A Common Space for Comparing Measures
Absolute continuity and singularity are properties of a pair of measures on the same measurable space. The tests can be simple, but their direction matters: \(\nu\ll\mu\) asks whether every \(\mu\)-null set is also \(\nu\)-null. A point mass, counting measure, and Cantor measure each behave differently from Lebesgue measure, so concrete null-set tests are especially useful.
Unless stated otherwise, all measures below are defined on \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\), the real line with its Borel sigma-algebra. Write \(m\) for Lebesgue measure restricted to Borel sets, \(\#\) for counting measure, \(\delta_a\) for the point mass at \(a\in\mathbb{R}\), and \(\kappa\) for the standard Cantor probability measure. The measure \(m\) is sigma-finite and infinite; \(\#\) is infinite and not sigma-finite; \(\delta_a\) and \(\kappa\) are finite probability measures. In particular, all comparisons use the same measurable space; no completion of the Borel sigma-algebra is being assumed.
A useful direct test for singularity is to find a measurable set that carries all of one measure and none of the other. Conversely, to disprove absolute continuity, it suffices to exhibit one null set for the proposed dominating measure that has positive measure for the other. These tests do not require either measure to be finite or sigma-finite.
The Cantor Measure and Its Support
Let \(C\) be the middle-thirds Cantor set: start with \([0,1]\), remove the open middle third, and repeat this removal in each remaining closed interval. At stage \(n\), the set consists of \(2^n\) closed intervals, each of length \(3^{-n}\). The standard Cantor probability measure \(\kappa\) can be defined as the distribution of the random series \(\sum_{j=1}^{\infty}2\xi_j/3^j\), where the digits \(\xi_j\) are independent and each takes the values \(0\) and \(1\) with probability \(1/2\). Equivalently, a specified sequence of the first \(n\) digits has probability \(2^{-n}\). This standard product-measure construction gives a Borel probability measure carried by \(C\).
Proof. The stage-\(n\) approximation to \(C\) covers \(C\) by \(2^n\) intervals of length \(3^{-n}\). Therefore its total length is \(2^n3^{-n}=(2/3)^n\), so \(0\le m(C)\le(2/3)^n\) for every \(n\). Since \((2/3)^n\) tends to zero, \(m(C)=0\). The random series defining \(\kappa\) takes values in \(C\), so \(\kappa(C)=1\). Finally, for any \(x\in C\), at each stage \(n\), the point \(x\) belongs to one stage-\(n\) interval, whose Cantor measure is \(2^{-n}\). Hence \(0\le\kappa(\{x\})\le2^{-n}\) for every \(n\), and \(\kappa(\{x\})=0\). For \(x\notin C\), \(\kappa(\{x\})=0\) because \(\kappa(C^c)=0\). Taking \(A=C\) gives \(\kappa(A^c)=0\) and \(m(A)=0\), which is exactly \(\kappa\perp m\). \(\square\)
Worked Comparisons
Worked Example: Lebesgue Measure and a Point Mass
Compare \(m\) and \(\delta_0\) in both directions, and test whether they are singular.
The Borel singleton \(\{0\}\) has \(m(\{0\})=0\) but \(\delta_0(\{0\})=1\). Thus \(\delta_0\not\ll m\). For the reverse direction, take the Borel set \(E=(1,2)\). It has \(\delta_0(E)=0\), while \(m(E)=2-1=1\), so \(m\not\ll\delta_0\). For singularity, take \(A=\{0\}\). Then \(\delta_0(A^c)=0\), and \(m(A)=0\), so \(\delta_0\perp m\).
Check: the set used to disprove \(m\ll\delta_0\) excludes the atom at \(0\), and its Lebesgue measure is exactly its interval length, \(1\). The singleton carrying the point mass has Lebesgue measure zero, as required for singularity.
Worked Example: Counting Measure and a Point Mass
Compare \(\#\) and \(\delta_0\), paying attention to the choice of test point.
A set has counting measure zero only if it is empty: any nonempty set contains a point, and its counting measure is at least \(1\). Therefore every \(\#\)-null Borel set is also \(\delta_0\)-null, and \(\delta_0\ll\#\). For the reverse direction, choose \(y=1\), which is different from the atom \(0\). Then \(\delta_0(\{1\})=0\), whereas \(\#(\{1\})=1\). Thus \(\#\not\ll\delta_0\).
These measures are not singular. If \(A\) had full \(\delta_0\)-measure, it would have to contain \(0\). If also \(\#(A)=0\), then \(A\) would have to be empty, a contradiction. Equivalently, in the definition \(\delta_0(A^c)=0\) forces \(0\in A\), while \(\#(A)=0\) forces \(A=\varnothing\).
Check: the singleton \(\{1\}\), not \(\{0\}\), witnesses the failure of \(\#\ll\delta_0\): its two measures are \(1\) and \(0\), respectively. At the atom itself, \(\delta_0(\{0\})=1\).
Worked Example: Lebesgue Measure and Cantor Measure
Show that \(m\) and \(\kappa\) are singular, and check that neither is absolutely continuous with respect to the other.
From the proposition, \(m(C)=0\) and \(\kappa(C)=1\), so \(\kappa\not\ll m\). Also, the open middle-third interval \((1/3,2/3)\) is disjoint from \(C\). Consequently \(\kappa((1/3,2/3))=0\), while its Lebesgue measure is \(2/3-1/3=1/3\). Thus \(m\not\ll\kappa\). The set \(C\) itself verifies singularity: \(\kappa(C^c)=0\) and \(m(C)=0\).
Check: the two tests have different roles. The null set \(C\) disproves \(\kappa\ll m\) and proves singularity; the removed interval, whose length is exactly \(1/3\), disproves \(m\ll\kappa\).
Worked Example: Cantor Measure and a Point Mass
For any fixed \(a\in\mathbb{R}\), compare \(\kappa\) and \(\delta_a\).
The proposition gives \(\kappa(\{a\})=0\), including when \(a\) lies in \(C\). Since \(\delta_a(\{a\})=1\), we have \(\delta_a\not\ll\kappa\). For the other direction, \(\delta_a(\mathbb{R}\setminus\{a\})=0\), while \(\kappa(\mathbb{R}\setminus\{a\})=1-\kappa(\{a\})=1\). Hence \(\kappa\not\ll\delta_a\). Nevertheless, the measures are singular: with \(A=\{a\}\), \(\delta_a(A^c)=0\) and \(\kappa(A)=0\).
Check: the point mass assigns \(1\) to its own singleton, but the Cantor measure assigns \(0\) to every singleton. The set \(\mathbb{R}\setminus\{a\}\) has Cantor measure \(1\) because \(\kappa\) has total mass \(1\) and no atom at \(a\).
Putting the Tests Together
The following table records the directional tests. In each row, “\(\nu\ll\mu\)” is read from left to right as written in the first column; “singular” means that the pair admits disjoint full-measure supports as in the definition.
| Pair on the Borel line | Absolute continuity | Mutually singular? |
|---|---|---|
| \(\delta_a\) and \(m\) | Neither direction | Yes |
| \(\kappa\) and \(m\) | Neither direction | Yes |
| \(\kappa\) and \(\delta_a\) | Neither direction | Yes |
| \(\delta_a\) and \(\#\) | \(\delta_a\ll\#\), but \(\#\not\ll\delta_a\) | No |
| \(\kappa\) and \(\#\) | \(\kappa\ll\#\), but \(\#\not\ll\kappa\) | No |
| \(m\) and \(\#\) | \(m\ll\#\), but \(\#\not\ll m\) | No |
To justify the counting-measure entries, \(\#\)-null sets are empty, so every measure on this space is absolutely continuous with respect to \(\#\). In particular, \(m\ll\#\), \(\delta_a\ll\#\), and \(\kappa\ll\#\). The reverse assertions fail on a singleton: \(m(\{y\})=0\) and \(\kappa(\{y\})=0\) for every \(y\), while \(\#(\{y\})=1\); for \(\delta_a\), choose \(y\ne a\), so \(\delta_a(\{y\})=0\) and again \(\#(\{y\})=1\).
None of these pairs involving \(\#\) and a nonzero measure is singular. Indeed, if \(\#(A)=0\), then \(A=\varnothing\). A nonzero measure \(\nu\) cannot satisfy \(\nu(A^c)=0\) for that choice, since then \(\nu(A^c)=\nu(\mathbb{R})>0\). This applies to \(m\), \(\delta_a\), and \(\kappa\). It also explains why absolute continuity and singularity are not alternatives that exhaust all possibilities: for example, \(\delta_a\ll\#\), but \(\delta_a\) and \(\#\) are not singular.
A Reliable Decision Procedure and a Common Pitfall
For a proposed relation \(\nu\ll\mu\), test Borel sets \(E\) with \(\mu(E)=0\) and check whether \(\nu(E)=0\). One counterexample settles failure; to establish the relation, the implication must hold for every measurable null set. For singularity, look instead for a single Borel set \(A\) satisfying both \(\mu(A)=0\) and \(\nu(A^c)=0\). A set on which one measure is positive is not automatically a witness to either property: the two required conditions must match their definitions.
A common error is to use the atom as the test point when disproving \(\#\ll\delta_a\). The incorrect calculation “\(\#(\{a\})=1\) and \(\delta_a(\{a\})=0\)” fails because the definition of \(\delta_a\) gives \(\delta_a(\{a\})=1\). The correction is to choose a different point \(y\ne a\). Then \(\#(\{y\})=1\) and \(\delta_a(\{y\})=0\), which is the required null-set witness. Keeping the proposed dominating measure on the right side of the null-set test prevents this kind of directional error.
Check Your Understanding
For each question, identify the measurable set used in the test and state which measure is zero on it.
- For fixed \(a\), give a Borel set proving that \(\delta_a\not\ll m\), and give one proving that \(m\not\ll\delta_a\).
- Why does \(\kappa(C)=1\) and \(m(C)=0\) prove both \(\kappa\not\ll m\) and \(\kappa\perp m\)?
- For fixed \(a\), choose a point \(y\ne a\) and use its singleton to show that \(\#\not\ll\delta_a\).
- Explain why \(\kappa\ll\#\), but \(\#\not\ll\kappa\).
- Can \(\delta_a\) and \(\#\) be singular even though \(\delta_a\ll\#\)? Justify your answer from the definition.