Upon reviewing some basic real analysis I have encountered two different definitions for Radon measure. Let the underlying space $X$ be locally compact and Hausdorff. Folland's Real Analysis gives the definition

A Radon measure is a Borel measure that is finite on all compact sets, outer regular on Borel sets, and inner regular on open sets.

Folland goes on to prove that a Radon measure is inner regular on $\sigma$-finite sets, and remarks that full inner regularity is too much to ask for, especially in the context of the Riesz representation theorem for positive linear functionals on $C_c(X)$. Folland's approach seems to match the approach taken by Rudin, if I recall.

However, I've heard from others, as well as Wikipedia, that a Radon measure is defined as a Borel measure that is locally finite (which means finite on compact sets for LCH spaces) and inner regular, and no mention of outer regularity.

Neither definition seems to connect well with Bourbaki's approach of defining Radon measures as positive linear functionals on $C_c(X)$, because, at least according to Wikipedia's article on the Riesz representation theorem, a positive linear functional on $C_c(X)$ uniquely corresponds to a regular Borel measure, which is stronger than Radon in either of the two definitions given above.

Sadly I do not have any more advanced analysis treatises to compare against, so I was hoping somebody could clear up this discrepancy.


One standard example is the reals numbers times the reals with the discrete topology: $X = \mathbb{R} \times \mathbb{R}_d$.

This is a locally compact metrizable space. The compact subsets intersect only finitely many horizontal lines and each of those non-empty intersections must be compact. A Borel set $E\subset X$ intersects each horizontal slice $E_y$ in a Borel set.

Consider the following Borel measure where $\lambda$ is Lebesgue measure on $\mathbb{R}$: $$ \mu(E) = \sum_{y} \lambda(E_y). $$ This is easily checked to define an inner regular Borel measure and its null sets are precisely those Borel sets that intersect each horizontal line in a null set. In particular, the diagonal $\Delta = \{(x,x) : x \in \mathbb{R}\}$ is a null set. However, every open set containing $\Delta$ must intersect each horizontal line in a set of positive measure, so it must have infinite measure and hence $\mu$ is not outer regular.

Now define $\nu$ by the same formula as $\mu$ if $E$ intersects only countably many horizontal lines, and set $\nu(E) = \infty$ if $E$ intersects uncountably many horizontal lines. Now this measure $\nu$ is inner regular on open sets and outer regular on Borel sets.

Finally, you can check that $\mu$ and $\nu$ assign the same integral to compactly supported continuous functions in $X$.

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    $\begingroup$ A good discussion of the problems you run into without $\sigma$-compactness in Arveson's notes and also Lanford's notes from which the above discussion is taken. $\endgroup$ – Hyperspace Oct 17 '12 at 0:29
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    $\begingroup$ I'm trying to understand how to resolve the discrepancy based on your answer. Does this mean that the two definitions of Radon measure as given above both lead to uniqueness results in the Riesz representation theorem, but that they are simply different? Is the result given on Wikipedia false? $\endgroup$ – Christopher A. Wong Oct 17 '12 at 1:41
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    $\begingroup$ @ChristopherA.Wong: There are two versions of uniqueness in the Riesz representation theorem: there is a unique inner regular measure and there is a unique quasi-regular measure (outer regular on Borel sets and inner regular on open sets). Quasi-regular measures are inner regular on measurable sets of $\sigma$-finite measure, so the measure $\nu$ here is the one you get from the Riesz representation theorem with regularity as stated on Wikipedia before the theorem. You can get distinct measures. The link to regular Borel measures in the theorem there is wrong without further assumptions on X. $\endgroup$ – commenter Oct 17 '12 at 7:16
  • $\begingroup$ @commenter, Perfect, this is exactly what I'm looking for. Now the only sad thing for me is that this means there are two different definitions of Radon measure floating around, each corresponding to a very similar representation theorem. $\endgroup$ – Christopher A. Wong Oct 17 '12 at 8:38
  • $\begingroup$ @Christopher: Both versions have their virtues and their drawbacks---(go beyond local compactness and things get really hairy, see e.g. Bogachev or Fremlin's volume 4, chapter 43... :-)) Lanford's notes construct first a quasi-regular version of a functional and then modify it to an inner regular one. A technical point to think about: for inner regular measures there is no difference between locally null sets ($\mu(N \cap K) = 0$ for compact $K$) and null sets. The diagonal $\Delta$ in Hyperspace's answer is $\mu$-null and locally $\nu$-null but not $\nu$-null, as $\nu(\Delta) = \infty$... $\endgroup$ – commenter Oct 17 '12 at 9:31

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