Classification Depends on the Reference Measure
A measure made from Lebesgue, point-mass, and Cantor components does not have one fixed classification as absolutely continuous or singular: the answer depends on the reference measure. A Cantor measure is singular with respect to Lebesgue measure, for instance, but is absolutely continuous with respect to a reference measure that includes that same Cantor measure. The practical task is to identify which components the reference measure sees, and in what proportions.
Throughout, measures are finite positive Borel measures on \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\). The symbol \(m\) denotes Lebesgue measure restricted to \(\mathcal{B}_{\mathbb{R}}\); this measure is sigma-finite and is not being treated as a completed measure here. For a measure \(\eta\), \(\nu\ll\eta\) means that every Borel \(\eta\)-null set is \(\nu\)-null. The notation \(\nu\perp\eta\) means that some Borel set \(N\) satisfies \(\eta(N)=0\) and \(\nu(N^c)=0\). A decomposition \(\nu=\nu_{\mathrm{ac}}+\nu_{\mathrm{s}}\), where \(\nu_{\mathrm{ac}}\ll\eta\) and \(\nu_{\mathrm{s}}\perp\eta\), is the Lebesgue decomposition relative to \(\eta\); \(\nu_{\mathrm{ac}}\) and \(\nu_{\mathrm{s}}\) are its absolutely continuous and singular parts.
Let \(K\) be the middle-thirds Cantor set, and let \(C\) be the standard Cantor probability measure on \(\mathcal{B}_{\mathbb{R}}\), with \(C(K)=1\). At stage \(n\) of the Cantor construction, \(K\) lies in \(2^n\) intervals of length \(3^{-n}\). Their total length is \((2/3)^n\), so \(m(K)=0\). The measure \(C\) assigns mass \(2^{-n}\) to each stage-\(n\) interval. Each point belongs to at most two such intervals, so its \(C\)-mass is at most \(2\cdot2^{-n}\) for every \(n\), and hence \(C(\{x\})=0\). Thus \(C\) is concentrated on a Lebesgue-null set but has no atoms. The Dirac measure \(\delta_a\) has mass one at \(a\) and zero outside \(\{a\}\).
Proof. Define \(f=\sum_{j=1}^r(d_j/c_j)\chi_{A_j}\). For any Borel set \(E\), disjointness and concentration give \(\int_E f\,d\eta=\sum_j(d_j/c_j)\eta(E\cap A_j)=\sum_j(d_j/c_j)c_j\lambda_j(E)=\sum_jd_j\lambda_j(E)\). Thus \(f\) is the Radon–Nikodym density of the displayed sum. If another measure is concentrated on an \(\eta\)-null set, it is singular by definition. \(\square\)
In applications, carriers can be made disjoint even when the measures’ topological supports overlap. For example, \(0\in K\), but \(C(\{0\})=0\), so the point mass at zero and the Cantor measure can be placed on the disjoint carriers \(\{0\}\) and \(K\setminus\{0\}\).
Three Explicit Classifications
Worked Example: Decomposing Relative to Lebesgue Measure
On \((\mathbb{R},\mathcal{B}_{\mathbb{R}},m)\), classify \(\nu(E)=\int_E(1+x)\chi_{[0,1]}(x)\,dm(x)+3\delta_0(E)+2C(E)\) for every Borel set \(E\).
The first term is absolutely continuous with density \((1+x)\chi_{[0,1]}(x)\). The other two terms are concentrated on \(K\): \(0\in K\) and \(C(K)=1\), while \(m(K)=0\). Their sum is therefore singular with respect to \(m\). Hence \(\nu_{\mathrm{ac}}(E)=\int_E(1+x)\chi_{[0,1]}(x)\,dm(x)\) and \(\nu_{\mathrm{s}}(E)=3\delta_0(E)+2C(E)\).
Check: the absolutely continuous mass is \(\int_0^1(1+x)\,dx=1+\tfrac12=\tfrac32\). The singular mass is \(3\cdot1+2\cdot1=5\), so \(\nu(\mathbb{R})=\tfrac32+5=\tfrac{13}{2}\). The original formula gives the same total: \(\int_0^1(1+x)\,dx+3+2=\tfrac32+3+2=\tfrac{13}{2}\).
Worked Example: A Reference Measure with Three Components
Define the finite reference measure \(\eta(E)=\int_E\chi_{[0,1]}\,dm+2\delta_0(E)+3C(E)\) and the target measure \(\nu(E)=2\int_E\chi_{[0,1]}\,dm+5\delta_0(E)+7C(E)+4\delta_2(E)\). Find the Lebesgue decomposition of \(\nu\) relative to \(\eta\).
Use the disjoint carriers \(A=[0,1]\setminus K\), \(B=\{0\}\), and \(D=K\setminus\{0\}\). The measure \(m|_{[0,1]}\) is concentrated on \(A\), because \(m(K)=0\); \(\delta_0\) is concentrated on \(B\); and \(C\) is concentrated on \(D\), because \(C(K)=1\) and \(C(\{0\})=0\). The three coefficients of \(\eta\) on these carriers are \(1,2,3\), while those of the corresponding target components are \(2,5,7\). The remaining measure \(4\delta_2\) is singular relative to \(\eta\), since \(\eta(\{2\})=0\).
Consequently, the density and the two parts are \(f=2\chi_A+\tfrac52\chi_B+\tfrac73\chi_D\), \(\nu_{\mathrm{ac}}(E)=\int_E f\,d\eta\), and \(\nu_{\mathrm{s}}=4\delta_2\). To check the density on every Borel set \(E\), split it across the carriers: \(\int_E f\,d\eta=2m(E\cap A)+\tfrac52\cdot2\delta_0(E)+\tfrac73\cdot3C(E)=2m(E\cap[0,1])+5\delta_0(E)+7C(E)\). Here \(m(E\cap A)=m(E\cap[0,1])\) and \(C(E\cap D)=C(E)\), by the null-set properties just established. Adding \(4\delta_2(E)\) recovers \(\nu(E)\).
Check: \(\eta(\mathbb{R})=1+2+3=6\). The density contributes \(2\cdot1+\tfrac52\cdot2+\tfrac73\cdot3=2+5+7=14\), and the singular part contributes \(4\), giving \(\nu(\mathbb{R})=18\). Directly, the target formula gives \(2+5+7+4=18\).
Worked Example: A Cantor Component Becomes Absolutely Continuous
Let \(\rho(E)=m(E\cap[0,1])+2\delta_0(E)\), and let \(\tau(E)=3m(E\cap[0,1])+4\delta_0(E)+5C(E)\). Classify \(\tau\) relative to \(\rho\).
The sets \(A=[0,1]\setminus\{0\}\) and \(B=\{0\}\) are disjoint carriers for the two components of \(\rho\). On \(A\), the target’s Lebesgue component has coefficient \(3\) and the reference has coefficient \(1\); on \(B\), the target atom has coefficient \(4\) and the reference atom has coefficient \(2\). Thus the density is \(g=3\chi_A+2\chi_B\), and \(\tau_{\mathrm{ac}}(E)=\int_Eg\,d\rho\). The Cantor term is singular relative to \(\rho\): the set \(K\setminus\{0\}\) has \(\rho\)-measure zero and \(C\)-measure one, so \(\tau_{\mathrm{s}}=5C\).
Check: \(\rho(\mathbb{R})=1+2=3\), and \(\int g\,d\rho=3\cdot1+2\cdot2=7\). The singular mass is \(5\), so the decomposition gives total mass \(12\); the original formula gives \(3+4+5=12\). In particular, \(C\) is singular relative to \(m\), but the target’s Cantor component is not automatically singular relative to every mixed measure: here it is singular because the chosen reference has no Cantor component.
Why the Classification Is Unique
Once a candidate density and a candidate singular part have been found, the setwise identity verifies that they sum to the target. The conditions \(\nu_{\mathrm{ac}}\ll\eta\) and \(\nu_{\mathrm{s}}\perp\eta\) also matter: they make this decomposition unique.
Proof. By singularity, there are \(\eta\)-null sets \(N\) and \(\widetilde{N}\) such that \(\beta\) is concentrated on \(N\) and \(\widetilde{\beta}\) is concentrated on \(\widetilde{N}\). Put \(S=N\cup\widetilde{N}\); then \(\eta(S)=0\), so \(\alpha(S)=\widetilde{\alpha}(S)=0\) by absolute continuity. For any Borel set \(E\), \(\beta(E\setminus S)=\widetilde{\beta}(E\setminus S)=0\), while \(\alpha(E\cap S)=\widetilde{\alpha}(E\cap S)=0\). Therefore \(\alpha(E)=\nu(E\setminus S)=\widetilde{\alpha}(E)\). This holds for every Borel \(E\), so \(\alpha=\widetilde{\alpha}\); subtracting from \(\nu\) gives \(\beta=\widetilde{\beta}\). \(\square\)
A Common Misclassification
A tempting but incorrect shortcut is to label a component singular as soon as it is singular with respect to Lebesgue measure. For the reference measure in the second example, that would incorrectly put \(7C\) into the singular part merely because \(C\perp m\). But the reference contains \(3C\), and \(7C\ll3C\): if \(\eta(E)=0\), then \(3C(E)\le\eta(E)=0\), so \(7C(E)=0\). Its density on the Cantor carrier is \(7/3\), not zero. Singularity and absolute continuity must always be tested against the chosen reference measure.
For a finite mixture, a reliable procedure is to list its component measures, select disjoint carriers up to null sets, and compare the target and reference coefficients on each carrier. The unmatched target components must then be tested for concentration on a reference-null set. Finally, verify the proposed identity on an arbitrary Borel set, not just by comparing total masses: equal totals alone do not establish equality of measures.
Check Your Understanding
For each question, state the reference measure before classifying the components.
- Relative to Lebesgue measure, which parts of \(m|_{[0,1]}+2\delta_1+3C\) are absolutely continuous and singular?
- For \(\eta=m|_{[0,1]}+4C\) and \(\nu=2m|_{[0,1]}+12C+\delta_3\), find the density of \(\nu_{\mathrm{ac}}\) relative to \(\eta\) and identify \(\nu_{\mathrm{s}}\).
- Why can the carriers \(\{0\}\) and \(K\setminus\{0\}\) distinguish a point mass at zero from the Cantor measure?
- Give a Borel set witnessing that \(C\) is singular relative to \(m\), and give one witnessing that \(C\) is singular relative to \(m|_{[0,1]}+2\delta_0\).
- Explain why checking only the total masses of a proposed decomposition does not prove that it holds on every Borel set.