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First recall the definition of Borel regular finite measure $\mu$ on $\mathcal{P}([0,1])$

  1. Every Borel set is $\mu$ measurable.
  2. For every subset $A\subset [0,1]$, there exists a Borel set $B$ such that $A\subset B$ and $$\mu(A) = \mu(B).$$ Measures only satisfy condition $(1)$ are called Borel measures, and if $(2)$ is satisfied as well are called Borel regular measures .

My question is how do we intuitively understand the "regularity" here?

I think it has to do with the idea of continuity.

Look at the space of bounded measurable functions $B([0,1])$ with the uniform norm, its continuous dual space is the space of finite additive finite valued Borel measures.

If we look at the space of bounded continuous functions $C_b([0,1])$ with the uniform norm, its continuous dual space is the space of finite additive finite valued Borel regular measures.

How did continuity play the roll here to force the measures to be regular? And is there any examples where a Borel non-regular measure is clearly a continuous linear functional on $B([0,1])$ but not on $C_b([0,1])$?

Edit: And reading from wiki, I just want to add:

without the assumption of regularity in general, the uniqueness will fail. That is two Borel non-regular measures can act as two different continous linear functionals on bounded measurable functions. But when restricted to the continuous functions, their integrations give the same continuous linear functional.

Is there any good examples where we can see this?

I don't quite understand the example given in wiki

Without the condition of regularity the Borel measure need not be unique. For example, let X be the set of ordinals at most equal to the first uncountable ordinal Ω, with the topology generated by "open intervals". The linear functional taking a continuous function to its value at Ω corresponds to the regular Borel measure with a point mass at Ω. However it also corresponds to the (non-regular) Borel measure that assigns measure 1 to any measurable subset of the space of ordinals less than Ω that is closed and unbounded, and assigns measure 0 to other measurable subsets.

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  • $\begingroup$ I don't think you are saying what you mean, when you say, in effect, that every subset $A$ of the interval is $\mu$-measurable... because this is not possible with a finite measure. (It would be possible with a counting measure or weighted counting measure, but the measure of the whole interval would be infinite, because an uncountable sum of positive reals is $+\infty$). $\endgroup$ – paul garrett Oct 22 '17 at 22:22
  • $\begingroup$ ... oop, unless the measure of only countably-many points is non-zero, positive. :) $\endgroup$ – paul garrett Oct 22 '17 at 22:28
  • $\begingroup$ The motivation may have been to extend the class of measurable sets to include all "small" sets, that is if $A\subset B$ and $\mu (B)=0$ then $A$ is measurable with $\mu (A)=0.$...... E.g. if $\mathcal B$ is the set of Borel subsets of $\Bbb R$ and $\mathcal C$ is the set of all subsets of members of $\mathcal B$ that have measure $0$, then the $\sigma$-algebra generated by $\mathcal B \cup \mathcal C$ is the set of Lebesgue-measurable subsets of $\Bbb R.$ $\endgroup$ – DanielWainfleet Oct 23 '17 at 19:29
  • $\begingroup$ It can be important to distinguish "is" from "corresponds to". Misleading: "the continuous dual space of $C_b[0,1]$ is the space of signed Borel regular measures." Correct: "the continuous dual space of $C_b[0,1]$ corresponds to the space of signed Borel regular measures." $\endgroup$ – Dap Oct 24 '17 at 6:28
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Regarding the non-uniqueness:

For finite measures, uniqueness holds in the Riesz representation if $X$ a perfectly normal space, i.e. every closed set $C$ is a zero set of a continuous function $f_C:X\to [0,1].$ This condition fixes the measure of $C$, because for finite measures we can use the monotone convergence theorem and pointwise convergence $\lim_{n\to\infty}(1-f_C(x))^n=1_C.$ The measure is then fixed on the whole Borel algebra. You can use a similar argument for $\sigma$-finite measures as well (e.g. $\sigma$-compact space + compact sets have finite measure).

In the example from Wikipedia, i.e. the order topology on $\Omega+1$ where $\Omega$ is the first uncountable ordinal, there aren't enough continuous functions to determine the measure of $\{\Omega\}.$ Every continuous function is constant on a neighborhood of $\Omega.$ If $f:\Omega+1\to\mathbb R$ is continuous, by continuity there is an open interval $(\alpha_n,\Omega]$ sent to $(f(\Omega)-1/n,f(\Omega)+1/n).$ A countable union of countable ordinals is countable, so $\sup \alpha_n<\Omega$, which means $f$ is $f(\Omega)$ on the whole interval $[\alpha_n,\Omega]$. In particular $\{\Omega\}$ is not a zero set. This explains why you can have two measures, like $\mu_1(S)=1$ iff $\Omega\in S,$ and $\mu_2(S)=1$ iff $S\setminus\{\Omega\}$ is closed and unbounded, that give identical integrals on $C_b[0,1]$ but different integrals on $B[0,1].$

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What you call a "measure" is more commonly called an "outer measure." When restricted to the Borel sets, a Borel regular outer measure becomes a genuine, countably additive, Borel measure. Conversely, the outer measure generated by a Borel measure is a Borel regular outer measure. So you are really just talking about Borel measures.

That the continuous dual of $C[0,1]$ consists finite signed (!) regular Borel measures is the Riesz representation theorem. Every finite signed Borel measure on $[0,1]$ is regular, so you are really just talking about finite signed Borel measures.

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  • $\begingroup$ Thank you for the reply! I guess I wanted to use $(0,1)$ or $\mathbb{R}$ so the classic Riesz representation could not apply here directly... And reading from wiki, I just want to add: without the assumption of regularity in general, the uniqueness will fail. That is two Borel non-regular measures can act differently on bounded measurable functions, but when restricted to the continuous functions, their integrations give the same linear functional. Is there any classical examples where we can see this? $\endgroup$ – Xiao Oct 24 '17 at 1:46

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