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Since I am reading some stuff about weak convergence of probability measures, I started to wonder what is the dual space of the space consisting of all the finite (signed) measures (which is well known to be a Banach space with the norm being total variation). Is there any characterization of it? We may impose extra assumptions on the underlying space if necessary.

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Dunford and Schwartz, Linear Operators I doesn't contain a description of that space. In a footnote on p.374 they remark that no satisfactory descriptions were known at the time of writing. This suggests that there is no easy characterization as you're asking for. – t.b. Oct 22 '11 at 17:41
@t.b.: Thanks for sharing. Perhaps the one GEdgar gives below is one the "various sorts of representations" they mentioned in the footnote? – Syang Chen Oct 23 '11 at 3:34
yes, I think so. It is unsatisfactory in that you can't say what you really get. Something humungous, in any case. – t.b. Oct 23 '11 at 3:42

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Well, your space of measures is isometric to $L^1(\mu)$ for some (probably very big, non-sigma-finite) measure $\mu$. So it is enough to know what is the dual of an $L^1$ space.

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Thanks, could you sketch how the "big space" is obtained? Or could you give some reference? – Syang Chen Oct 22 '11 at 20:42
Take a maximal family $\mathcal A$ of mutually singular probability measures. (Use Zorn's Lemma.) The space of measures is isometrically the $l_1$-sum of $L^1(\nu)$ as $\nu$ ranges over the family $\mathcal A$. – GEdgar Oct 22 '11 at 22:13
So there is no canonical representation for it? What's dual of $L^1$ when the space is not $\sigma$-finite? – Syang Chen Oct 23 '11 at 4:03
@Xianghong: Note that you have an $l_1$-sum of $L^1(\nu)$ where $\nu$ are probabilities. Its dual space is the $l_\infty$-product of the corresponding $L^\infty(\nu)$-spaces. – t.b. Oct 23 '11 at 10:02

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