What Mutual Singularity Means
Two measures can assign their mass to disjoint measurable parts of the same space. Mutual singularity records exactly this situation: there is a measurable set carrying one measure and a disjoint measurable set carrying the other. The criterion is useful only when both measures are defined on the same measurable space, so the choice of sigma-algebra matters as much as the choice of sets.
In this tutorial, \((X,\mathcal{M})\) is a measurable space, and \(\mu\) and \(\nu\) are positive measures on it. They need not be finite or sigma-finite for the definition below. A measurable set \(E\) carries a measure \(\mu\) if \(\mu(E^c)=0\); this means the measure is concentrated on \(E\). When \(\nu\) is a finite signed or complex measure, its total variation \(|\nu|\) is a positive measure, and the corresponding definition uses \(|\nu|\).
The measures must live on the same sigma-algebra: every set used to establish singularity must be measurable for both measures. In particular, the usual Cantor probability measure is naturally a Borel measure. We will compare it with Lebesgue measure restricted to the Borel sets, not assume that the Cantor measure is defined on the full Lebesgue sigma-algebra \(\mathcal{L}\).
The Null-Set Criterion
- \(\nu\perp\mu\).
- There is an \(A\in\mathcal{M}\) such that \(\mu(A)=0\) and \(\nu(A^c)=0\).
Proof. Suppose first that \(\nu\perp\mu\). By definition, there are disjoint \(E,F\in\mathcal{M}\) with \(\mu(E^c)=0\) and \(\nu(F^c)=0\). Since \(E\cap F=\varnothing\), we have \(F\subseteq E^c\). Monotonicity of \(\mu\) gives \(0\le\mu(F)\le\mu(E^c)=0\), so \(\mu(F)=0\). Taking \(A=F\) gives \(\mu(A)=0\) and \(\nu(A^c)=\nu(F^c)=0\).
Conversely, suppose \(A\in\mathcal{M}\), \(\mu(A)=0\), and \(\nu(A^c)=0\). Set \(E=A^c\) and \(F=A\). These sets are measurable and disjoint; also \(\mu(E^c)=\mu(A)=0\) and \(\nu(F^c)=\nu(A^c)=0\). Thus they carry \(\mu\) and \(\nu\), respectively, and \(\nu\perp\mu\). This argument also covers zero measures and the empty space: it uses only the stated null-set equalities. \(\square\)
The criterion says that one measure can be confined to a set that the other measure regards as null. It also clarifies why “full measure” here means that the complement has measure zero. It does not require the set itself to have finite measure; it may have infinite measure.
Proof. If \(\nu\perp\mu\), the null-set criterion supplies \(A\in\mathcal{M}\) with \(\mu(A)=0\) and \(\nu(A^c)=0\). To apply the criterion with the measures reversed, take \(A^c\): then \(\nu(A^c)=0\) and \(\mu((A^c)^c)=\mu(A)=0\). Hence \(\mu\perp\nu\). The same argument with the measures interchanged proves the reverse implication. \(\square\)
Worked Examples
Worked Example: A Dirac Measure and Borel Lebesgue Measure
Let the measurable space be \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\), where \(\mathcal{B}_{\mathbb{R}}\) is the Borel sigma-algebra. Write \(m_{\mathcal{B}}\) for Lebesgue measure restricted to Borel sets, so \(m_{\mathcal{B}}(B)=m(B)\) for \(B\in\mathcal{B}_{\mathbb{R}}\). For a fixed \(a\in\mathbb{R}\), the Dirac measure is \(\delta_a(B)=1\) if \(a\in B\), and \(0\) otherwise. Show that \(\delta_a\perp m_{\mathcal{B}}\).
Use \(A=\{a\}\), which is Borel. Its Lebesgue measure is zero: for every \(r>0\), \(\{a\}\subseteq(a-r,a+r)\), so \(0\le m_{\mathcal{B}}(\{a\})\le 2r\); letting \(r\) decrease to zero gives \(m_{\mathcal{B}}(\{a\})=0\). Since \(a\notin A^c\), the definition of the Dirac measure gives \(\delta_a(A^c)=0\). The null-set criterion, with \(\mu=m_{\mathcal{B}}\) and \(\nu=\delta_a\), now yields \(\delta_a\perp m_{\mathcal{B}}\). Check: \(m_{\mathcal{B}}(A)=0\) and \(\delta_a(A^c)=0\), exactly the two required equalities.
Worked Example: Cantor Measure and Borel Lebesgue Measure
Let \(C\) be the middle-thirds Cantor set in \(\mathbb{R}\), and let \(\kappa\) be the standard Cantor probability measure on \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\). Verify that \(\kappa\perp m_{\mathcal{B}}\), taking care that both measures are defined on this same Borel space.
At stage \(n\), the Cantor construction leaves \(2^n\) closed intervals, each of length \(3^{-n}\), and \(C\) is contained in their union. Therefore the total length of this cover is \(2^n3^{-n}=(2/3)^n\), and \(0\le m_{\mathcal{B}}(C)\le(2/3)^n\) for every positive integer \(n\). Since \((2/3)^n\to0\), \(m_{\mathcal{B}}(C)=0\).
The Cantor probability measure is the distribution of \(T=\sum_{j=1}^{\infty}2\xi_j/3^j\), where the digits \(\xi_j\) are independent and each takes the values \(0\) and \(1\) with probability \(1/2\). This distribution is a Borel measure. Every value of \(T\) lies in \(C\), so \(\kappa(C)=1\), equivalently \(\kappa(C^c)=0\). Taking \(A=C\) gives \(m_{\mathcal{B}}(A)=0\) and \(\kappa(A^c)=0\); hence \(\kappa\perp m_{\mathcal{B}}\). Check: the chosen set is Borel, has \(m_{\mathcal{B}}\)-measure \(0\), and has \(\kappa\)-measure \(1\). No claim that \(\kappa\) is defined on all of \(\mathcal{L}\) is needed.
Worked Example: A Dirac Measure and Cantor Measure
On \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\), verify that \(\delta_0\perp\kappa\). This example is useful because both measures give full mass to the Cantor set: \(0\in C\) gives \(\delta_0(C)=1\), and the definition of \(\kappa\) gives \(\kappa(C)=1\).
We need a different separating set. Take \(A=\{0\}\). To compute its Cantor measure, use the digit representation of \(T\). Since every term \(2\xi_j/3^j\) is nonnegative, \(T=0\) holds exactly when every digit \(\xi_j\) is zero. In particular, \(T=0\) requires the first \(n\) digits to be zero, an event of probability \(2^{-n}\). Conversely, all the digits being zero implies \(T=0\). Thus \(\kappa(\{0\})=\lim_{n\to\infty}2^{-n}=0\); the limit follows from continuity from above of probability on the decreasing events that the first \(n\) digits are zero. Also, \(\delta_0(\{0\}^c)=0\), since \(0\notin\{0\}^c\). The criterion with \(\mu=\kappa\), \(\nu=\delta_0\) proves \(\delta_0\perp\kappa\). Check: \(\kappa(A)=0\) and \(\delta_0(A^c)=0\), despite both measures assigning mass \(1\) to \(C\).
Common Pitfall: Overlap Does Not Decide Singularity
A misleading test is to look for a set to which both measures assign positive mass and conclude they cannot be singular. In the previous example, the calculation \(\delta_0(C)=1\) and \(\kappa(C)=1\) is correct, but the conclusion “therefore \(\delta_0\) and \(\kappa\) are not singular” is wrong. Singularity asks whether there exist disjoint sets carrying the measures, not whether there is a set to which both assign positive mass. Here the correction is \(A=\{0\}\): \(\kappa(A)=0\), while \(\delta_0(A^c)=0\), which proves singularity by the criterion.
A second error is to use a set that is measurable for only one measure. The Cantor set \(C\) is Borel, so it belongs to the common sigma-algebra in the Cantor example. More generally, when comparing measures, first specify their common measurable space; only then can a proposed separating set certify singularity.
For a finite signed or complex measure \(\nu\), the same null-set test is applied to its positive total variation \(|\nu|\): \(\nu\perp\mu\) means there is an \(A\in\mathcal{M}\) with \(\mu(A)=0\) and \(|\nu|(A^c)=0\). Using \(|\nu|\), rather than the possibly cancelling signed value \(\nu(A^c)\), ensures that the condition expresses concentration of the entire signed or complex measure.
Check Your Understanding
In each question, name the common measurable space and use the null-set criterion where appropriate.
- For a fixed \(a\in\mathbb{R}\), identify a Borel set that proves \(\delta_a\perp m_{\mathcal{B}}\), and state both required measure values.
- Why does the Cantor-measure example use \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\) rather than assume that \(\kappa\) is defined on \((\mathbb{R},\mathcal{L})\)?
- Give the two measure values that establish \(\kappa\perp m_{\mathcal{B}}\) using the Cantor set \(C\).
- For \(\delta_0\) and \(\kappa\), explain why their both assigning mass \(1\) to \(C\) does not disprove singularity. Identify a separating set.
- State the null-set criterion for a finite signed measure \(\nu\) and a positive measure \(\mu\), using the total variation of \(\nu\).