Two Different Relations Between Measures
Absolute continuity and singularity both compare measures through sets of measure zero, but they ask different questions. Absolute continuity asks whether one measure can detect a set that the other considers null. Singularity asks whether the measures can be concentrated on disjoint parts of the space. These conditions are not complementary: some pairs satisfy neither, and a zero measure can satisfy both.
Throughout, \((X,\mathcal{M})\) is a measurable space, and \(\mu\) and \(\nu\) are nonnegative measures on it, not necessarily finite. The notation \(\nu\ll\mu\) is directional: it means that every \(\mu\)-null measurable set is also \(\nu\)-null. Singularity is written \(\nu\perp\mu\) and is symmetric. The definitions below require neither finiteness nor sigma-finiteness.
For singularity, the witness set separates the measures: \(\mu\) is concentrated on \(A\), while \(\nu\) is concentrated on its complement. This condition is symmetric. Indeed, the same witness can be replaced by \(A^c\) when the roles of \(\mu\) and \(\nu\) are exchanged. In contrast, \(\nu\ll\mu\) need not imply \(\mu\ll\nu\).
What the Two Conditions Do—and Do Not—Imply
Proof. Suppose \(\nu\perp\mu\), and take \(A\in\mathcal{M}\) such that \(\mu(A^c)=0\) and \(\nu(A)=0\). Since \(\nu\ll\mu\), the \(\mu\)-null set \(A^c\) is also \(\nu\)-null, so \(\nu(A^c)=0\). The sets \(A\) and \(A^c\) are disjoint and have union \(X\), so countable additivity gives \(\nu(X)=\nu(A)+\nu(A^c)=0\). For every \(E\in\mathcal{M}\), monotonicity gives \(0\le\nu(E)\le\nu(X)=0\), hence \(\nu(E)=0\). Conversely, the zero measure assigns zero to every measurable set, so it satisfies the implication defining \(\nu\ll\mu\). For singularity, take \(A=X\); then \(\mu(A^c)=\mu(\varnothing)=0\) and \(\nu(A)=0\). \(\square\)
Consequently, a nonzero \(\nu\) cannot be both absolutely continuous with respect to \(\mu\) and singular to \(\mu\). But the proposition does not say that every nonzero \(\nu\) must satisfy one of those relations. A measure may fail absolute continuity because it assigns positive measure to a \(\mu\)-null set, and fail singularity because no set that is full for \(\mu\) is null for \(\nu\).
Worked Example: Both Relations Hold for a Zero Measure
Let \(X=\{a,b\}\), let \(\mathcal{M}\) be the collection of all subsets of \(X\), and define \(\mu(\{a\})=2\), \(\mu(\{b\})=3\). Let \(\nu(E)=0\) for every \(E\in\mathcal{M}\). Classify \(\nu\) relative to \(\mu\).
The total mass of \(\mu\) is \(\mu(X)=2+3=5\), while \(\nu(X)=0\). If \(E\in\mathcal{M}\) has \(\mu(E)=0\), then \(\nu(E)=0\) because \(\nu\) is zero on every subset. Thus \(\nu\ll\mu\). For singularity choose \(A=X\). Then \(A^c=\varnothing\), so \(\mu(A^c)=0\), and \(\nu(A)=\nu(X)=0\). Therefore \(\nu\perp\mu\) as well. Check: the two defining tests give \(\nu(E)=0\) for every \(\mu\)-null \(E\), and the explicit witness \(A=X\) has \(\mu(A^c)=0\) and \(\nu(A)=0\).
Worked Example: Singular but Not Absolutely Continuous
Let \(X=\{a,b\}\) with \(\mathcal{M}\) consisting of all subsets. Set \(\mu(\{a\})=2\), \(\mu(\{b\})=0\), \(\nu(\{a\})=0\), and \(\nu(\{b\})=3\). Classify the pair.
For any \(E\subseteq X\), additivity gives \(\mu(E)=2\) if \(a\in E\), and \(0\) otherwise; likewise, \(\nu(E)=3\) if \(b\in E\), and \(0\) otherwise. The set \(A=\{a\}\) satisfies \(\mu(A^c)=\mu(\{b\})=0\) and \(\nu(A)=0\), so \(\nu\perp\mu\). But \(\{b\}\) is \(\mu\)-null and has \(\nu(\{b\})=3\), so \(\nu\not\ll\mu\). Check: the total masses are \(\mu(X)=2+0=2\) and \(\nu(X)=0+3=3\); the witness gives disjoint concentration, while the null-set test fails exactly at \(\{b\}\).
A Support Test on Finite Spaces
On a finite space, the classification can be read directly from the points carrying positive mass. For a nonnegative measure \(\rho\) on a finite set \(X\), define its support here by \(S_\rho=\{x\in X:\rho(\{x\})>0\}\). This definition is specific to the finite setting; it does not require a topological notion of support.
Proof. Because \(X\) is finite, additivity gives \(\rho(E)=\sum_{x\in E}\rho(\{x\})\) for every \(E\subseteq X\) and either measure \(\rho\). All summands are nonnegative. Hence \(\mu(E)=0\) exactly when \(E\cap S_\mu=\varnothing\). If \(S_\nu\subseteq S_\mu\), every \(\mu\)-null \(E\) avoids \(S_\mu\), and therefore avoids \(S_\nu\); thus \(\nu(E)=0\) and \(\nu\ll\mu\). Conversely, if \(x\in S_\nu\) but \(x\notin S_\mu\), then \(\mu(\{x\})=0\) and \(\nu(\{x\})>0\), contradicting \(\nu\ll\mu\). This proves the first equivalence.
For the second, suppose \(S_\mu\cap S_\nu=\varnothing\). Choose \(A=S_\mu\). Every point outside \(S_\mu\) has zero \(\mu\)-mass, so \(\mu(A^c)=0\); and \(A\) contains no point of \(S_\nu\), so \(\nu(A)=0\). Thus \(\nu\perp\mu\). Conversely, if \(\nu\perp\mu\), choose \(A\subseteq X\) with \(\mu(A^c)=0\) and \(\nu(A)=0\). Every point of \(S_\mu\) must lie in \(A\), since a point in \(A^c\) with positive \(\mu\)-mass would make \(\mu(A^c)>0\). Every point of \(S_\nu\) must lie in \(A^c\), since a point in \(A\) with positive \(\nu\)-mass would make \(\nu(A)>0\). Thus \(S_\mu\cap S_\nu=\varnothing\). The arguments include empty supports, so they also cover zero measures. \(\square\)
Worked Example: Neither Absolutely Continuous nor Singular
Let \(X=\{a,b,c\}\), \(\mathcal{M}\) be its power set, and define \(\mu(\{a\})=2\), \(\mu(\{b\})=3\), \(\mu(\{c\})=0\), while \(\nu(\{a\})=0\), \(\nu(\{b\})=5\), \(\nu(\{c\})=7\). Classify the pair using the support test and verify the definitions directly.
The supports are \(S_\mu=\{a,b\}\) and \(S_\nu=\{b,c\}\). Since \(c\in S_\nu\) but \(c\notin S_\mu\), \(S_\nu\not\subseteq S_\mu\), so \(\nu\not\ll\mu\). Directly, \(\mu(\{c\})=0\) but \(\nu(\{c\})=7\). Also \(S_\mu\cap S_\nu=\{b\}\ne\varnothing\), so the measures are not singular. To check this from the definition, any set \(A\) with \(\mu(A^c)=0\) must contain both \(a\) and \(b\): omitting either point leaves positive \(\mu\)-mass in \(A^c\). It follows that \(\nu(A)\ge\nu(\{b\})=5\), so no such \(A\) can have \(\nu(A)=0\). Check: \(\mu(X)=2+3+0=5\) and \(\nu(X)=0+5+7=12\); the failing null set \(\{c\}\) and the unavoidable shared mass at \(b\) establish the two failures separately.
How to Classify a Pair
For general measurable spaces, the definitions give a reliable order of tests. First test \(\nu\ll\mu\) by looking for a measurable \(E\) with \(\mu(E)=0\) but \(\nu(E)>0\). Such a set disproves absolute continuity. Then test singularity by seeking a measurable \(A\) with \(\mu(A^c)=0\) and \(\nu(A)=0\). Failure of one test does not decide the other. In particular, a counterexample to absolute continuity is not automatically a witness to singularity.
| Relation | Test | What it says |
|---|---|---|
| \(\nu\ll\mu\) | Every \(\mu\)-null set is \(\nu\)-null | Null-set preservation in the direction from \(\mu\) to \(\nu\) |
| \(\nu\perp\mu\) | Some \(A\) has \(\mu(A^c)=0\) and \(\nu(A)=0\) | The measures are concentrated on complementary sets |
| Neither | A null-set counterexample and no singularity witness | The measures overlap in a way that also leaves mass outside the support of \(\mu\) |
| Both | Possible when \(\nu=0\) | Absolute continuity and singularity are not mutually exclusive for all measures |
Common mistake. A tempting but false claim is that every pair of measures must be either absolutely continuous or singular. In the last finite-space example, one might notice that \(\{c\}\) is \(\mu\)-null and conclude that it witnesses singularity. That is incorrect: singularity requires a set that is full for \(\mu\) and null for \(\nu\), not merely a \(\mu\)-null set on which \(\nu\) is positive. The correction is to check both conditions for a proposed witness. Every \(\mu\)-full set in that example contains \(b\), and therefore has \(\nu\)-measure at least \(5\); no singularity witness exists.
The distinction matters whenever a theorem or decomposition uses one of these relations as a hypothesis. Absolute continuity controls the behavior on null sets and has an indicated direction; singularity describes separation and is symmetric. They are useful, different properties—not two exhaustive labels for all pairs of measures.
Check Your Understanding
For each question, state the relevant null-set or full-measure condition explicitly, including which measure assigns zero mass.
- Give a measurable set that disproves \(\nu\ll\mu\) in the finite example with supports \(\{a,b\}\) and \(\{b,c\}\).
- Why does a witness \(A\) for \(\nu\perp\mu\) also show \(\mu\perp\nu\)?
- On a finite set, what support condition is equivalent to \(\nu\ll\mu\)?
- If \(\nu\ll\mu\) and \(\nu\perp\mu\), what must \(\nu\) be? Give the key two sets whose \(\nu\)-measures must vanish.
- Can two nonzero measures on a finite set be both absolutely continuous in the stated direction and singular? Explain using their supports.