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Let $C_b(\Omega,V )=$ { $ f:\Omega\rightarrow V $ } is the Banach space of all bounded continuous functions in Banach space $V$ with a norm $\|\cdot\|$ defined as $\|f\|_\infty=\sup _{x\in\Omega}\|f(x)\|$. Let $C_b(\Omega)=C_b(\Omega,\mathbb R)$. For a normal topological space $\Omega$ ( $T_4$-space) it holds that

$$ C_b(\Omega)^*=rba(\Omega), $$

where $rba(\Omega)$ is the space of regular bounded finitely additive measures, and also $$ x^*f=\int\limits_{\Omega}f(\omega)\mu(d\omega),\quad f\in C_b(\Omega),\quad x^*\in C_b(\Omega)^*,\quad \mu\in rba(\Omega) $$

Are there more precise results for the case $C_b (\mathbb R)$? Particularly I'm interested in more "beautiful" presentation of the measure $\mu$.

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3  
The dual of $C_b({\mathbb R})$ is the space of Radon measures on the Stone-Cech compactification of ${\mathbb R}$ -- I don't know if that is beautiful enough. What kinds of result were you hoping to find? –  Yemon Choi Dec 16 '11 at 5:14
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A related question: mathoverflow.net/questions/44183/… In particular, Grey Kuperberg's answer makes me skeptical you will get a "good" answer... –  Matthew Daws Dec 16 '11 at 7:50
    
Thank you. Could you please clarify the form of the functional? Will it be the same? $$ x^*f=\int\limits_{\beta\mathbb R}f(\omega)\mu(d\omega) $$ And how we understand $f(\omega)$ when $\omega\in\beta\mathbb R\setminus \mathbb R$? By the "beautiful" I meant, for example in our case, that functional consists of two parts: integral on $\mathbb R$ and on $\beta\mathbb R\setminus \mathbb R$ (because I don't know about measurability in $\beta\mathbb R$) –  Mariarty Dec 16 '11 at 8:38
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It follows from the universal properties of $\beta{\mathbb R}$ that any continuous bounded function on ${\mathbb R}$ extends uniquely to a continuous function on $\beta{\mathbb R}$, so that is how you define $f(\omega)$. One gets the representing measure $\mu$ by applying the Riesz representation theorem for $C(\beta{\mathbb R})$. If you are asking for an "explicit" description of the representing measure, then I think the question that Matthew Daws links to suggests that this is unlikely –  Yemon Choi Dec 16 '11 at 9:49
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Since the OP is satisfied with the answers given in the comments, I vote to close. –  Bill Johnson Dec 16 '11 at 23:14
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