When Does a Density Preserve the Same Null Sets?
Absolute continuity says that one measure assigns zero to every set that the other measure regards as negligible. Equivalence is stronger: the two measures have exactly the same null sets. When a Radon–Nikodym density exists, this distinction has a precise expression. A density that is merely nonnegative can make some sets invisible to the measure it defines; a density that is positive almost everywhere cannot.
All measures in this tutorial are nonnegative measures on a measurable space \((X,\mathcal{M})\). A measure \(\mu\) is sigma-finite if \(X\) is a countable union of sets of finite \(\mu\)-measure. For measures \(\mu,\nu\) on \((X,\mathcal{M})\), the notation \(\nu\ll\mu\) means that \(\mu(E)=0\) implies \(\nu(E)=0\) for every \(E\in\mathcal{M}\). The measures are equivalent, written \(\mu\sim\nu\), if both \(\nu\ll\mu\) and \(\mu\ll\nu\). Thus equivalence means preservation of null sets in both directions, not equality of the measures.
We use the Radon–Nikodym theorem in its sigma-finite form: if \(\mu\) and \(\nu\) are sigma-finite measures on \((X,\mathcal{M})\) and \(\nu\ll\mu\), then there is a nonnegative measurable function \(f\) such that \(\nu(E)=\int_E f\,d\mu\) for every \(E\in\mathcal{M}\). This function, unique up to equality \(\mu\)-almost everywhere, is denoted \(d\nu/d\mu\). Here “positive \(\mu\)-almost everywhere” means \(\mu(\{x:f(x)=0\})=0\); the density is finite \(\mu\)-almost everywhere under these hypotheses. We do not require the measure space to be complete.
A Zero-Integral Fact
We first prove the fact that turns an integral representation into a statement about null sets. It applies on any measure space \((X,\mathcal{M},\mu)\), without a finiteness assumption.
Proof. If \(g=0\) almost everywhere on \(E\), then the nonnegative function \(g\chi_E\) is zero almost everywhere, so its integral is zero. Conversely, suppose \(\int_E g\,d\mu=0\). For each integer \(k\ge1\), let \(A_k=E\cap\{g\ge1/k\}\). The pointwise inequality \(g\chi_E\ge (1/k)\chi_{A_k}\) gives \(0\le\mu(A_k)/k\le\int_Eg\,d\mu=0\), so \(\mu(A_k)=0\). Every point of \(E\) where \(g>0\), including a point where \(g=+\infty\), belongs to some \(A_k\). Therefore \(E\cap\{g>0\}=\bigcup_{k\ge1}A_k\) is null, as required. \(\square\)
Proof of the Equivalence Criterion
Proof of the theorem. Suppose first that \(\mu\sim\nu\). The set \(Z=\{x\in X:f(x)=0\}\) is measurable. By the density identity, \(\nu(Z)=\int_Z f\,d\mu=0\). Since \(\mu\ll\nu\), this implies \(\mu(Z)=0\). Hence \(f>0\) \(\mu\)-almost everywhere.
Conversely, suppose \(f>0\) \(\mu\)-almost everywhere. We already have \(\nu\ll\mu\) as a hypothesis. To prove the other direction, take any \(E\in\mathcal{M}\) with \(\nu(E)=0\). For each integer \(k\ge1\), set \(A_k=\{x:f(x)\ge1/k\}\). The density formula and nonnegativity give
It follows that \(\mu(E\cap A_k)=0\) for every \(k\). Since \(\bigcup_{k\ge1}A_k=\{f>0\}\), countable subadditivity gives \(\mu(E\cap\{f>0\})=0\). The remaining part \(E\cap\{f=0\}\) is null by the assumed almost-everywhere positivity of \(f\). Thus \(\mu(E)=0\), proving \(\mu\ll\nu\) and hence \(\mu\sim\nu\). \(\square\)
The sigma-finiteness assumptions serve to ensure that the stated Radon–Nikodym density exists under the theorem used here. Once its integral representation is available, the proof of the equivalence criterion uses only that representation and the definition of absolute continuity. The density is determined only up to a \(\mu\)-null set, so changing its values on such a set does not change the criterion.
Examples: Positive Densities and Failure of Equivalence
Worked Example: A Density That Vanishes at One Point
On \(X=[0,1]\) with its Borel sigma-algebra, let \(\mu\) be Lebesgue measure restricted to \([0,1]\), and define \(\nu(E)=\int_E2x\,dm(x)\) for every Borel set \(E\). Both measures are finite, hence sigma-finite. The density is \(f(x)=2x\), which vanishes at \(x=0\) but is positive for every \(x\in(0,1]\). Since \(\mu(\{0\})=0\), it is positive \(\mu\)-almost everywhere, so the criterion gives \(\mu\sim\nu\).
For an exact calculation, take \(E=[1/4,3/4]\). Its Lebesgue measure is \(3/4-1/4=1/2\), while
As a check on the normalization, \(\nu([0,1])=\int_0^1 2x\,dx=1^2-0^2=1=\mu([0,1])\). For any Borel \(E\) with \(\mu(E)=0\), the integral of \(2x\) over \(E\) is zero, so \(\nu(E)=0\). In the reverse direction, if \(\nu(E)=0\), the theorem gives \(\mu(E)=0\). The single point where the density vanishes causes no failure of equivalence.
Worked Example: Equivalent Measures on a Countable Space
Let \(X=\mathbb{N}=\{1,2,\ldots\}\), let \(\mathcal{M}\) be the power set of \(\mathbb{N}\), and let \(\mu\) be counting measure, so \(\mu(E)\) is the number of elements of \(E\), possibly \(+\infty\). It is sigma-finite because \(\mathbb{N}\) is a countable union of singletons, each of measure \(1\). Define \(\nu(E)=\sum_{n\in E}2^{-n}\). This is a finite measure. For every \(E\subseteq\mathbb{N}\),
Every value \(f(n)=2^{-n}\) is strictly positive, so \(f>0\) everywhere and the measures are equivalent. For example, for \(E=\{2,4\}\), \(\mu(E)=2\) and \(\nu(E)=2^{-2}+2^{-4}=1/4+1/16=5/16\). The density formula gives the same result: \(\int_Ef\,d\mu=f(2)+f(4)=1/4+1/16=5/16\).
As an independent check of the total mass, the geometric series gives \(\nu(\mathbb{N})=\sum_{n=1}^{\infty}2^{-n}=(1/2)/(1-1/2)=1\). Meanwhile \(\int_{\mathbb{N}}f\,d\mu=\sum_{n=1}^{\infty}2^{-n}=1\), as required by the density representation. Every nonempty subset has positive \(\nu\)-measure because it contains some \(n\) and therefore its sum includes the positive term \(2^{-n}\); the only \(\nu\)-null set is the empty set. Counting measure also has only the empty set as a null set. This verifies the same null-set conclusion directly.
Worked Example: A Density That Vanishes on a Set of Positive Measure
On \(X=[0,1]\) with its Borel sigma-algebra, let \(\mu\) be restricted Lebesgue measure and define \(\nu(E)=\int_E2\chi_{[0,1/2]}(x)\,dm(x)\). Both measures are finite. The density \(f=2\chi_{[0,1/2]}\) is nonnegative, so \(\nu\ll\mu\). But it vanishes on \((1/2,1]\), a set of \(\mu\)-measure \(1-1/2=1/2\); it is not positive \(\mu\)-almost everywhere.
The mass of \(\nu\) is \(\nu([0,1])=\int_0^{1/2}2\,dx=2(1/2-0)=1\). However, for \(E=(1/2,1]\), whose endpoints are excluded exactly as written, \(\mu(E)=1/2\), while \(\nu(E)=\int_E2\chi_{[0,1/2]}\,dm=0\). Thus \(\mu\not\ll\nu\), and the measures are not equivalent. The calculation illustrates why nonnegativity of the density alone is insufficient: it must be strictly positive almost everywhere for equivalence.
Check: the set where \(f=0\) in this example has measure \(1/2\), not zero. In the first example, the zero set was \(\{0\}\), which has measure zero. The criterion therefore predicts different outcomes, in agreement with the explicit null-set calculations.
A Common Mistake
A tempting but incorrect argument is: “The Radon–Nikodym density is nonnegative, so the measures are equivalent.” Nonnegativity proves only the direction \(\nu\ll\mu\) in the density representation: if \(\mu(E)=0\), then \(\int_E f\,d\mu=0\), hence \(\nu(E)=0\). It does not prove \(\mu\ll\nu\).
The third example supplies a concrete counterexample to that claim. There \(f=2\chi_{[0,1/2]}\) is nonnegative, but for \(E=(1/2,1]\) we have \(\mu(E)=1/2\) and \(\nu(E)=0\). The corrected test is strict positivity almost everywhere, not mere nonnegativity. Conversely, strict positivity need not hold at every point: the first example has \(f(0)=0\) and still gives equivalent measures because that exceptional point is \(\mu\)-null.
Check Your Understanding
In each question, distinguish the one-way absolute continuity supplied by a density from equivalence, which requires the same null sets in both directions.
- Let \(\mu=m\) on \([0,1]\) and define \(\nu(E)=\int_E x\,dm(x)\). Are \(\mu\) and \(\nu\) equivalent? Explain how the zero set of the density affects your answer.
- Suppose \(\nu(E)=\int_E f\,d\mu\), where \(\mu,\nu\) are sigma-finite and \(f\ge0\) is measurable. Which absolute-continuity direction follows directly from the representation?
- On \(\mathbb{N}\) with counting measure, let \(f(1)=0\) and \(f(n)=2^{-n}\) for \(n\ge2\). Is the measure defined by \(\nu(E)=\sum_{n\in E}f(n)\) equivalent to counting measure? Identify a set that decides the question.
- In the theorem’s converse, why do the sets \(\{f\ge1/k\}\) suffice to show that \(\nu(E)=0\) implies \(\mu(E)=0\) when \(f>0\) almost everywhere?
- Give an example of a density that is zero at some points but positive almost everywhere, and explain why those zero values do not prevent equivalence.