From Open-Set Content to Measurable Sets
Let \(X\) be a locally compact Hausdorff space, and let \(C_c(X)\) denote the real-valued continuous functions on \(X\) with compact support. Suppose \(L:C_c(X)\to\mathbb{R}\) is a positive linear functional: linearity holds over the real numbers, and \(L(f)\ge 0\) whenever \(f\ge 0\). For an open set \(V\subseteq X\), its content records how large \(L(f)\) can be when \(f\) is supported inside \(V\) and bounded by \(1\). This content is defined on open sets; it is not yet a measure on all subsets.
Here \(\operatorname{supp} f\) is the closure of \(\{x:f(x)\ne0\}\). We use the cutoff property for locally compact Hausdorff spaces: if \(K\subseteq V\), with \(K\) compact and \(V\) open, there is a function \(\eta\in C_c(X)\) such that \(0\le\eta\le1\), \(\eta=1\) on \(K\), and \(\operatorname{supp}\eta\subseteq V\). This is the Urysohn cutoff property. In the proof below, the essential use of compactness is that a test function admissible for \(\lambda(V)\) has compact support that can be enclosed in a compact subset of the open set being considered.
The outer-measure construction from open covers gives countable subadditivity, and \(\mu^*(A)\le\lambda(V)\) whenever \(A\subseteq V\) and \(V\) is open, by using the single-set cover. The key additional step is to prove that splitting a set across an open set cannot increase its outer measure.
The Compact-Cutoff Splitting Argument
Proof. The empty open set causes no difficulty: if \(U=\varnothing\), then \(A\cap U=\varnothing\), so the asserted inequality is equality. Now fix an open \(U\). If \(\mu^*(A)=+\infty\), the inequality holds because its right-hand side is at most \(+\infty\). It remains to consider \(\mu^*(A)<\infty\).
Let \(\varepsilon>0\). By the definition of the infimum, choose a countable open cover \((V_i)_{i\ge1}\) of \(A\) such that $$ \sum_{i=1}^{\infty}\lambda(V_i)<\mu^*(A)+\varepsilon. $$ In particular, every \(\lambda(V_i)\) is finite. Choose positive numbers \(\varepsilon_i\) with \(\sum_i\varepsilon_i<\varepsilon\). For each \(i\), the set \(V_i\cap U\) is open. By the definition of its content, choose \(f_i\in C_c(X)\) with \(0\le f_i\le1\), \(\operatorname{supp}f_i\subseteq V_i\cap U\), and $$ L(f_i)>\lambda(V_i\cap U)-\varepsilon_i. $$ If \(\lambda(V_i\cap U)=0\), the zero function can be used instead; the non-strict inequality \(L(f_i)\ge\lambda(V_i\cap U)-\varepsilon_i\) is all that is needed.
Put \(K_i=\operatorname{supp}f_i\). This is a compact subset of \(U\), and \(V_i\setminus K_i\) is open because \(K_i\) is closed. If \(g\in C_c(X)\), \(0\le g\le1\), and \(\operatorname{supp}g\subseteq V_i\setminus K_i\), then \(f_i\) and \(g\) have disjoint supports. Thus \(0\le f_i+g\le1\), with compact support contained in \(V_i\). The definition of \(\lambda(V_i)\) and linearity give $$ L(f_i)+L(g)=L(f_i+g)\le\lambda(V_i). $$ Taking the supremum over all such \(g\) yields \(L(f_i)+\lambda(V_i\setminus K_i)\le\lambda(V_i)\). Combining this with the choice of \(f_i\), we obtain $$ \lambda(V_i\cap U)+\lambda(V_i\setminus K_i)\le\lambda(V_i)+\varepsilon_i. $$
The open sets \(V_i\cap U\) cover \(A\cap U\). The open sets \(V_i\setminus K_i\) cover \(A\setminus U\): if \(x\in A\setminus U\) and \(x\in V_i\), then \(x\notin K_i\) because \(K_i\subseteq U\). Therefore the definition of \(\mu^*\) and countable subadditivity of sums imply $$ \begin{aligned} \mu^*(A\cap U)+\mu^*(A\setminus U) &\le \sum_i\lambda(V_i\cap U)+\sum_i\lambda(V_i\setminus K_i)\\ &\le \sum_i\lambda(V_i)+\sum_i\varepsilon_i\\ &<\mu^*(A)+2\varepsilon. \end{aligned} $$ As this holds for every \(\varepsilon>0\), the desired inequality follows. For the Carathéodory criterion, the reverse inequality \(\mu^*(A)\le\mu^*(A\cap U)+\mu^*(A\setminus U)\) follows from \(A=(A\cap U)\cup(A\setminus U)\) and countable subadditivity. Hence equality holds for every \(A\subseteq X\), which is precisely measurability of \(U\). \(\square\)
The proof does not add the contents of \(V_i\cap U\) and \(V_i\setminus U\) as if they were disjoint open pieces. Instead, it selects one test function nearly attaining the first content. Its compact support is then excluded from the second open set, so the two test functions have disjoint supports and can be added without exceeding \(1\).
From Open Sets to Borel Sets
Proof. The Carathéodory outer-measure theorem states that the sets \(E\subseteq X\) satisfying $$ \mu^*(A)=\mu^*(A\cap E)+\mu^*(A\setminus E)\qquad\text{for every }A\subseteq X $$ form a sigma-algebra, and that restricting \(\mu^*\) to this sigma-algebra gives a measure. This is the standard outer-measure theorem from the earlier material on outer measures. The splitting theorem proves that every open set belongs to this sigma-algebra. Since \(\mathcal{B}_X\) is generated by open sets, it is contained in the Carathéodory sigma-algebra. Restriction therefore gives a measure on \(\mathcal{B}_X\). \(\square\)
This conclusion concerns Borel measurability; it does not assert that every subset of a Borel null set is measurable, which would require completeness. Nor does it assert finiteness or local finiteness of the resulting measure. Those are separate properties, not consequences of Borel measurability alone.
Worked Example: A Functional Supported at Two Points
Take \(X=\mathbb{R}\) and define \(L(f)=2f(0)+3f(2)\) for \(f\in C_c(\mathbb{R})\). This is positive: if \(f\ge0\), then \(2f(0)+3f(2)\ge0\). For an open set \(V\), the content is $$ \lambda(V)=2\chi_V(0)+3\chi_V(2). $$ Indeed, every admissible \(f\) has \(f(0)\le1\) and \(f(2)\le1\), and \(f(x)=0\) at either point outside \(V\). These give the upper bound. If either or both points belong to \(V\), compactly supported cutoffs inside \(V\) can equal \(1\) at those points, giving the matching lower bound.
For every \(A\subseteq\mathbb{R}\), the resulting outer measure is $$ \mu^*(A)=2\chi_A(0)+3\chi_A(2). $$ For the lower bound, any open cover of \(A\) must cover each of its points among \(0\) and \(2\), and the sum of the cover contents must therefore include the corresponding weight or weights. For the upper bound, cover each included point by an open neighborhood that excludes the other point, and cover all remaining points using the open set \(\mathbb{R}\setminus\{0,2\}\), whose content is zero.
For \(A=\{0,1,2\}\) and \(U=(-1,1)\), strict endpoint membership gives \(A\cap U=\{0,1\}\) and \(A\setminus U=\{2\}\). Thus \(\mu^*(A)=2+3=5\), \(\mu^*(A\cap U)=2\), and \(\mu^*(A\setminus U)=3\). The split adds to \(2+3=5\), exactly as required. Directly, the formula for \(\mu^*\) also gives \(2\chi_A(0)+3\chi_A(2)=5\), checking the result.
Worked Example: Open-Set Content from Integration on an Interval
Let \(X=\mathbb{R}\) and let \(L(f)=\int_0^1 f(x)\,dx\) for \(f\in C_c(\mathbb{R})\). Positivity follows because a nonnegative function has nonnegative integral. For \(U=(1/4,3/4)\), we compute \(\lambda(U)\). Every admissible \(f\) satisfies \(0\le f\le1\) and vanishes outside \(U\), so $$ L(f)\le \int_{1/4}^{3/4}1\,dx=\frac34-\frac14=\frac12. $$ For \(0<\delta<1/4\), choose a continuous, piecewise linear cutoff supported in \(U\), bounded by \(1\), and equal to \(1\) on \([1/4+\delta,3/4-\delta]\). Its integral is at least the length of that interval: $$ L(f)\ge(3/4-\delta)-(1/4+\delta)=\frac12-2\delta. $$ Letting \(\delta\) decrease to zero shows that the supremum is at least \(1/2\). The upper and lower bounds give \(\lambda(U)=1/2\).
As a second calculation for this functional, take \(A=(1/3,2/3)\). The single open set \(A\) covers itself, so \(\mu^*(A)\le\lambda(A)\). The same upper-bound argument gives \(\lambda(A)\le 2/3-1/3=1/3\), and cutoffs equal to \(1\) on \([1/3+\delta,2/3-\delta]\) give \(\lambda(A)\ge 1/3-2\delta\). Hence \(\lambda(A)=1/3\) and \(\mu^*(A)\le1/3\). Conversely, for any countable open cover \((V_i)\) of \(A\), the sets \(V_i\cap[0,1]\) cover \(A\), and $$ \sum_i\lambda(V_i)\ge\sum_i m(V_i\cap[0,1])\ge m(A)=\frac13. $$ Here \(\lambda(V_i)\ge m(V_i\cap[0,1])\) follows by the same compact-cutoff approximation on open subsets of \([0,1]\), and the second inequality is countable subadditivity of Lebesgue outer measure. Taking the infimum over covers proves \(\mu^*(A)\ge1/3\). Thus \(\mu^*(A)=1/3\), agreeing with the interval length.
Worked Example: Checking the Split for an Interval
Continue with \(L(f)=\int_0^1 f(x)\,dx\), and take \(A=[0,1]\) and the open set \(U=(1/4,3/4)\). The same cover argument as in the previous example, now applied to \([0,1]\), gives \(\mu^*(A)=1\): the open set \((- \delta,1+\delta)\) has content at most \(1\), while every open cover has total content at least the Lebesgue outer measure \(m([0,1])=1\).
Since \(U\) has strict endpoints, \(A\cap U=(1/4,3/4)\), whose outer measure is \(3/4-1/4=1/2\). The complement within \(A\) is \(A\setminus U=[0,1/4]\cup[3/4,1]\). Its two intervals have lengths \(1/4\) and \(1-3/4=1/4\); their shared accounting does not overlap, so its outer measure is \(1/4+1/4=1/2\). Therefore $$ \mu^*(A\cap U)+\mu^*(A\setminus U)=\frac12+\frac12=1=\mu^*(A). $$ The total is also checked by adding the two component lengths of \(A\): \(1/4+1/2+1/4=1\).
A Common Pitfall: Adding Overlapping Contents
It is tempting to argue that the two pieces of an open cover set can simply be charged at their full contents. That is not generally valid. For Lebesgue measure on \(\mathbb{R}\), let \(V=(-1,1)\), \(U=(-1,1)\), and \(K=[-1/2,1/2]\). Then \(V\cap U=V\) has content \(2\), while \(V\setminus K=(-1,-1/2)\cup(1/2,1)\) has content \(1\). The incorrect estimate $$ \lambda(V\cap U)+\lambda(V\setminus K)=2+1=3\le\lambda(V)=2 $$ fails: the sets overlap on the two outer intervals.
The proof does not make that estimate. It chooses a particular function \(f\) whose value \(L(f)\) is close to \(\lambda(V\cap U)\), then takes \(K\) to contain the support of that function. Every admissible function \(g\) for \(V\setminus K\) has support disjoint from \(\operatorname{supp}f\). The valid bound is \(L(f)+L(g)\le\lambda(V)\), because \(f+g\le1\) pointwise. Taking the supremum over \(g\) and allowing the approximation error for \(f\) to tend to zero gives the required cover-cost estimate.
Check Your Understanding
For each question, distinguish the open-set content from the outer measure on arbitrary subsets.
- Why is \(V_i\setminus K_i\) open in the splitting proof, and why does it cover every point of \(A\setminus U\) lying in \(V_i\)?
- For \(L(f)=f(0)\) on \(C_c(\mathbb{R})\), compute \(\lambda(V)\) for an open set \(V\) containing \(0\), and for one not containing \(0\).
- Explain why \(f_i+g\le1\) in the proof, even though both \(f_i\) and \(g\) may individually take the value \(1\).
- For the functional \(L(f)=\int_0^1f(x)\,dx\), calculate the content of \((1/2,3/4)\) using an upper bound and cutoffs supported inside the interval.
- Which standard outer-measure result turns measurability of every open set into measurability of every Borel set?