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I am trying to understand the notion of Radon measure, but I am a little bit lost with the different conventions used in the litterature. More precisely, I have a doubt about the very definition of Borel measure.

Suppose that $(X,\mathcal{B},\mu)$ is a measure space, where $X$ is a topological space. I have find two different definitions for "$\mu$ is a Borel measure":

-Def 1 : $\mu$ is a Borel measure if $\mathcal{B}$ contains the Borel $\sigma$-algebra of $X$,

-Def 2: $\mu$ is a Borel measure if $\mathcal{B}$ is exactly the Borel $\sigma$-algebra of $X$.

The same thing happens for the notion of Radon measure, as it can be either considered as Borel measure in the sense of Def 1, or in the sense of Def 2.

Of course, Def 1 gives a more general notion of Borel or Radon measure. For example the Lebesgue measure (defined on the Lebesgue $\sigma$-algebra of $\mathbb{R}^n$) is Radon in the sense of Def 1, but not in the sense of Def 2.

Are there (other) reasons as to why one may prefer Def 1 to Def 2 or vice versa ?

Apparently, Def 2 makes it quite difficult to have a "complete Radon measure", which makes me think that it is a little bit artificial or restrictive. But maybe many results hold only for Radon measures in the sense of Def 2, without possible extension to Radon measures in the sense of Def 1 ? Or maybe there is a trivial way to transfer any result involving a Borel measure in the sense of Def 2 to a result involving a Borel measure in the sense of Def 1 ?

A related question is the following : if $\mu$ is Radon in the sense of Def 2, will its completion be Radon in the sense of Def 1 ? Same question when you replace "Radon" by "inner regular", "outer regular", and "locally finite".

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I'm not familiar with definition 1 for Borel measures, but the main poin of Radon measures is that they are inner regular, so the measurable sets can be approximated by Borel sets anyways. –  Michael Greinecker Oct 13 '12 at 10:20
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According to Bourbaki's definition, a Radon Measure is a certain kind of linear functional on a certain kind of space of continuous functions. So to start with it is not even defined on Borel sets. –  Gerald Edgar Oct 13 '12 at 16:10
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2 Answers

Let $(X,\mathcal M, \mu)$ be a measure space, where $\mu$ is a positive measure and $X$ is topological space. Let $\mathcal B$ the Borel $\sigma$-algebra on $X$.

The measure $\mu$ is called a Borel measure whenever $\mathcal M\supset \mathcal B$ and $\mu$ is finite on compact sets.

A Radon measure $L$ on $X$ is a continuous linear form on the vector space $C_c(X;\mathbb R)$ (real-valued continuous functions with compact support). The celebrated Riesz-Markov representation theorem establishes that if $X$ is a locally compact space and $L$ is positive (i.e. non-negative on non-negative functions) then there exists a complete outer regular measure space $(X,\mathcal M, \mu)$ such that $\mu$ is a Borel measure and $$ Lf=\int_Xfd\mu,\quad\text{for $f\in C_c(X;\mathbb R)$}. $$ Inner regularity is true when $X$ is $\sigma$-compact. Walter Rudin classical book, Real and Complex Analysis remains the best reference in the literature.

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From a geometric measure theory perspective, it is standard to define Radon measures $\mu$ to be Borel regular measures that give finite measure to any compact set. Of course, their connection with linear functionals is very important, but in all the references I know, they start with a notion of a Radon measure and then prove representation theorems that represent linear functionals by integration against Radon measures.

Here are some examples:

$\color{blue}{I:}$ Evans and Gariepy's Measure Theory and Fine Properties of Functions states it this way:

  1. A [outer] measure $\mu$ on $X$ is regular if for each set $A \subset X$ there exists a $\mu$-measurable set $B$ such that $A\subset B$ and $\mu(A)=\mu(B)$.
  2. A measure $\mu$ on $\Bbb{R}^n$ is called Borel if every Borel set is $\mu$-measurable.
  3. A measure $\mu$ on $\Bbb{R}^n$ is Borel regular if $\mu$ is Borel and for each $A\subset\Bbb{R}^n$ there exists a Borel set $B$ such that $A\subset B$ and $\mu(A) = \mu(B)$.
  4. A measure $\mu$ on $\Bbb{R}^n$ is a Radon measure if $\mu$ is Borel regular and $\mu(K) < \infty$ for each compact set $K\subset \Bbb{R}^n$.

$\color{blue}{II:}$ In De Lellis' very nice exposition of Preiss' big paper, he doesn't even define Radon explicitly, but rather talks about Borel Regular measures that are also locally finite, by which he means $\mu(K) < \infty$ for all compact $K$. His Borel regular is a bit different in that he only considers measurable sets -- $\mu$ is Borel regular if any measurable set $A$ is contained in a Borel set $B$ such that $\mu(A) = \mu(B)$. (I am referring to Rectifiable Sets, Densities and Tangent Measures by Camillo De Lellis.)

$\color{blue}{III:}$ In Leon Simon's Lectures on Geometric Measure Theory, he defines Radon measures on locally compact and separable spaces to be those that are Borel Regular and finite on compact subests.

$\color{blue}{IV:}$ Federer 2.2.5 defines Radon Measures to be measure a $\mu$, over a locally compact Hausdorff spaces, that satisfy the following three properties:

  1. If $K\subset X$ is compact, then $\mu(K) < \infty$.
  2. If $V\subset X$ is open, then $V$ is $\mu$ measurable and

    $\hspace{1in} \mu(V) = \sup\mu(K): K\text{ is compact, } K\subset V$

  3. If $A\subset X$, then

    $\hspace{1in} \mu(A) = \inf\mu(V): V\text{ is open, } A\subset V$

Note: it is a theorem (actually, a Corollary 1.11 in Mattila's Geometry of Sets and Measures in Euclidean Spaces) that a measure is a Radon a la Federer if and only if it is Borel Regular and locally finite. I.e {Federer Radon} $\Leftrightarrow$ {Simon or Evans and Gariepy Radon}. (I am referring of course to Herbert Federer's 1969 text Geometric Measure Theory.)

$\color{blue}{V:}$ For comparison, Folland (in his real analysis book) defines things a bit differently. For example, he defines regularity differently than the first, third and fourth texts above. In those, a measure $\mu$ is regular if for any $A\subset X$ there is a $\mu$-measurable set $B$ such that $A\subset B$ and $\mu(A) = \mu(B)$. In Folland, a Borel measure $\mu$ is regular if all Borel sets are approximated from the outside by open sets and from the inside by compact sets. I.e. if

$\hspace{1in}\mu(B) = \inf \mu(V): V\text{ is open, } B\subset V$

and

$\hspace{1in}\mu(B) = \sup \mu(K): K\text{ is compact, } K\subset B$

for all Borel $B\subset X$.

Folland's definition of Radon is very similar to Federer's but not quite the same:

A measure $\mu$ is Radon if it is a Borel measure that satisfies:

  1. If $K\subset X$ is compact, then $\mu(K) < \infty$.
  2. If $V\subset X$ is open, then

    $\hspace{1in} \mu(V) = \sup\mu(K): K\text{ is compact, } K\subset V$

  3. If $A\subset X$ and $A$ is Borel then

    $\hspace{1in} \mu(A) = \inf\mu(V): V\text{ is open, } A\subset V$

... and by Borel measure, Folland means a measure whose measuralbe sets are exactly the Borel sets.

Discussion: Why choose one definition over another? Partly personal preference -- I prefer the typical approach taken in geometric measure theory, starting with an outer measure and progressing to Radon measures a la Evans and Gariepy or Simon or Federer or Mattila. It seems, somehow, more natural and harmonious with the Caratheodory criterion and Caratheodory construction used to generate measures, like the Hausdorff measures.

With this approach, for example, sets with an outer measure of 0 are automatically measurable.

Another reason not to use the more restrictive definition 2 (in the question above) it makes sense to require that continuous images of Borel sets be measurable. But all we know is that continuous maps map Borel to Suslin sets. And there are Suslin sets which are not Borel! If we use the definition of Borel regular, as in I,III and IV above, then Suslin sets are measurable. There is a very nice discussion of this in section 1.7 of Krantz and Parks' Geometric Integration Theory -- see that reference for the definition of Suslin sets. (Krantz and Parks is yet another text I could have added to the above list that agrees with I, III, and IV as far as Radon, Borel regular, etc. goes.

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What would be a good example of a non-Borel regular Borel measure on a second countable metric space? All the standard procedures for constructing measures I'm aware of seem to yield measures satisfying this condition. The examples I was able to produce either live on large (i.e. non-separable) spaces or fail to measure all Borel sets. –  Theo Buehler Jan 1 '13 at 22:51
    
Quick answer is that I don't know (haven't thought about it). I have student who loves counterexamples and is something of an expert on them. I will ask him. –  Kevin R. Vixie Jan 2 '13 at 1:55
    
Thanks! Meanwhile, I remembered an old construction due to Oxtoby dx.doi.org/10.1090/S0002-9947-1946-0018188-5 which produces a non-trivial (invariant and inner regular) Borel measure on every separable completely metrizable group. If the group is not locally compact then every open set has infinite measure, but there are always sets of finite measure, thus Borel regularity fails. This construction can be adapted to give a non-Borel regular Borel measure even on $\mathbb{R}$, by working on the set of irrationals and using that they are homeomorphic to $\Bbb{Z^N}$. –  Theo Buehler Jan 4 '13 at 16:42
    
To get your (Oxtoby's) example are you using the version of regularity used by Folland? I assume so ... –  Kevin R. Vixie Jan 4 '13 at 19:00
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