I need to use the version of this theorem about the dual space of $C_0(X)$, which is the set of continuous functions on X which vanish at infinity. I found in Wikipedia that: Thm from Wiki

Here the space $X$ is needed to be LCH space. But I also found in Page 383 of Conway's book A course in functional analysis 2nd that:

Thm from Conway's book

Here $M(X)$ denotes the space of regular Borel measures and the space $X$ is not need to be Hausdorff. Now my puzzles are:

1)Does the theorem apply for a general locally compact space which is not necessarily Hausdorff?

2)Conway said "$M(X)$ is a normed space, the norm $\left\|\mu\right\|=\left|\mu\right|(X)$". But I think not every $\mu$ in $M(X)$ is finite, such as the Lebesgue measure of the real line. So should the definition $M(X)$ be modified to "the space of finite regular Borel measures"? In this sens, $M(X)$ is a well-defined normed space and the isometrically isomorphism gets its meaning.

Is there somebody who has read Conway's Book and could you help me? Thanks very much!

  • $\begingroup$ De Groot duality is a duality between compact Haudorff partially ordered spaces and stably compact spaces. While stably compact spaces $X$ do not have enough continuous functions from $X$ to $\mathbb{R}$, they certainly have enough upper semicontinuous and lower semicontinuous functions that are all measurable and from which one can recover the measure. $\endgroup$ Jan 23 at 14:21

1 Answer 1


Theorem 1 is essentially the representation theorem (or the "representation definition" in Bourbaki) for the Alexandroff one point compactification of a T$_2$ locally compact space (the difference between continuous functions with value 0 at infinity and all continuous functions with any finite value at infinity does not change the dual, except for the delta measures concentrated on the point at infinity). And yes, the dual (for a compact T$_2$ space) is that of finite regular Borel measures or equivalently that of finite measures on the Baire $\sigma$-algebra, the minimum one that makes measurable all continuous functions (all such Baire measures are regular; Baire is smaller, being generated by the 0-sets of continuous functions i.e. closed=compact G$_\delta$ sets instead of all closed=compact sets).

It does not apply when there are not enough continuous (real / complex) valued functions, so you can apply it to a space whose associated T$_0$ is T$_2$ locally compact, but not in general. I think that Bourbaki has something in exercises for (possibly non T$_1$) quasi-compact and normal spaces.

To see easy examples of quasi-compact T$_0$ spaces with too few real or complex continuous functions, consider finite T$_0$ spaces, which are the same thing as finite posets (with principal order filters as basis for the topology).

Edit. Incorporating bathalf15320 absolutely correct comment: if you want the dual of the space of continuous bounded functions on an arbitrary space, you note that by Gelfand duality you are considering the maximum, universal compactification of the T$_0$ complete regularization of the initial space, hence the dual is given by the Radon measures on that compact T$_2$ space. But the universal compactification is much more difficult to understand than the one-point compactification (the minimum one, which exists only for locally compact T$_2$ spaces).

As for T$_2$, already the old general topology book by Kelly noted (essentially) that what one needs in analysis is not T$_2$, but that the associated T$_0$ space (identify points that are in the same closed sets i.e. in the same open sets; example: in spaces of measurable functions, identify functions a.e. equal; more generally, for topologies and uniformities defined by a family of pseudo-metrics, one identifies points at 0 distance for all such pseudo-metrics) has a series of properties, in ascending order:

T$_2$: equivalent to unicity of limits;

T$_3$ (regular and T$_0$): equivalent to the previous one plus "the unique possible way to extend by continuity a function (i.e. extend taking the limit) always works giving a continuous function"

T$_{3+1/2}$ (T$_0$ and completely regular: points and closed sets are separated by a continuous real valued function, equivalently: topology induced by a family of pseudo-metrics, equivalently: by a uniformity): equivalent to embeddable as subspace of a compact T$_2$ space (or also equivalent to "subspace of a normal T$_1$ space").

[Having more open sets than a T$_{3+1/2}$ topology is equivalent to "distinct points are separated by a continuous real valued function"; it is independent from T$_3$ and lies between T$_2$ and T$_3+1/2$, but usually when one talks about "sufficiently many continuous functions" means the stronger condition T$_3+1/2$.]

Now, sometimes one defines "locally compact" as "each point has a basis of quasi-compact closed neighborhoods", which is the same as "regular and each point has a quasi-compact neighborhood" or also "the associated T$_0$ space is locally compact T$_2$". [Kelly's book used very much the "regular" trick to avoid Hausdorff].

  • $\begingroup$ Thanks a lot. But I have a question: Do you mean that there are not enough continuous functions in a non−Hausdorff space in the last paragraph of your answer? $\endgroup$
    – shepherd
    Jan 23 at 10:55
  • 1
    $\begingroup$ As you surmise the problem in the non Hausdorff case is the lack of continuous functions but note that many authors embed the former condition in their definition of local compactness. In fact, the result can be extended to completely regular spaces but working with the space of bounded continuous functions, not with the norm but with Buck's strict topology, the finest l.c. topology which agrees with compact convergence on the unit ball. There are various ploys to work in the non $T_2$ case but you would have to be highly motivated to want to try them. $\endgroup$ Jan 23 at 12:34

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