Absolute Continuity as Preservation of Null Sets

Absolute Continuity Means Preserving Null Sets

Two measures can assign very different amounts of mass to a set while agreeing about which sets are negligible. Absolute continuity expresses precisely this one-way agreement: a measure \(\nu\) is absolutely continuous with respect to a measure \(\mu\) when every measurable set of \(\mu\)-measure zero also has \(\nu\)-measure zero. The notation is \(\nu\ll\mu\). The order matters: it is \(\nu\) that must preserve the null sets of \(\mu\), and this condition does not require \(\mu\) and \(\nu\) to have the same null sets.

Definition: On a measurable space \((X,\mathcal{M})\), let \(\mu\) and \(\nu\) be positive measures, not necessarily finite. We write \(\nu\ll\mu\) if, for every \(E\in\mathcal{M}\), \(\mu(E)=0\) implies \(\nu(E)=0\). This is absolute continuity of a measure; it is distinct from absolute continuity of a function on an interval.

No finiteness or sigma-finiteness assumption is needed for this definition. In particular, absolute continuity is a condition only on null sets; it is not a bound comparing the sizes of \(\nu(E)\) and \(\mu(E)\) for sets of positive \(\mu\)-measure. When \(\nu\) is a finite signed measure, the appropriate definition uses its total variation, so that cancellation between positive and negative mass cannot conceal mass on a \(\mu\)-null set.

Definition: Let \((X,\mathcal{M})\) be a measurable space, let \(\mu\) be a positive measure, and let \(\nu\) be a finite signed measure. We write \(\nu\ll\mu\) if \(|\nu|\ll\mu\), meaning that \(\mu(E)=0\) implies \(|\nu|(E)=0\) for every \(E\in\mathcal{M}\). Here \(|\nu|\) is the total variation measure of \(\nu\).

For a finite signed measure, \(|\nu|(E)\) measures the total amount of positive and negative mass on \(E\), without cancellation. It can be defined by taking the supremum of \(\sum_{j=1}^k|\nu(E_j)|\) over all finite measurable partitions \(E=\bigcup_{j=1}^k E_j\) into pairwise disjoint sets. In particular, \(|\nu(E)|\le|\nu|(E)\). The signed-measure definition is equivalent to requiring \(\nu(E)=0\) on every \(\mu\)-null measurable set, as the following proposition proves.

Proposition: Let \((X,\mathcal{M})\) be a measurable space, let \(\mu\) be any positive measure, and let \(\nu\) be a finite signed measure. Then \(\nu\ll\mu\) if and only if \(\nu(E)=0\) for every \(E\in\mathcal{M}\) with \(\mu(E)=0\).

Proof. Suppose first that \(\nu\ll\mu\). By definition, if \(\mu(E)=0\), then \(|\nu|(E)=0\). Since \(|\nu(E)|\le|\nu|(E)\), it follows that \(\nu(E)=0\). Conversely, suppose that \(\nu(A)=0\) for every measurable \(\mu\)-null set \(A\). Fix \(E\in\mathcal{M}\) with \(\mu(E)=0\). Every measurable subset \(A\subseteq E\) is also \(\mu\)-null, since \(0\le\mu(A)\le\mu(E)=0\). Thus \(\nu(A)=0\) for every such \(A\). For any finite measurable partition \(E=\bigcup_{j=1}^k E_j\), each \(E_j\) is a measurable subset of \(E\), so \(\sum_{j=1}^k|\nu(E_j)|=0\). Taking the supremum over these partitions gives \(|\nu|(E)=0\). This holds for every \(\mu\)-null \(E\), proving \(\nu\ll\mu\). \(\square\)

Finite Spaces: Find the Null Sets First

Worked Example: A Finite Reference Measure

Let \(X=\{a,b,c\}\), \(\mathcal{M}=\mathcal{P}(X)\), and define the finite positive measures \(\mu\) and \(\nu\) by their singleton masses: \(\mu(\{a\})=2,\ \mu(\{b\})=0,\ \mu(\{c\})=3\), and \(\nu(\{a\})=1,\ \nu(\{b\})=0,\ \nu(\{c\})=5\). Determine whether \(\nu\ll\mu\).

For any \(E\subseteq X\), finite additivity gives \(\mu(E)=2\chi_E(a)+0\chi_E(b)+3\chi_E(c)\). All three nonzero terms are nonnegative, so \(\mu(E)=0\) exactly when \(a\notin E\) and \(c\notin E\). Therefore the only \(\mu\)-null sets are \(\varnothing\) and \(\{b\}\). Their \(\nu\)-measures are \(\nu(\varnothing)=0\) and \(\nu(\{b\})=0\). Hence \(\nu\ll\mu\).

For a check on a set that is not null, take \(E=\{a,c\}\). Direct calculation gives \(\mu(E)=2+3=5\) and \(\nu(E)=1+5=6\). The measures differ there, which is allowed: absolute continuity requires preservation of zero measure, not equality of measures. Conversely, the fact that \(\nu(E)=6\) while \(\mu(E)=5\) says nothing against \(\nu\ll\mu\).

This finite example illustrates a useful test: determine all \(\mu\)-null sets, then check whether each has \(\nu\)-measure zero. It is not enough to compare the measures on a few sets of positive measure. For signed measures, the same test applies to total variation; simply checking a signed measure's value on a single null set can miss cancellation among its subsets.

Counting Measure and the Vacuous Case

Worked Example: A Finite Measure Compared with Counting Measure

Let \(X=\mathbb{N}=\{1,2,\ldots\}\), \(\mathcal{M}=\mathcal{P}(\mathbb{N})\), and let \(\mu\) be counting measure, so \(\mu(E)\) is the number of elements of \(E\), or \(+\infty\) if \(E\) is infinite. This measure is sigma-finite but not finite. Define a finite positive measure \(\nu\) by \(\nu(E)=\sum_{n\in E}2^{-n}\). Determine whether \(\nu\ll\mu\), and compute \(\nu(\{1,3\})\).

Every nonempty set \(E\subseteq\mathbb{N}\) contains at least one integer, so \(\mu(E)\ge1\); in particular, the only \(\mu\)-null set is \(\varnothing\). Since \(\nu(\varnothing)=0\), the null-set condition holds and \(\nu\ll\mu\). For the requested set, its elements are exactly \(1\) and \(3\), and therefore \(\nu(\{1,3\})=2^{-1}+2^{-3}=\tfrac12+\tfrac18=\tfrac58\), while \(\mu(\{1,3\})=2\).

The total mass of \(\nu\) is \(\nu(\mathbb{N})=\sum_{n=1}^{\infty}2^{-n}=1\), by the geometric-series formula with first term \(\tfrac12\) and ratio \(\tfrac12\). Thus \(\nu\) is finite even though the reference measure \(\mu\) is infinite. As an independent check for the computed set, the weights omitted from \(\{1,3\}\) sum to \(\sum_{n=2,\,n\ne3}^{\infty}2^{-n}=\tfrac14+\tfrac1{16}+\tfrac1{32}+\cdots=\tfrac38\); together with \(\tfrac58\), this gives the total mass \(1\).

More generally, on a countable set with its full power-set sigma-algebra, counting measure has no nonempty null set. Consequently every positive measure on that measurable space is absolutely continuous with respect to counting measure: the implication in the definition has no nonempty set to test. This is a genuine, though sometimes uninformative, instance of absolute continuity. It does not say that the other measure has a density, nor that the two measures have the same null sets.

Dirac Measures: Check Whether the Point Can Be Detected

Proposition: Let \((X,\mathcal{M})\) be any measurable space, let \(\mu\) be a positive measure, and let \(x_0\in X\). Define the Dirac measure by \(\delta_{x_0}(E)=1\) if \(x_0\in E\), and \(\delta_{x_0}(E)=0\) if \(x_0\notin E\), for \(E\in\mathcal{M}\). Then \(\delta_{x_0}\ll\mu\) if and only if no \(\mu\)-null measurable set contains \(x_0\).

Proof. By the definition of absolute continuity, \(\delta_{x_0}\ll\mu\) means that every measurable \(E\) with \(\mu(E)=0\) satisfies \(\delta_{x_0}(E)=0\). By the definition of the Dirac measure, \(\delta_{x_0}(E)=0\) exactly when \(x_0\notin E\). Thus the required condition is exactly that no \(\mu\)-null measurable set contains \(x_0\). \(\square\)

Worked Example: Lebesgue Measure and a Dirac Measure

On \((\mathbb{R},\mathcal{L})\), let \(m\) be Lebesgue measure and let \(\delta_0\) be the Dirac measure at \(0\). The singleton \(\{0\}\) is Lebesgue measurable and \(m(\{0\})=0\), while \(\delta_0(\{0\})=1\). Thus \(\{0\}\) is a null set for \(m\) that is not a null set for \(\delta_0\), and \(\delta_0\not\ll m\). In the reverse direction, \(m\not\ll\delta_0\): the measurable set \(\{1\}\) satisfies \(\delta_0(\{1\})=0\) and \(m(\{1\})=0\), so this set does not witness failure; instead use \(E=(1,2)\), for which \(\delta_0(E)=0\) and \(m(E)=1\). Therefore \(m\not\ll\delta_0\).

As a contrasting case, on \((\mathbb{R},\mathcal{L})\) with reference measure \(\mu=\delta_0\), the only \(\delta_0\)-null measurable sets are those not containing \(0\). For every such set \(E\), \(\delta_0(E)=0\), so \(\delta_0\ll\delta_0\). This agrees with the proposition: no \(\delta_0\)-null set contains \(0\).

The criterion also works when \(\{x_0\}\) is not measurable, so one must not use \(\mu(\{x_0\})\) unless the singleton belongs to \(\mathcal{M}\). For example, take \(X=\{a,b\}\), \(\mathcal{M}=\{\varnothing,X\}\), and the finite measure \(\mu\) with \(\mu(X)=1\). The only \(\mu\)-null measurable set is \(\varnothing\), which does not contain \(a\), so \(\delta_a\ll\mu\). But \(\{a\}\notin\mathcal{M}\); the expression \(\mu(\{a\})\) is undefined. This is why the general criterion is stated in terms of measurable null sets containing the point.

A Common Pitfall: Signed Cancellation

For a signed measure, a zero value on one null set does not by itself show that the set has zero total variation. Consider the finite measurable space \(X=\{a,b\}\), \(\mathcal{M}=\mathcal{P}(X)\), with the zero measure \(\mu\), and the finite signed measure \(\nu\) given by \(\nu(\{a\})=1\), \(\nu(\{b\})=-1\). The calculation \(\nu(X)=1+(-1)=0\) is sometimes incorrectly treated as evidence that \(\nu\ll\mu\). But \(\mu(\{a\})=0\), whereas \(|\nu|(\{a\})=|\nu(\{a\})|=1\). Hence \(\nu\not\ll\mu\). The correction is to test every \(\mu\)-null measurable set, or equivalently to use total variation: here \(\mu\) is zero on every set, but \(|\nu|\) is not.

Absolute continuity is useful because it records which null sets remain negligible under a change of measure. Its definition makes no assertion about the existence of a density; that is a separate question. For any proposed claim \(\nu\ll\mu\), begin with the sets \(E\) satisfying \(\mu(E)=0\), and verify that \(\nu(E)=0\) for positive \(\nu\), or \(|\nu|(E)=0\) for finite signed \(\nu\). For a Dirac measure, the decisive question is whether a \(\mu\)-null measurable set can contain its point of concentration.

Check Your Understanding

For each question, identify the relevant measurable null sets before deciding absolute continuity.

On \(\{a,b,c\}\) with the full power set, let \(\mu\) have masses \(1,0,4\) and let a positive measure \(\nu\) have masses \(2,3,1\), respectively. Is \(\nu\ll\mu\)?
On \(\mathbb{N}\) with counting measure, can a positive measure of infinite total mass be absolutely continuous with respect to counting measure? Explain using the null-set definition.
On a measurable space \((X,\mathcal{M})\), state the criterion for \(\delta_{x_0}\ll\mu\) without assuming that \(\{x_0\}\) is measurable.
For a finite signed measure \(\nu\), why does \(\nu(E)=0\) on a single \(\mu\)-null set not suffice to show \(\nu\ll\mu\)?
On \((\mathbb{R},\mathcal{L})\), which of \(m\ll\delta_0\) and \(\delta_0\ll m\) holds? Give a measurable witness for each failed implication.