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Let $X$ be a compact Hausdorff space. Let $\mu$ be a Borel, probability measure on $X$. Does it automatically follow that $\mu$ is regular? That is, for all Borel $E \subset X$, must we have $$\mu(E) = \inf \mu(U) = \sup \mu(K)$$ where the infimum is taken over open $U \supset E$ and the supremum is taken over compact $K \subset E$? I know this is true when $X$ is a metric space or, more generally, when every open subset of $X$ is $\sigma$-compact. I imagine this fails in general though.

I have a proposed counterexample which is a long way from being complete, but I'll put it here anyway. Let $I$ be an uncountable index set. Denote the product of $I$ many copies of the discrete space $\{0,1\}$ by $2^I$ and give it the (compact) product topology. I would like to give $2^I$ it's "product measure" as well, but I don't understand products of infinite families of measure spaces. Luckily, $2^I$ is, in a natural way, a compact group, so we can use the Haar measure which does what I wanted the product measure to do. Let $E$ be the set of elements of $2^I$ with countable support.

Issue 1: Is $E$ Borel? I don't see why it should be...

Assuming $E$ is Borel, it must have measure zero. Consider the translates $$xE = \{ y \in 2^I : y_i = x_i \text{ for all but countably many } i \in I\}$$ of $E$. By choosing a sequence of points $x_1,x_2,x_3,\ldots \in X$ such that any two differ in uncountably many coordinates (this is possible), we see that the translates $x_1E,x_2E,x_3E,\ldots$ are disjoint. So, if $E$ had positive measure, we get that $\mu(x_1E \cup x_2E \cup \ldots) = \mu(E) + \mu(E) + \ldots = \infty$ by translation invariance of the measure. This contradicts $\mu(2^I) =1$.

Now, I would like to use $E$ to contradict outer regularity. So how is it that one can find open $U \supset E$? For $F \subset I$ finite, let $U_F = \{x \in 2^I : x_i = 0 \text{ for all } i \in F\}$ (basic open set). I think essentially the only way to cover $E$ by an open set is to choose an uncountable family $F_j$ of disjoint finite subsets of $I$ and to consider $U = \bigcup_j U_{F_j}$. This covers $E$ since, if $x \in E$, then the support of $x$ is countable, so $x$ is nonzero on countably many of the $F_j$, so $x$ is zero on some particular $F_j$ and $x \in U$.

Issue 2: Is it true that $U$ should have $\mu(U)=1$? For some reason I feel maybe it should.

If both of the above issues are resolved, then $2^I$ is not regular since $\mu(E) = 0$ and $\inf \mu(U) = 1$.

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Products of infinite families (of arbitrary cardinality) of probability spaces do work right, see e.g. Fremlin's Measure Theory. –  tomasz Jun 21 '12 at 22:00
    
Also, if I read my notes from measure theory lectures right, any finite Baire measure on a compact space extends uniquely to a Radon (so regular) measure. –  tomasz Jun 21 '12 at 22:08
    
It's occured to me now that Haar measure is Radon by definition (or, if you take a wider definition, it is Radon for locally compact groups), so it cannot be the counterexample you sought. –  tomasz Jun 22 '12 at 1:34
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1 Answer

up vote 3 down vote accepted

Here's a famous example of the counter-intuitive behaviour that you want.

The Dieudonné measure $\mu$ is a Borel probability measure on the compact space $X=[0,\omega_1]$, where $\omega_1$ is the first uncountable ordinal. It has the property that $\mu(K)=0$ for every compact subset of $X\setminus\{\omega_1\}$, yet $\mu(X\setminus\{\omega_1\})=1$.

You can find more details in Volume 2 of Bogachev's Measure Theory Example 7.1.3 (page 68).

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The Borel sets in $[0,\omega_1)$ are exactly the sets containing a closed cofinal subset or are the complement of such a set. Dieudonné's measure takes value $1$ on the former and $0$ on the latter. Now extend this to $[0,\omega_1]$ by setting $\mu(\omega_1) = 0$. Note also that $\mu$ is complete and purely atomic. –  t.b. Jun 21 '12 at 21:59
    
@t.b. Thanks! I was too lazy to give all the details... –  Byron Schmuland Jun 21 '12 at 22:00
    
@ByronSchmuland : I think you're wrong in the first statement. For example, a singleton set is closed and thus Borel, but it's certainly not dependent on countably many axes. You're confusing Borel sets with Baire sets. For large products, the product of Borel algebras is NOT the Borel algebra of the product, but rather its proper subalgebra. –  tomasz Jun 21 '12 at 22:14
    
@tomasz I guess I didn't think that through very carefully. I will remove the offending claim. –  Byron Schmuland Jun 21 '12 at 22:22
    
Thanks Byron and @tomasz. It's a very nice example. I made a few attempts to pare it down to something simpler, but it seems to be optimal. –  Mike Jun 21 '12 at 23:53
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