# Dual of $C_0(X)$, space of all continuous functions vanishing at infinity

Let $$X$$ be locally compact, Hausdorff space. We know that the dual of $$C_0(X)$$( continuous functions vanishing at infinity) is the set of all complex regular Borel measure on $$X$$.

Now consider $$X= \mathbb{N}$$ with usual topology. Then $$X$$ is locally compact, Hausdorff. Let us consider N with counting measure. Then $$C_0(\mathbb{N})$$ is the space of all sequences converging to 0, denoted by $$c_0$$. We know that the dual of $$c_0$$ is $$l_1$$( space of all summable sequences).

My question: (1)Are there any other special type of space for which dual of $$C_0(X)$$ will be $$L_1(X)$$? (2) Is the above true for $$\mathbb{R}$$ with usual topology and Lebesgue measure?

• This basically amounts to asking for which locally compact Hausdorff spaces you have the property that there is some (positive, locally finite) regular Borel measure with respect to which all finite regular Borel measures are absolutely continuous. My guess is that this is true only for discrete spaces. Reals certainly do not satisfy it with the Lebesgue measure. For instance, no discrete measure is absolutely continuous. – tomasz Jun 20 '19 at 17:41
• Well, I may have gone a bit too far with claiming that it is discrete. But this property implies that the measure in question is basically a counting measure (every point has nonzero measure). – tomasz Jun 20 '19 at 17:55

## 1 Answer

Suppose $$\mu$$ is a (regular Borel) measure such that $$L^1(\mu)=C_0(X)^*$$, i.e. for every regular Borel measure $$\nu$$ on $$X$$ we have a function $$f\in L^1(\mu)$$ such that for every Borel $$A\subseteq X$$ we have $$\nu(A)=\int_Af(x)\,\mathrm{d}\mu(x)$$.

This implies that in particular, this holds for all Dirac measures $$\delta_x$$ for $$x\in X$$, which implies that for each $$x\in X$$ we must have $$\mu(\{x\})>0$$ and $$\mu(\{x\})<\infty$$.

Since $$\mu(\{x\})<\infty$$, by regularity, we conclude that the space $$X$$ must be locally countable: if every neighbourhood of a point was uncountable, then they would all have infinite measure, and so by regularity, so would $$\{x\}$$.

I suspect that if $$X$$ is locally compact and locally countable, then you can find such a measure, though I don't see any obvious construction in general.