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For $X$ locally compact (let's take $X=\mathbb{R}^d$), we know that the dual of $C_0(X)$ is $M(X)$, the space of regular borel measures on X. $C_0(X)$ is separable but is $M(X)$ separable? I have tried searching but haven't seen the result nowhere, so I think not.

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As long as $X$ is countable, $M(X)$ is separable. If $X$ is countable, the dual is isomorphic to $\ell_1$ hence separable. If $X$ is uncountable, Dirac measures form an uncountable, discrete set. It cannot happen in a separable metric space.

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I'm thinking of X=R^d typically, so not countable. –  user44670 Jan 14 '13 at 1:21
    
For the countable case look at the Mazurkiewicz-Sierpiński theorem. For the uncountable one, this is folklore. –  J. Lund Jan 14 '13 at 1:23
    
Ok, so it's not true that $M(\mathbb{R}^d)$ is separable. –  user44670 Jan 14 '13 at 1:24
    
The "Dirac measures" argument shows that $M(\mathbb{R}^n)$ contains $\ell_1(\mathfrak{c})$ hence it is very inseparable. –  J. Lund Jan 14 '13 at 1:28
    
Mazurkiewicz-Sierpiński shows much more than what you need here: points are closed, hence for countable $X$ we have $\mathcal{P}(X) = \mathcal{B}(X)$ and you know the measures on $\mathbb{N}$. –  Martin Jan 14 '13 at 1:30

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