\mathbb{N})$ where $\beta\mathbb{N}$ is the space of ultrafilters (or the Stone-Cech compactification of $\mathbb{N}$), by Riesz–Markov–Kakutani representation theorem the dual of $\ell^{\infty}=C(\beta\mathbb{N})$ consists of regular Borel measure on $\beta\mathbb{N}$. My questions are:

  1. This means a finitely additive measure on $\mathbb{N}$ corresponds to a countably additive measure on $\beta\mathbb{N}$. How can that be? How to understand this? Is there a direct discription of this correspondence?

  2. It seems the space of finitely additive measures is given the norm of total variation. Can we view it as a subspace of $B(\mathcal{P}(\mathbb{N}),\mathbb{R})$ (all bounded maps from $\mathcal{P}(\mathbb{N})$ to $\mathbb{R}$) and give it the sup norm?

  3. The obvious (countably additive) measures on $\beta\mathbb{N}$ are the point measures (either at some natural number or at some ultrafilter) or their countable weighted sum (I guess these already contain the space $\ell^1$). What are some non-trivial examples? Can we "classify" these measures, whatever that means?

Couchy
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Lxm
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  • See my remarks of 8 years ago at https://math.stackexchange.com/a/392719/442 – GEdgar Dec 16 '21 at 00:51
  • As $\ell^\infty$ is the dual of $\ell^1$, the latter is also canonically a subspace of the dual of $\ell^\infty$. There is no hope at all of classifying measures as asked. You cannot give concrete examples of the extra measures (due to AC etc). – Henno Brandsma Dec 16 '21 at 06:40
  • @HennoBrandsma I agree that it doesn't seem plausible, since classifying these measures is probably more difficult than classifying ultrafilters (there is some work on that topic though). Can we at least understand these measures in terms of ultrafilters? – Lxm Dec 16 '21 at 07:03
  • Dear @GEdgar, a similar question came up here. As the OP here states, there are elements in $\ell^_\infty$ which are not countably additive (Banach limits for example) while all elements in $C^(\beta\mathbb{N})$ are $\sigma$-aditive. I guess that the confusion is that one my think (erroneously) that an element $\mu$ in $\ell^*$ is the restriction to $2^{\mathbb{N}}$ of a (signed) Radon measure on $\mathscr{B}(\beta\mathbb{N})$. Do you have any idea on how a signed-Borelmeasure on $\beta\mathbb{N}$ relates to an element in $\ell_\infty$? – Mittens Mar 10 '23 at 22:48

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