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Let $X$ be a compact metric space and $C(X)=\{ f:X\rightarrow \mathbb{R} \ | \ \ f \ continuous\}$ with the uniform norm. It is a separable Banach space.

1) I'm aware of the fact that $C(X)^*$, the space of continuous linear functionals $C(X)\rightarrow \mathbb{R}$ coincide with $M(X)$, the space of signed regular borel measures on $X$.

This intuitively makes sense because $\mu\in M(X)$ can be naturally be seen as an evaluation map $f\mapsto \mathbb{R}$ by means of integration.

2) I'm also aware that $(C(X)^*)^*$, the dual of the dual, coincide with the set of bounded Borel measurable functions $F:X\rightarrow\mathbb{R}$ (source: P. Lax, Functional Analysis).

UPDATE Landscape provided a counterexample in the comments. My source was: P. Lax "Functional Analysis" 2002, Page 82, Theorem 14(ii). I guess this might be a mistake, perhaps fixed in some errata somewhere.

New question: C.f. Landscape's example, let $g_A$ be the extension (given by Hahn-Banach) of $f_A$ to $C([0,1])^{**}$. Looking at $g_A$ as a function $X\rightarrow\mathbb{R}$, $g_A$ must satisfy by construction the equation $\int_{[0,1]}g_A \ d \ \delta_x= 1$ if $x\in A$ and $0$ otherwise. This seems to imply that $g_A$ is the characteristic function of $A$. Hence not Borel if $A$ is not Borel.

Thus, $g_A$ as a function $(X\rightarrow\mathbb{R})$ is uniquely determined by the construction starting from $f_A$. But is it $g_A$ as an element of $C([0,1])^{**}$ uniquely determined? If so, it looks to me we would have a reasonably well defined notion of integration for non-measurable sets.

Thanks!

Screen shot of Lax's theorem 14 on page 82:

Lax, Theorem 14, page 82

Martin
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OldFella
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    Things in $C(X)^{**}$ are things that morally can be integrated with respect to every Borel measure. Does that give you intuition about Q1? – Mariano Suárez-Álvarez May 15 '13 at 08:48
  • Mariano thanks for the answer. This is helpful. But also, e.g., analytic functions can be integrated and are not borel. But beside this observation... Why is this the right way to look at C(X)*? After all, we do not look at $C(X)^$ as things that can be "integrated" with respect to every continuous $f\in C(X)$, don't we? Please bear with my ignorance here. – OldFella May 15 '13 at 08:56
  • Let me also ask, just to be sure. it correct to write that the action of $F\in C(X)^{**}$ on measures is expressed as $F(\mu)=\int F d \mu$, right? – OldFella May 15 '13 at 08:58
  • The assertion in your question 1 is incorrect. Please the comment given by commenter here. – 23rd May 15 '13 at 10:20
  • Hello Landscape, thanks! this is quite surprising as I think Lex's book is a very solid reference. Regarding the counterexample proposed in the page you linked. To apply Hahn-Banach it is required for $A$ not just to be subset of $[0,1]$ but to be a sub-linear space. Does this make any difference? – OldFella May 15 '13 at 11:42
  • Space $C(X)^{**}$ contains the space of bounded Borel functions. But it contains other objects in addition. What is the exact quote in Lax? Is there an on-line errata for that text? – GEdgar May 15 '13 at 13:48
  • Page 82, Theorem 14(ii). "Let Q be a compact Hausdorff space, C(Q) the space of continuous real valued functions on Q normed by the max norm. Then (ii) $C^{\prime\prime}$ is $L^\infty(Q)$, the space of all bounded Borel measurable functions on Q".

    It this is wrong it means that some elements in $C(X)^{**}$ by the process of taking countable limits of continuous functions. Borel measurable functions forms the smallest set arising from the continuous functions by closing under limits of sequences.

     – OldFella May 15 '13 at 15:41
  • GEdgar, is there a precise characterization of $C(X)^{**}$ available somewhere? – OldFella May 15 '13 at 15:43
  • Sorry I don't know why my second last comment is corrupted. In the sentence "it means ... $C(X)^{**}$ by the process" the intermediate subsentence "can not be obtained by" has been lost. – OldFella May 15 '13 at 15:52
  • @RaMa2013: I think you misunderstood the comment in the linked page. Given $A\subset[0,1]$, $A$ defines a linear functional, say $f_A$, on the linear subspace $\ell^1[0,1]$ of $C[0,1]^$. $\ell^1[0,1]$ is the collection of $\mu=\sum_{n}a_n\delta_{x_n}$, where $a_n\in\mathbb{R}$, $\delta_{x_n}$ is the Dirac measure at $x_n\in[0,1]$ and $|\mu|=\sum_n|a_n|<+\infty$. By definition, $f_A(\mu)=\sum_na_n\delta_{x_n}(A)$. You may check that $f_A$ is bounded on $\ell^1[0,1]$ and by Hahn-Banach, it can be extended to an element in $C[0,1]^{*}$. Note that $A$ could be not Borel. – 23rd May 15 '13 at 15:58
  • @Landscape The problem is that the topology on $l^1$ is not the topology of $C(X)^*$. So I am not sure that H-B can be used. In the induced topology of $C(X)^$, I think that $l^1$ is dense.. – N. S. May 15 '13 at 18:31
  • The norm topology on $l^1$ is the norm topology on $C(X)^\ast$. – GEdgar May 15 '13 at 18:34
  • @N.S.: Why not? Please note that topology here we are talking about should be strong topology, i.e. the topology induced by norm. The norm on $C(X)^$ is the total variation of sign measures, which, when restricted to $\ell^1$, coincides the explicit formula given in my last comment. By the way, I think maybe you mean $\ell^1$ is dense in weak topology or something like that. – 23rd May 15 '13 at 18:37
  • @GEdgar: Thank you for your comment. I didn't notice that when inputting. – 23rd May 15 '13 at 18:38
  • @Landscape Yes I mean the weak-star topology (aka the vague topology). The space of measures is the dual of $C(X)$, even if $X$ is not compact (i use this often in my research), but then the measures can be infinite... In that case, the weak-star topology is the only one you can use... Maybe Lax was using this setting. – N. S. May 15 '13 at 18:40
  • @Landscape Ups I said somethings stupid.. What I meant is: For $X$ locally compact, the space of measures is the dual of $C_C(X)$, compactly supported continuous functions, which in the case of $X$ compact is the same as $C(X)$.... And if $X$ is not compact, the standard topology considered is the weak-$^*$ topology.... So my question is: is the statement true if $C_C(X)$ is equipped with the weak star topology instead? – N. S. May 15 '13 at 18:48
  • I added a screen shot with the precise statement of the theorem in Lax to avoid further speculation. @N.S. No it is not true (assuming you mean the dual of $C_c(X)$ in your question). The dual space of $(X^\ast, \mathrm{weak}^\ast)$ is $X$ itself. See this answer for a proof. – Martin May 15 '13 at 18:51
  • @Martin: Nice work! – 23rd May 15 '13 at 19:05
  • @N.S.: Please never mind. Martin has already answered your question. – 23rd May 15 '13 at 19:06
  • @RaMa2013: You are welcome! Sorry for so many comments under your question. – 23rd May 15 '13 at 19:08
  • It says the proof is in Appendix A, but I think that is only the proof of (i). – GEdgar May 15 '13 at 19:24

1 Answers1

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Let $X$ be a compact metric space. Let $\mathcal B$ be the collection of Borel sets on $X$. Banach space $C(X)$ is the set of all (necessarily bounded) continuous real-valued functions $f : X \to \mathbb R$, with norm $$ \|f\|_\infty = \max\{|f(x)| : x \in X\} $$

Banach space $M(X)$ is the set of all countably-additive signed ($\mathbb R$-valued) measures on the sigma-algebra $\mathcal B$. The norm is the total variation: for $\mu \in M(X)$, write $\mu = \mu^+ - \mu^{-}$ where $\mu^+$ and $\mu^-$ are positive measures, singular to each other, and let $$ \|\mu\|_1 = \mu^+(X)+\mu^-(X) $$

The pairing $M(X) \times C(X) \to \mathbb R$ defined by $$ \langle\mu,f\rangle = \int_X f \;d\mu $$ identifies $C(X)^\ast$ isometrically with $M(X)$ in these norms.

Another description of $M(X)$ may be obtained as follows. Let $\{\tau_i : i \in I\}$ be a maximal (under inclusion) family of mutually singular probability measures on $(X,\mathcal B)$. (Such a family exists by Zorn's Lemma. But it is not unique.) Identify $M(X)$ isometrically with the $l^1$-direct sum of the family of all the $L^1(\tau_i)$ spaces: $$ M(X) \approx \left(\bigoplus_{i\in I} L^1(X,\mathcal B,\tau_i)\right)_{1} $$

Once we describe $M(X)$ in this way, we get the corresponding description of the dual as the $l^\infty$-sum of the spaces $L^\infty(\tau_i)$. $$ C(X)^{\ast\ast}\approx \left(\bigoplus_{i\in I} L^\infty(X,\mathcal B,\tau_i)\right)_{\infty} $$

GEdgar
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