13
$\begingroup$

I see that there are various definitions for Radon measure and they are NOT equivalent, but they are equivalent on locally compact Hausdorff spaces. I think this is the reason why Radon measure has several different definitions, but I'm curious what is the standard one. On locally compact Hausdorff spaces, representation theorem let Radon measure play an important role. However, I think there would be some areas in mathematics that Radon measure still plays an important role when the given space is NOT locally compact hausdorff space. In that situation, what is the definition of Radon measure? (For example, under the definition given in Folland's Real Analysis, I found that Vitali-Carathéodory theorem is true for Radon measure on a Hausdorff space. (Not necessarily locally compact))

$\endgroup$

1 Answer 1

11
$\begingroup$

You don't give explicitly the options you are choosing between, so I am not a priori certain if I am answering your question. However, I do want to give it a try.

For all of the details, see: https://en.wikipedia.org/wiki/Radon_measure (honestly this is mostly paraphrasing from the relevant articles there, although with some commentary of my own included and large parts of the content from there omitted).

Radon measure - a measure $\mu$ which satisfies two properties:

  1. $\mu$ is locally finite

  2. $\mu$ is inner regular

Radon measures are only defined for Hausdorff spaces. This is because of both the local finiteness condition as well as the inner regularity condition. Just to make sure we are on the same page:

locally finite measure - a measure $\mu$ on a $\sigma-$algebra $\Sigma$ of a Hausdorff topological space $(X, \tau)$ such that:

  1. $\tau \subset \Sigma$

  2. For every $x \in X$, there exists an open neighborhood (i.e. open set) $U_x \in \tau$ such that $x\in U_x$ and $$\mu(U_x) < \infty$$

Note that the second condition relies upon the first, because otherwise we are not guaranteed that the expression $\mu(U_x)$ is even defined.

inner regular measure - a measure $\mu$ on a $\sigma$-algebra $\Sigma$ of a Hausdorff topological space $(X, \tau)$ such that:

  1. $\tau \subset \Sigma$

  2. For every set $A \in \Sigma$, and for every $\varepsilon > 0$, there exists a compact subset $K \subset X$ such that $$\mu(A \setminus K):=\mu(A-K) < \varepsilon$$

Note that an equivalent formulation of the second condition for inner regularity is:

$$\mu(A) = \sup\{\mu(K) :\ K\text{ is compact}, K \subset A \}$$

Also note that because $(X,\tau)$ is by assumption Hausdorff, all compact sets $K$ are closed (this is not true for general topological spaces). This makes inner regularity a true dual notion to outer regularity, at least in those spaces satisfying the Heine-Borel theorem, since then it means "every measurable set can have its measure be approximated from inside by closed (and bounded) sets", while outer regularity means "every measurable set can have its measure be approximated from outside by open sets". Both are true for Lebesgue measure on the real line (for which Heine-Borel also holds), but one probably sees outer regularity used more often in general in measure theory (for example in the Caratheodory extension theorem) because the open-ness condition extends more elegantly to general spaces which don't satisfy the Heine-Borel theorem, in contrast to inner regularity, for which one must switch explicitly to compact sets from closed sets in order to not lose generality; but one loses a certain degree of intuition in the process.

Radon measures are meant to some extent to be generalizations of both the Lebesgue and Dirac measures on the real line, since they interact well with the underlying topology of the space and because the measure of points does not have to be zero (in contrast to the Lebesgue measure).

Technically we can extend the definition to non-Hausdorff spaces by changing "compact" to "closed-compact" in the inner regularity property, but this seems not to be useful, so it is almost never done. (Note that even in the non-Hausdorff case a Radon measure has to be defined on a $\sigma-$algebra containing the Borel $\sigma-$algebra, and thus the underlying topology, in order for both the local finiteness as well as the inner regular conditions to make sense. We need "closed compact" and not just compact, so that $X \setminus K$ is an open set, and thus $A \setminus K = A \cap U$ for some $U \in \tau$. )

Locally Compact Hausdorff spaces: On these spaces (but these spaces only, not on general Hausdorff spaces), Radon measures are particularly nice. In this case, Radon measures can be expressed in terms of continuous linear functionals on the space of continuous functions of compact support. (Note that since our space is Hausdorff having compact support means having closed support, which is good because one usually wants the support to be closed, or to define the support to be the closure of the set on which the function is non-zero.)

Specifically, in the case of a locally compact Hausdorff space, the space of continuous functions with compact support $C_C (X, \tau)$ is a vector space, and every Radon measure $\mu$ on $(X, \tau)$ defines a continuous linear functional on $C_C(X, \tau)$ of the form:

$$ f \mapsto \int f \mathrm{d}\mu$$

By the Riesz-Markov-Kakutani theorem, every (positive) continuous linear functional on $(X, \tau)$ can be written in the above form. Thus some authors define Radon measures in this manner for locally compact Hausdorff spaces only, and ignore the case of Radon measures on general Hausdorff spaces entirely (e.g. Bourbaki, Hewitt & Stromberg, Dieudonne).

$\endgroup$
3
  • 2
    $\begingroup$ I think a Radon measure only needs to be inner regular on open sets, not all measurable sets. Otherwise, the Haar measure of $\Bbb{R}/\Bbb{Z}\times \Bbb{R}_d$ (where $\Bbb{R}_d$ is the additive group of real numbers with the discrete topology, and $\Bbb{R}/\Bbb{Z}$ is the circle group with the usual topology) is not Radon. $\endgroup$
    – Cronus
    Nov 30, 2017 at 21:54
  • 1
    $\begingroup$ @Cronus You may very well be right. Radon measures are not my research area. The term may also not be used to consistently to mean the same thing throughout the literature. $\endgroup$ Dec 1, 2017 at 0:27
  • 1
    $\begingroup$ Yeah, you're right, I think it's really not used consistently (which is too bad!). $\endgroup$
    – Cronus
    Dec 1, 2017 at 16:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .