Null Sets as a Test for a Density
A density represents a measure by integrating a function against a reference measure. This representation imposes an immediate restriction: a set of reference measure zero must also have measure zero under the represented measure. Thus, a single null set carrying positive mass rules out every possible density, not just a particular candidate. This test detects both point masses and measures spread continuously over a null set.
Throughout, \((X,\mathcal{M},\mu)\) is a measure space and \(\nu\) is a positive measure on the same measurable space. Unless stated otherwise, the measures in examples are finite. A function \(g:X\to[0,\infty]\) is a Radon–Nikodym density of \(\nu\) with respect to \(\mu\) if it is \(\mathcal{M}\)-measurable and \(\nu(E)=\int_E g\,d\mu\) for every \(E\in\mathcal{M}\). When \(\nu\) is finite, such a density is integrable, since \(\int_X g\,d\mu=\nu(X)<\infty\). Equality of densities is understood \(\mu\)-almost everywhere.
Absolute continuity is a condition about all measurable null sets. Singularity says that the measure is concentrated on one particular null set. A nonzero measure singular to \(\mu\) cannot be absolutely continuous with respect to \(\mu\): its concentrating set has \(\mu\)-measure zero but carries its entire positive mass.
The Criterion for Nonexistence
Proof. If a density \(g\) exists and \(\mu(E)=0\), then the integral of any nonnegative measurable function over \(E\) is zero. Therefore \(\nu(E)=\int_Eg\,d\mu=0\), proving \(\nu\ll\mu\). Conversely, if \(\nu\ll\mu\), the Radon–Nikodym theorem for sigma-finite measures gives a nonnegative measurable \(g\) such that \(\nu(E)=\int_Eg\,d\mu\) for every \(E\in\mathcal{M}\). Thus the two conditions are equivalent. The final statement is the contrapositive of the definition of absolute continuity: failure of \(\nu\ll\mu\) means that there is a measurable \(E\) with \(\mu(E)=0\) and \(\nu(E)>0\). \(\square\)
The Radon–Nikodym theorem used in the converse is a standard result established in the earlier material on the Radon–Nikodym theorem; its proof is not repeated here. The diagnostic direction is especially direct: to disprove existence, it is enough to find one null set with positive \(\nu\)-measure.
Worked Example: A Point Mass Has No Lebesgue Density
Let \(X=[0,1]\), let \(\mathcal{M}=\mathcal{B}_{[0,1]}\) be its Borel sigma-algebra, and let \(\mu\) be Lebesgue measure restricted to \(\mathcal{B}_{[0,1]}\). Set \(\nu=\delta_{1/2}\), the Dirac measure at \(1/2\), so \(\nu(E)=1\) if \(1/2\in E\) and \(\nu(E)=0\) otherwise, for Borel \(E\).
The singleton \(E=\{1/2\}\) is Borel and \(\mu(E)=0\), because a singleton has Lebesgue measure zero. But \(\nu(E)=1\), since \(1/2\in E\). Hence \(\nu\not\ll\mu\), and the theorem rules out a density. More explicitly, if \(g\) were a nonnegative measurable density, then \(\nu(\{1/2\})=\int_{\{1/2\}}g\,d\mu=0\), contradicting \(\nu(\{1/2\})=1\). The integral over a \(\mu\)-null set is zero even if \(g\) takes the value \(+\infty\) there.
Check: \(\nu(X)=1\), as \(1/2\in[0,1]\); the contradiction is not a mismatch in total mass but the mass assigned to a null singleton.
A Singular Measure Without Atoms
A point mass is concentrated at one point. Failure of absolute continuity is not limited to this atomic case: a measure can assign zero mass to every singleton and still place positive mass on a Lebesgue-null set. The standard example is the Cantor probability measure.
Worked Example: The Cantor Probability Measure Has No Lebesgue Density
Take \(X=[0,1]\), \(\mathcal{M}=\mathcal{B}_{[0,1]}\), and let \(\mu\) be Lebesgue measure restricted to the Borel sets. Let \(C\) be the middle-thirds Cantor set. At stage \(n\), the construction leaves \(2^n\) closed intervals, each of length \(3^{-n}\). Their union \(C_n\) is Borel, \(C=\bigcap_{n\geq 0}C_n\), and \(m(C_n)=2^n3^{-n}=(2/3)^n\). Since \(C\subseteq C_n\) for every \(n\), \(m(C)\leq(2/3)^n\) for every \(n\), so \(m(C)=0\).
Let \(\nu\) be the standard Cantor probability measure: each of the \(2^n\) stage-\(n\) intervals has mass \(2^{-n}\), and \(\nu(C)=1\). It is concentrated on \(C\), so \(\nu(C)=1\) while \(\mu(C)=0\). Consequently \(\nu\not\ll\mu\) and there is no density with respect to Lebesgue measure.
This example is continuous in the measure-theoretic sense of having no atoms. Indeed, for any \(x\in C\), at each stage \(n\), \(x\) lies in a stage-\(n\) interval of mass \(2^{-n}\). Therefore \(0\leq\nu(\{x\})\leq2^{-n}\) for every \(n\), which forces \(\nu(\{x\})=0\). If \(x\notin C\), then \(\nu(\{x\})=0\) because \(\nu\) is concentrated on \(C\). Thus every singleton has \(\nu\)-measure zero, even though the null set \(C\) has measure one under \(\nu\).
Check: at stage \(2\), there are \(2^2=4\) intervals, each of length \(3^{-2}=1/9\), so their total length is \(4/9=(2/3)^2\). Each carries Cantor measure \(2^{-2}=1/4\), giving total mass \(4(1/4)=1\). The shrinking total lengths and unchanged total mass show why this measure cannot have a Lebesgue density.
A Density Can Vanish Without Failing to Exist
Not every zero of a density signals nonexistence. A density may vanish on a set of positive reference measure and still represent a nonzero measure. Absolute continuity only requires that reference-null sets have measure zero; it does not require the density to be strictly positive. Strict positivity almost everywhere is relevant to equivalence of measures, not to the existence of a density.
Worked Example: A Nonzero Density That Vanishes on Half the Space
Again take \(X=[0,1]\), \(\mathcal{M}=\mathcal{B}_{[0,1]}\), and let \(\mu\) be Lebesgue measure restricted to Borel sets. Define \(g(x)=2\chi_{[0,1/2]}(x)\) and, for Borel \(E\subseteq[0,1]\), define \(\nu(E)=\int_Eg\,d\mu\). This explicitly defines \(\nu\) only on Borel sets.
The total mass is \(\nu(X)=2m([0,1/2])=2(1/2)=1\). For \(E=[0,1/4]\), \(\nu(E)=2m([0,1/4])=2(1/4)=1/2\). For \(F=(1/2,1]\), \(\nu(F)=0\), since \(g=0\) throughout \(F\). The density vanishes on \(F\), which has \(\mu(F)=1/2>0\), but it is not zero almost everywhere: \(g=2\) on \([0,1/2]\), a set of measure \(1/2\). The measure is nonzero and absolutely continuous, and its density exists by its defining formula.
Check: the disjoint sets \([0,1/2]\) and \((1/2,1]\) partition \(X\). Their \(\nu\)-masses are \(2(1/2)=1\) and \(0\), respectively, adding to the previously computed total mass \(1\).
Proof. Suppose first that \(g=0\) \(\mu\)-almost everywhere. There is a measurable null set \(N\) outside which \(g=0\). For every \(E\in\mathcal{M}\), the integral \(\int_Eg\,d\mu\) is zero, since \(g\) vanishes outside \(N\) and the integral over \(E\cap N\) is zero. Thus \(\nu(E)=0\) for every \(E\). Conversely, suppose that \(\nu\) is the zero measure. For each positive integer \(n\), let \(E_n=\{x:g(x)>1/n\}\), which is measurable. On \(E_n\), \(g\geq1/n\), so \(\nu(E_n)=\int_{E_n}g\,d\mu\geq(1/n)\mu(E_n)\). Since \(\nu(E_n)=0\), it follows that \(\mu(E_n)=0\). The set where \(g>0\) is \(\bigcup_{n=1}^{\infty}E_n\): any positive value of \(g\), including \(+\infty\), exceeds \(1/n\) for some \(n\). Countable subadditivity gives \(\mu(\{g>0\})=0\), hence \(g=0\) \(\mu\)-almost everywhere. \(\square\)
Common Mistakes and a Reliable Test
Mistake: putting an atom into a density by assigning it an infinite value. For the point mass at \(1/2\), one might propose \(g=\infty\chi_{\{1/2\}}\). This does not work: since \(\mu(\{1/2\})=0\), \(\int_{\{1/2\}}g\,d\mu=0\), not \(1\). An infinite function value at a null point does not create mass there. The correction is to test the defining identity on the singleton itself; its integral is zero for every nonnegative measurable candidate density.
Mistake: confusing a zero region with nonexistence. In the preceding density example, \(\nu((1/2,1])=0\) while \(\mu((1/2,1])=1/2\). This does not violate \(\nu\ll\mu\), whose implication runs from \(\mu(E)=0\) to \(\nu(E)=0\), not in the reverse direction. The measure is nonzero because \(\nu([0,1/2])=1\).
Domain check. In the examples on \([0,1]\), the measures were defined on \(\mathcal{B}_{[0,1]}\). The formulas for \(\nu(E)\) therefore apply to Borel \(E\). They do not, without further definition, assert values on every Lebesgue-measurable subset of \([0,1]\). If instead a measure is defined on the completed Lebesgue sigma-algebra, then its domain includes those additional sets; the domain must be specified before using a formula.
In practice, when asked whether a density exists, first identify the reference measure \(\mu\) and the sigma-algebra on which both measures are defined. Then search for a measurable \(E\) with \(\mu(E)=0\) and \(\nu(E)>0\). Such a set proves at once that \(\nu\not\ll\mu\) and that no Radon–Nikodym density can represent \(\nu\). If no such set exists and the measures satisfy the sigma-finiteness hypotheses, the Radon–Nikodym theorem supplies a density. Once a density exists, its vanishing on a positive-measure set is compatible with absolute continuity; vanishing almost everywhere on the whole space, by contrast, represents only the zero measure.
Check Your Understanding
For each question, keep the reference measure and the domain of the measures explicit.
- On the Borel subsets of \([0,1]\), compare Lebesgue measure with \(\delta_0\). Which measurable null set rules out a Lebesgue density for \(\delta_0\)?
- Why does the fact that the Cantor probability measure gives every singleton mass zero not imply that it is absolutely continuous with respect to Lebesgue measure?
- For \(g=3\chi_{[0,1/3]}\) on \([0,1]\) with Lebesgue measure, compute \(\int_0^1g\,dm\). On what positive-measure set does \(g\) vanish, and does this prevent a density from existing?
- Prove that if \(\nu(E)=\int_Eg\,d\mu\) for all measurable \(E\), then \(\mu(E)=0\) implies \(\nu(E)=0\).
- If a nonnegative density \(g\) represents the zero measure, what can be concluded about \(g\) almost everywhere, and which sets \(\{g>1/n\}\) establish the conclusion?