Local Finiteness and Regularity of the Constructed Measure

From the Construction to Regularity

Let \(X\) be a locally compact Hausdorff space, and let \(C_c(X)\) denote the real-valued continuous functions on \(X\) with compact support. A positive linear functional \(L:C_c(X)\to\mathbb{R}\) satisfies \(L(f)\ge0\) whenever \(f\ge0\). The open-set content used in constructing its measure is

\lambda(U)=\sup\{L(f): f\in C_c(X),\ 0\le f\le1,\ \operatorname{supp} f\subseteq U\}, \qquad U\text{ open}.

The associated outer measure is defined by covering arbitrary sets with open sets:

\mu^*(A)=\inf\left\{\sum_{j=1}^{\infty}\lambda(U_j): A\subseteq\bigcup_{j=1}^{\infty}U_j,\ U_j\text{ open}\right\}.

By the preceding construction, the restriction of \(\mu^*\) to the Borel sigma-algebra \(\mathcal{B}_X\) is a measure \(\mu\), and the open-set content is countably subadditive: if \(U\subseteq\bigcup_j U_j\) with all sets open, then \(\lambda(U)\le\sum_j\lambda(U_j)\). We will use these established properties, together with positivity of \(L\) and the existence of compactly supported continuous cutoffs on locally compact Hausdorff spaces.

Definition: A Borel measure \(\mu\) on \(X\) is locally finite if \(\mu(K)<\infty\) for every compact set \(K\subseteq X\). It is outer regular on Borel sets if \(\mu(E)=\inf\{\mu(U): E\subseteq U,\ U\text{ open}\}\) for every \(E\in\mathcal{B}_X\). It is inner regular on open sets if \(\mu(U)=\sup\{\mu(K):K\subseteq U,\ K\text{ compact}\}\) for every open \(U\).

These are distinct assertions. Local finiteness does not say that \(\mu(X)\) is finite. Outer regularity approximates a Borel set from outside by open sets; inner regularity here approximates an open set from inside by compact sets. The latter is not, by definition alone, a statement about every Borel set.

Compact Sets Have Finite Measure

Theorem: For the measure \(\mu\) constructed from \(L\) on a locally compact Hausdorff space \(X\), every compact set \(K\subseteq X\) has \(\mu(K)<\infty\).

Proof. If \(K=\varnothing\), then \(\mu(K)=0\). Otherwise, local compactness and the Hausdorff property give an open set \(V\) containing \(K\) whose closure is compact. A compactly supported continuous cutoff \(\eta\in C_c(X)\) can be chosen with \(0\le\eta\le1\) and \(\eta=1\) on \(\overline V\). For any \(f\in C_c(X)\) with \(0\le f\le1\) and \(\operatorname{supp}f\subseteq V\), we have \(f\le\eta\) everywhere: on \(V\), \(\eta=1\), and off \(V\), \(f=0\). Positivity of \(L\) gives \(L(f)\le L(\eta)\), so taking the supremum over such \(f\) yields \(\lambda(V)\le L(\eta)<\infty\). The construction gives \(\mu(V)=\mu^*(V)=\lambda(V)\), and monotonicity of the measure gives \(\mu(K)\le\mu(V)<\infty\). \(\square\)

The cutoff is the key step: it bounds every admissible test function for \(V\) by one fixed function on which \(L\) has a finite value. Local compactness provides the relatively compact neighborhood and the cutoff. Without compact support, \(L\) is not even defined on an arbitrary continuous function such as the constant function \(1\) on a noncompact space.

Worked Example: Counting Measure on the Discrete Natural Numbers

Take \(X=\mathbb{N}=\{1,2,\ldots\}\) with the discrete topology. It is locally compact and Hausdorff; its compact sets are exactly its finite sets. Every function with finite support belongs to \(C_c(X)\). Define \(L(f)=\sum_{n\ge1}f(n)\), a finite sum for \(f\in C_c(X)\). This is a positive linear functional.

Every subset \(A\subseteq\mathbb{N}\) is open. For finite \(F\subseteq A\), the function \(\chi_F\) is continuous, has compact support in \(A\), and satisfies \(0\le\chi_F\le1\). Thus \(\lambda(A)\ge L(\chi_F)=|F|\). Conversely, if \(f\) is admissible for \(\lambda(A)\), then it has finite support in \(A\) and \(L(f)\le|\operatorname{supp}f|\). It follows that \(\lambda(A)=|A|\), with value \(+\infty\) when \(A\) is infinite.

For a finite set \(K=\{2,5,8,11\}\), \(\mu(K)=4\). Indeed, the single open set \(K\) covers \(K\) at cost \(\lambda(K)=4\), while any open cover has total cost at least \(4\), since it must cover four distinct points. For \(U=\mathbb{N}\), the finite sets \(\{1,\ldots,N\}\subseteq U\) have measures \(N\); hence \(\sup_{K\subseteq U,\ K\text{ compact}}\mu(K)=+\infty=\mu(U)\). This checks inner regularity in this case and also shows that local finiteness does not imply finite total measure.

Outer Regularity on Borel Sets

Theorem: For every Borel set \(E\in\mathcal{B}_X\), the constructed measure satisfies \(\displaystyle\mu(E)=\inf\{\mu(U):E\subseteq U,\ U\text{ open}\}\).

Proof. For each open \(U\), the one-set cover of \(U\) gives \(\mu^*(U)\le\lambda(U)\). Conversely, if \((U_j)\) is any countable open cover of \(U\), countable subadditivity of the open-set content gives \(\lambda(U)\le\sum_j\lambda(U_j)\). Taking the infimum over all such covers shows \(\lambda(U)\le\mu^*(U)\). Therefore \(\mu(U)=\mu^*(U)=\lambda(U)\) for every open \(U\).

Now fix a Borel set \(E\). Since \(\mu\) is a measure, it is monotone, so \(\mu(E)\le\mu(U)\) whenever \(E\subseteq U\) and \(U\) is open. This gives \(\mu(E)\le\inf_{U\supseteq E}\mu(U)\). In the other direction, consider any countable open cover \(E\subseteq\bigcup_j U_j\), and put \(W=\bigcup_j U_j\). Then \(W\) is open and \(E\subseteq W\). Using \(\mu(W)=\lambda(W)\) and countable subadditivity, we obtain \(\inf_{U\supseteq E}\mu(U)\le\mu(W)=\lambda(W)\le\sum_j\lambda(U_j)\). Taking the infimum over all countable open covers gives \(\inf_{U\supseteq E}\mu(U)\le\mu^*(E)=\mu(E)\). The two inequalities prove the assertion. \(\square\)

Worked Example: A Point Mass on the Real Line

Let \(X=\mathbb{R}\) with its usual topology and let \(L(f)=f(0)\) for \(f\in C_c(\mathbb{R})\). This is positive. If an open set \(U\) contains \(0\), a compactly supported continuous cutoff \(f\) can be chosen with \(0\le f\le1\), support in \(U\), and \(f(0)=1\). Every admissible \(f\) has \(f(0)\le1\), so \(\lambda(U)=1\). If \(0\notin U\), every admissible \(f\) vanishes at \(0\), so \(\lambda(U)=0\).

For any \(A\subseteq\mathbb{R}\), if \(0\notin A\), the open set \(\mathbb{R}\setminus\{0\}\) covers \(A\) at cost zero, and therefore \(\mu^*(A)=0\). If \(0\in A\), every open cover of \(A\) contains at least one set that contains \(0\), so its total cost is at least \(1\). The one-set cover \(\mathbb{R}\) has cost \(1\), giving \(\mu^*(A)=1\). Thus the constructed Borel measure is \(\delta_0\).

For \(U=(-1,2)\), \(\mu(U)=1\). The compact set \(K=\{0\}\subseteq U\) has \(\mu(K)=1\), so the compact approximation reaches the full measure. For \(V=(3,4)\), \(\mu(V)=0\), and the empty compact set has measure \(0\). These computations check inner regularity for both a positive-measure open set and a zero-measure open set.

Inner Regularity on Open Sets

To prove inner regularity, we need to connect a test function supported in an open set to the measure of a compact subset. The crucial estimate is that if \(f\in C_c(X)\), \(0\le f\le1\), and \(K=\operatorname{supp}f\), then \(L(f)\le\mu(K)\). We give the cover argument in detail.

Proof of the estimate. If \(K=\varnothing\), then \(f=0\) and the claim follows. Otherwise, let \((U_j)_{j\ge1}\) be any countable open cover of \(K\). Compactness gives a finite subcover \(U_1,\ldots,U_r\) after relabelling. By local compactness and the Hausdorff property, one can choose finitely many compactly supported cutoffs whose supports lie in members of this finite cover and whose sum is positive on \(K\). Group the cutoffs according to the corresponding \(U_i\), and divide each grouped sum by the total sum on a neighborhood of \(K\). Multiplying these ratios by \(f\) gives functions \(f_i\in C_c(X)\) with \(0\le f_i\le1\), \(\operatorname{supp}f_i\subseteq U_i\), and \(f=\sum_{i=1}^r f_i\). The ratios are only needed near \(K\); since \(f=0\) off \(K\), the products extend continuously by zero where the total sum vanishes. By linearity and the definition of \(\lambda\),

L(f)=\sum_{i=1}^r L(f_i)\le\sum_{i=1}^r\lambda(U_i)\le\sum_{j=1}^{\infty}\lambda(U_j).

This inequality holds for every countable open cover of \(K\). Taking the infimum over those covers gives \(L(f)\le\mu^*(K)=\mu(K)\), since \(K\) is compact and hence Borel. \(\square\)

Theorem: For every open set \(U\subseteq X\), \(\displaystyle\mu(U)=\sup\{\mu(K):K\subseteq U,\ K\text{ compact}\}\).

Proof. If \(K\subseteq U\) is compact, monotonicity gives \(\mu(K)\le\mu(U)\); therefore the supremum on the right is at most \(\mu(U)\). For the reverse inequality, \(\mu(U)=\lambda(U)\) by the open-set identity proved above. Every function \(f\) in the defining supremum for \(\lambda(U)\) has compact support \(K=\operatorname{supp}f\subseteq U\), and the estimate just proved gives \(L(f)\le\mu(K)\le\sup_{K'\subseteq U,\ K'\text{ compact}}\mu(K')\). Taking the supremum over all such \(f\) yields \(\mu(U)\le\sup_{K'\subseteq U,\ K'\text{ compact}}\mu(K')\). Combining the inequalities proves the result, including \(U=\varnothing\), for which both sides are zero. \(\square\)

Local compactness enters twice: it supplies compactly supported cutoffs for the local-finiteness bound, and it supplies cutoffs subordinate to a finite cover in the compact-support estimate. Compactness then reduces an open cover of the support to finitely many sets. Together these facts explain why the measure is regular in the stated directions.

A Common Pitfall

A common mistake is to infer that local finiteness implies \(\mu(X)<\infty\). In the discrete example, each compact set \(K\subseteq\mathbb{N}\) is finite and \(\mu(K)=|K|<\infty\), but the compact sets \(\{1,\ldots,N\}\) have measures \(N\), so \(\mu(\mathbb{N})=\sup_N N=+\infty\). The correct conclusion is only finiteness on each compact set. A separate finite-total-mass hypothesis would be needed to conclude that \(\mu(X)<\infty\).

Check Your Understanding

For each question, identify the relevant assumption before applying a regularity property.

Why does a compactly supported cutoff bound \(\lambda(V)\) when \(V\) has compact closure?
In the proof of outer regularity, where is countable subadditivity of \(\lambda\) used?
For \(L(f)=f(0)\) on \(C_c(\mathbb{R})\), compute the measure of an open set \(U\) that contains \(0\), and of one that does not.
Give a locally finite Borel measure on a locally compact space whose total mass is infinite, and verify finiteness on each compact set.
In the inner-regularity proof, why can the cover of \(\operatorname{supp}f\) be reduced to finitely many open sets?

Local Finiteness and Regularity of the Constructed Measure

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