Why the Measure Hypotheses Matter
The Radon–Nikodym theorem says that, under suitable hypotheses, one measure can be represented by integrating a density against another. The finiteness conditions are not interchangeable: a finite measure is automatically sigma-finite, while a sigma-finite measure may have infinite total mass. If sigma-finiteness is dropped, absolute continuity alone need not produce a density.
Throughout, measures are positive measures on the stated measurable space and may take the value \(+\infty\). A measure \(\mu\) is finite if \(\mu(X)<\infty\). It is sigma-finite if there are measurable sets \(E_n\) with \(X=\bigcup_{n=1}^{\infty}E_n\) and \(\mu(E_n)<\infty\) for every \(n\). The space itself need not have finite measure in the sigma-finite case. Neither definition requires the measure space to be complete.
A density in the results below is a measurable function \(f:X\to[0,\infty]\). The representation \(\nu(E)=\int_E f\,d\mu\) is required for every measurable set \(E\); the integral may be infinite. In the sigma-finite theorem, \(f\) is finite \(\mu\)-a.e. In the finite theorem, finiteness of \(\nu(X)\) also forces \(f\) to be integrable.
The Two Radon–Nikodym Statements
These are standard forms of the Radon–Nikodym theorem; their proofs are given in the later treatment of the finite theorem and its sigma-finite extension. The finite statement is a special case of the sigma-finite one: if a measure is finite, taking \(E_1=X\) and \(E_n=\varnothing\) for \(n\ge2\) gives a sigma-finite cover. Conversely, the sigma-finite theorem allows infinite total mass, but it still assumes both measures are sigma-finite.
The following elementary criterion explains the role of absolute continuity in either statement.
Proof. Let \(E\in\mathcal{M}\) satisfy \(\mu(E)=0\). Every nonnegative simple function supported on \(E\) has integral zero, because each of its measurable level sets is contained in \(E\) and therefore has \(\mu\)-measure zero. By the definition of the nonnegative Lebesgue integral as a supremum over simple functions, \(\int_E f\,d\mu=0\). Thus \(\nu(E)=0\), which is precisely \(\nu\ll\mu\). \(\square\)
So absolute continuity is necessary for a density representation. The Radon–Nikodym theorems assert that, under their stated finiteness hypotheses, it is also sufficient.
Worked Examples: Checking the Hypotheses
Worked Example: A Finite Atomic Measure
Let \(X=\{a,b,c\}\), \(\mathcal{M}\) be the collection of all subsets of \(X\), and define \(\mu(\{a\})=\tfrac12\), \(\mu(\{b\})=\tfrac13\), and \(\mu(\{c\})=\tfrac16\). Define \(\nu\) by \(\nu(\{a\})=1\), \(\nu(\{b\})=0\), and \(\nu(\{c\})=\tfrac12\). Both measures are finite: \(\mu(X)=\tfrac12+\tfrac13+\tfrac16=1\), and \(\nu(X)=1+0+\tfrac12=\tfrac32\). Since every singleton of positive \(\mu\)-measure, and hence every nonempty set, has positive \(\mu\)-measure, \(\nu\ll\mu\).
Set \(f(a)=2\), \(f(b)=0\), and \(f(c)=3\). On this finite space, integration is a weighted sum. For the three singletons,
Every subset is a disjoint union of these singletons, so additivity gives \(\nu(E)=\int_E f\,d\mu\) for every \(E\in\mathcal{M}\). Check: for \(E=\{a,c\}\), the integral is \(2\cdot\tfrac12+3\cdot\tfrac16=1+\tfrac12=\tfrac32\), and \(\nu(E)=1+\tfrac12=\tfrac32\). For \(E=X\), the integral is \(\tfrac32=\nu(X)\).
Worked Example: Infinite but Sigma-Finite Lebesgue Measure
Take \((\mathbb{R},\mathcal{L},m)\), where \(m\) is Lebesgue measure, and let \(\nu=m\). The measures are not finite because \(m(\mathbb{R})=\infty\). They are sigma-finite: the measurable sets \(E_n=[-n,n]\), for \(n\ge1\), cover \(\mathbb{R}\), and \(m(E_n)=2n<\infty\) for every \(n\). Also \(\nu\ll m\), since \(\nu=m\).
The density \(f(x)=1\) is measurable and finite everywhere. For every \(E\in\mathcal{L}\), the integral of the constant function \(1\) over \(E\) is \(m(E)\), including when this value is infinite. Therefore
Check: for \(E=[-2,3]\), the interval has length \(3-(-2)=5\), so \(\nu(E)=5\) and \(\int_E1\,dm=5\). For \(E=\mathbb{R}\), both sides equal \(+\infty\). This example satisfies the sigma-finite theorem, not the finite theorem.
In a sigma-finite space, one can always choose a disjoint measurable partition into sets of finite measure. Indeed, given a cover \(E_n\), set \(F_1=E_1\) and \(F_n=E_n\setminus\bigcup_{k=1}^{n-1}E_k\) for \(n\ge2\). Then the \(F_n\) are disjoint, cover \(X\), and satisfy \(\mu(F_n)\le\mu(E_n)<\infty\). This elementary localization is one reason sigma-finiteness is useful when extending a finite-measure result.
When Sigma-Finiteness Is Missing
Worked Example: Absolute Continuity Without a Density
Let \(X\) be an uncountable set, \(\mathcal{M}=\mathcal{P}(X)\), and let \(\mu\) be counting measure: \(\mu(A)\) is the number of elements of \(A\) if \(A\) is finite, and \(+\infty\) if \(A\) is infinite. Define \(\nu(A)=0\) when \(A\) is countable and \(\nu(A)=+\infty\) when \(A\) is uncountable.
First, \(\nu\) is a measure. For pairwise disjoint sets \(A_n\), if \(\bigcup_n A_n\) is countable, then every \(A_n\) is countable and both \(\nu(\bigcup_n A_n)\) and \(\sum_n\nu(A_n)\) are zero. If \(\bigcup_n A_n\) is uncountable, at least one \(A_n\) must be uncountable; otherwise a countable union of countable sets would be countable. In that case both \(\nu(\bigcup_n A_n)\) and \(\sum_n\nu(A_n)\) equal \(+\infty\). Thus countable additivity holds.
The only \(\mu\)-null set is the empty set, so \(\nu\ll\mu\). But \(\mu\) is not sigma-finite: every set of finite \(\mu\)-measure is finite, and a countable union of finite sets is countable, so such sets cannot cover the uncountable \(X\). In fact, \(\nu\) is not sigma-finite either, since every set of finite \(\nu\)-measure is countable.
Suppose a measurable \(f:X\to[0,\infty]\) represented \(\nu\), so \(\nu(A)=\int_A f\,d\mu\) for every \(A\). For each \(x\in X\), applying the representation to the singleton \(\{x\}\) gives \(0=\nu(\{x\})=\int_{\{x\}}f\,d\mu=f(x)\), because \(\mu(\{x\})=1\). Hence \(f(x)=0\) at every point, and consequently \(\int_X f\,d\mu=0\). This contradicts \(\nu(X)=+\infty\). No density exists. Check: the contradiction uses both the singleton equalities and the value on \(X\); the first forces the integral on \(X\) to be zero, while the definition of \(\nu\) gives \(+\infty\).
This counterexample shows that \(\nu\ll\mu\) alone does not guarantee a density when \(\mu\) is not sigma-finite. Sigma-finiteness is not a cosmetic way to ensure finite total mass; it rules out this kind of failure.
Worked Example: Why the Represented Measure Is Also Assumed Sigma-Finite
Let \((X,\mathcal{M},\mu)=([0,1],\mathcal{L},m)\), and define \(\nu(E)=0\) if \(m(E)=0\), and \(\nu(E)=+\infty\) if \(m(E)>0\). This is a measure: a countable union of null sets is null, and if a disjoint union has positive measure, at least one member has positive measure. In the latter case both the measure of the union and the sum of the measures are infinite. Also \(\nu\ll m\).
The measure \(\nu\) is not sigma-finite, because its finite-measure sets are all Lebesgue null, and countably many null sets cannot cover \([0,1]\), whose measure is \(1\). The extended-valued function \(f(x)=+\infty\) for every \(x\) represents it: \(\int_E f\,dm=0\) if \(m(E)=0\), and \(\int_E f\,dm=+\infty\) if \(m(E)>0\).
However, there cannot be a representing density that is finite \(m\)-a.e. Suppose \(f\) were finite outside a null set \(N\). The measurable sets \(\{f\le n\}\), \(n\ge1\), together cover \(X\setminus N\). Since \(m(X)=1\), at least one set \(C=\{f\le n\}\) has positive measure: if every such set were null, their countable union and \(N\) would have measure zero, contradicting \(m(X)=1\). On this \(C\), \(\int_C f\,dm\le n\,m(C)\le n<\infty\), whereas \(\nu(C)=+\infty\). This contradicts the representation. Thus the usual conclusion that the density is finite a.e. relies on the sigma-finiteness hypothesis for \(\nu\).
Common Pitfall: Confusing Finite with Sigma-Finite
A frequent error is to treat “sigma-finite” as if it meant “finite.” The Lebesgue measure example on \(\mathbb{R}\) disproves that: \(m(\mathbb{R})=\infty\), but \(\mathbb{R}=\bigcup_{n\ge1}[-n,n]\) with \(m([-n,n])=2n<\infty\). The reverse error is to assume that absolute continuity alone always gives a density. In the uncountable counting-measure example, \(\nu\ll\mu\) holds, but the singleton calculations prove that no density can represent \(\nu\).
When applying the theorem, check the two measures separately: determine whether each is finite or sigma-finite, and verify \(\nu\ll\mu\). For signed or complex measures, the corresponding versions impose sigma-finiteness on the total variation of the measure being represented; the positive-measure statements above are the basic forms considered here.
Check Your Understanding
For each question, identify the measure hypotheses before deciding whether a Radon–Nikodym conclusion applies.
- Why does a finite measure automatically satisfy the sigma-finite condition? Give an explicit cover.
- On \((\mathbb{R},\mathcal{L},m)\), verify that \(m\) is not finite but is sigma-finite, and state a density representing \(m\) relative to itself.
- In the uncountable counting-measure example, why does any proposed density have to vanish at every point?
- For the measure \(\nu\) on \([0,1]\) that is infinite on every set of positive Lebesgue measure, why does the extended-valued density \(f=+\infty\) fail the finite-a.e. conclusion?
- State the two measure hypotheses and the absolute-continuity hypothesis in the sigma-finite Radon–Nikodym theorem.