Are Borel Measures on Compact Metric Spaces Regular?

Yes—under the standard assumption that the Borel measure is finite. If $X$ is a compact metric space, $[?]$ and $E$ is a Borel subset of $X$, then $E$ can be enclosed by open sets whose measures approach $[?]$, and it contains compact sets whose measures approach $[?]$. This is the regularity property used throughout measure theory and analysis.

The word “Borel measure” does not by itself always imply finiteness. The theorem below is stated for finite measures; a locally finite Borel measure on a compact space is finite, since the whole space is compact. If a text uses “regular Borel measure” as a defined term, check its convention and assumptions.

What regularity means

A measure $\mu$ on the Borel sets of $X$ is regular if every Borel set $E$ satisfies both

$$\mu(E)=\inf\{\mu(U): E\subseteq U,\ U\text{ open}\}$$

and

$$\mu(E)=\sup\{\mu(K): K\subseteq E,\ K\text{ compact}\}.$$

These are called outer regularity and inner regularity, respectively. The equalities mean that for every $\varepsilon>0$, there are an open $U$ and a compact $K$ such that $K\subseteq E\subseteq U$ and

$$\mu(U\setminus E)<\varepsilon,\qquad \mu(E\setminus K)<\varepsilon.$$

The sets $U$ and $K$ need not equal $E$. Regularity says they can approximate it in measure.

Why open sets have compact inner approximations

Let $U\subseteq X$ be open, and suppose first that $U\ne X$. For each positive integer $n$, define

$$K_n=\{x\in X: d(x,X\setminus U)\ge 1/n\}.$$

The distance-to-a-set function is continuous, so $K_n$ is closed in $X$. Since $X$ is compact, each $K_n$ is compact. Also, $K_n\subseteq U$, the sets increase with $n$, and every point of $U$ has positive distance from the complement in some sufficiently small neighborhood. Therefore $U=\bigcup_{n=1}^{\infty}K_n$.

Continuity of measure from below now gives

$$\mu(U)=\lim_{n\to\infty}\mu(K_n)=\sup_n\mu(K_n).$$

If $U=X$, simply take $K_n=X$. Thus every open set is inner regular. It is outer regular immediately, because the set itself is an open superset.

Extending the result to all Borel sets

Consider the collection of sets that have both compact inner approximations and open outer approximations. It contains every open set by the preceding argument. For a finite measure, this collection is closed under complements: a compact approximation to a set gives an open approximation to its complement by taking the complement of the compact set, and an open approximation gives a compact approximation to the complement in the same way.

It is also closed under countable unions. For outer approximation, choose an open superset for each set with errors whose sum is less than a chosen $\varepsilon$, then take their union. For inner approximation, first approximate the countable union in measure by a finite partial union, using continuity from below. Approximate each of those finitely many sets from inside by a compact set; the finite union of the compact approximations is compact. These arguments use finiteness of $\mu(X)$, in particular when passing between a set and its complement.

The collection is therefore a sigma-algebra containing the open sets, so it contains the Borel sigma-algebra they generate. This proves regularity for every Borel set.

Worked example on the unit interval

Take $X=[0,1]$ with the usual metric and Lebesgue measure, and let $E=[0,1/2)$. For $0<\delta<1/4$, choose

$$K_\delta=[\delta,1/2-\delta],\qquad U_\delta=[0,1/2+\delta).$$

These are, respectively, compact and relatively open in $[0,1]$, and $K_\delta\subseteq E\subseteq U_\delta$. Their measure errors are

$$\mu(E\setminus K_\delta)=2\delta,\qquad \mu(U_\delta\setminus E)=\delta.$$

Given any $\varepsilon>0$, choosing $\delta<\min(\varepsilon/2,1/4)$ makes both errors less than $\varepsilon$. The compact set misses only a short interval near the left endpoint of $E$ and a short interval near its right endpoint; the open set adds only a short interval beyond $1/2$.

Common mistakes

• Assuming every Borel set is compact or open. Regularity provides approximating sets; it does not say the Borel set itself has either property.
• Omitting the finiteness assumption. Continuity from above and the complement argument require care for infinite measures. A general infinite Borel measure should not automatically be treated as covered by the finite-measure theorem.
• Confusing compactness with boundedness. Compactness here follows because the approximating sets are closed subsets of the compact space $X$. In a general metric space, closed sets need not be compact.
• Reversing the inclusions. Inner approximation uses $K\subseteq E$ with $K$ compact; outer approximation uses $E\subseteq U$ with $U$ open.