Equivalent Measures and Reciprocal Densities
When two measures have exactly the same null sets, each is absolutely continuous with respect to the other. Their Radon–Nikodym derivatives describe the change from one measure to the other, and the two derivatives undo one another almost everywhere. The qualification “almost everywhere” matters: a chosen version of a density can be zero or infinite on an exceptional null set, where taking its pointwise reciprocal may not make sense.
Throughout, \((X,\mathcal{M})\) is a measurable space, and \(\mu\) and \(\nu\) are sigma-finite measures on it. Sigma-finite means that \(X\) is a countable union of measurable sets of finite measure. The measures are equivalent, written \(\mu\sim\nu\), if \(\mu\ll\nu\) and \(\nu\ll\mu\); here \(\mu\ll\nu\) means that \(\nu(E)=0\) implies \(\mu(E)=0\) for every \(E\in\mathcal{M}\). All densities below are nonnegative measurable functions, defined up to almost-everywhere equality.
We use the chain rule for Radon–Nikodym derivatives, established in the earlier material on the chain rule. In precise form: for sigma-finite measures \(\alpha,\beta,\gamma\) on the same measurable space, if \(\alpha\ll\beta\) and \(\beta\ll\gamma\), then \(\alpha\ll\gamma\) and \(\dfrac{d\alpha}{d\gamma}=\dfrac{d\alpha}{d\beta}\dfrac{d\beta}{d\gamma}\) \(\gamma\)-almost everywhere. The result applies here with \(\alpha=\gamma=\mu\) and \(\beta=\nu\).
The Reciprocal-Density Theorem
Proof. First, \(f\) is finite \(\mu\)-almost everywhere. To check this, cover \(X\) by measurable sets \(X_j\) with \(\mu(X_j)<\infty\). Since \(\int_{X_j}f\,d\mu=\nu(X_j)\), the integral is finite for each \(j\) after using a finite-measure cover for \(\nu\) as well, or restricting \(X_j\) to intersections with such a cover. More explicitly, the sets \(X_j\cap Y_k\), where \(\mu(X_j)<\infty\) and \(\nu(Y_k)<\infty\), cover \(X\); on each, \(\int_{X_j\cap Y_k}f\,d\mu=\nu(X_j\cap Y_k)\le\nu(Y_k)<\infty\). A nonnegative function with finite integral is finite almost everywhere on that set. Countable union gives the claim on \(X\).
Let \(Z=\{x:f(x)=0\}\). The density identity gives \(\nu(Z)=\int_Z f\,d\mu=0\). Since \(\mu\ll\nu\), it follows that \(\mu(Z)=0\). Thus \(f>0\) \(\mu\)-almost everywhere. By equivalence, every \(\mu\)-null set is also \(\nu\)-null, so \(f\) is finite and strictly positive \(\nu\)-almost everywhere as well.
Now apply the chain rule to \(\mu\ll\nu\ll\mu\). Since \(\dfrac{d\mu}{d\mu}=1\) \(\mu\)-almost everywhere, it gives $$ 1=\frac{d\mu}{d\mu} =\frac{d\mu}{d\nu}\frac{d\nu}{d\mu} =gf \qquad \mu\text{-almost everywhere}. $$ On the \(\mu\)-full-measure set where \(f\) is finite and positive, division by \(f\) yields \(g=1/f\). Because \(\mu\) and \(\nu\) have the same null sets, this equality also holds \(\nu\)-almost everywhere. \(\square\)
To make a reciprocal function defined everywhere, take any measurable version of \(f\) and set $$ h(x)= \begin{cases} 1/f(x),&0<f(x)<\infty,\\ 0,&\text{otherwise}. \end{cases} $$ The exceptional set \(N=\{f=0\}\cup\{f=\infty\}\) is null for both measures, so this convention gives \(h=g\) almost everywhere for both \(\mu\) and \(\nu\). The value zero is a convenient choice, not an additional requirement on the Radon–Nikodym derivative.
Worked Example: Two Equivalent Measures on a Finite Set
Let \(X=\{a,b,c\}\) with all subsets measurable. Give \(\mu\) the point masses \(2,3,1\) at \(a,b,c\), respectively, and give \(\nu\) the point masses \(4,\tfrac32,5\). All masses are positive, so the measures have the same null sets.
On a finite set with counting measure as reference, the Radon–Nikodym derivative at a point is the ratio of the two point masses. Thus $$ f(a)=\frac{4}{2}=2,\qquad f(b)=\frac{3/2}{3}=\frac12,\qquad f(c)=\frac{5}{1}=5. $$ The candidate reciprocal values are \(g(a)=\tfrac12\), \(g(b)=2\), and \(g(c)=\tfrac15\). Check the density identities on \(E=\{a,c\}\): \(\nu(E)=4+5=9\), while \(\int_E f\,d\mu=2\cdot2+5\cdot1=9\). In the reverse direction, \(\mu(E)=2+1=3\), while \(\int_E g\,d\nu=\tfrac12\cdot4+\tfrac15\cdot5=3\). Finally, the pointwise products are \(2\cdot\tfrac12=1\), \(\tfrac12\cdot2=1\), and \(5\cdot\tfrac15=1\), as required.
Worked Example: A Density That Vanishes at a Null Point
On \([0,1]\) with its Borel sets, let \(\mu=m\) be Lebesgue measure and define \(\nu(E)=\int_E 2x\,dm(x)\). The density is \(f(x)=2x\), including \(f(0)=0\). Both measures are finite: \(\mu([0,1])=1\), and \(\nu([0,1])=\int_0^1 2x\,dx=1\).
They are equivalent. If \(\mu(E)=0\), then \(\nu(E)=\int_E2x\,dm=0\). Conversely, if \(\nu(E)=0\), then for each positive integer \(k\), \(2x\ge 1/k\) on \(E\cap[1/(2k),1]\). Hence \(0=\nu(E)\ge (1/k)m(E\cap[1/(2k),1])\), so each such intersection is null. Their union over \(k\) is \(E\cap(0,1]\), and the remaining point \(\{0\}\) is also Lebesgue-null. Therefore \(\mu(E)=0\).
The reciprocal density is \(g(x)=1/(2x)\) for \(x>0\); define \(g(0)=0\). For a direct check on \(E=(1/4,1/2]\), first compute \(\nu(E)=\int_{1/4}^{1/2}2x\,dx=(1/2)^2-(1/4)^2=3/16\). Then \(\int_E g\,d\nu=\int_{1/4}^{1/2}(1/(2x))\,2x\,dx=1/2-1/4=1/4=m(E)\). The reciprocal relation also checks pointwise on \(x>0\): \((2x)(1/(2x))=1\). At \(x=0\), the reciprocal is not defined by the quotient, but that point is null for both measures.
Worked Example: Equivalent Measures on a Countable Space
Let \(X=\mathbb{N}=\{1,2,\ldots\}\), with all subsets measurable. Set \(\mu(\{n\})=2^{-n}\) and \(\nu(\{n\})=4^{-n}\). Both are finite, since \(\sum_{n\ge1}2^{-n}=1\) and \(\sum_{n\ge1}4^{-n}=(1/4)/(1-1/4)=1/3\). Every singleton has positive mass for both, so they are equivalent.
For each \(n\), the density ratios are \(f(n)=4^{-n}/2^{-n}=2^{-n}\) and \(g(n)=2^{-n}/4^{-n}=2^n\). For every subset \(A\subseteq\mathbb{N}\), $$ \int_A f\,d\mu=\sum_{n\in A}2^{-n}2^{-n} =\sum_{n\in A}4^{-n}=\nu(A), \qquad \int_A g\,d\nu=\sum_{n\in A}2^n4^{-n} =\sum_{n\in A}2^{-n}=\mu(A). $$ For example, if \(A=\{1,3\}\), then \(\nu(A)=1/4+1/64=17/64\), and \(\int_A f\,d\mu=1/4+1/64=17/64\). Also \(\mu(A)=1/2+1/8=5/8\), and \(\int_Ag\,d\nu=1/2+1/8=5/8\). The reciprocal \(g(n)=2^n\) is unbounded, but it is finite at every point and is a valid density.
Almost Everywhere, Not Everywhere
A common mistake is to require a Radon–Nikodym density to be strictly positive at every point before taking its reciprocal. In the continuous example, \(f(0)=0\), so the pointwise calculation \(1/f(0)=1/0\) is undefined. It would be wrong to conclude that the reciprocal-density theorem fails or that the measures are not equivalent. The correction is that \(\{0\}\) is null for both measures: the density can be redefined there, for instance by setting \(f(0)=1\) before taking reciprocals, or by defining \(h(0)=0\) as above. The resulting reciprocal identity is an almost-everywhere identity, not an everywhere identity for arbitrary versions.
The null-set qualification also accounts for changing versions. If \(f\) is changed on a \(\mu\)-null set, it remains a version of \(d\nu/d\mu\). Since \(\mu\sim\nu\), the same set is \(\nu\)-null. Thus no choice on such a set affects the theorem. In applications, it is safest to state the identity as \(g=1/f\) almost everywhere and specify a convention only when a pointwise-defined reciprocal function is needed.
Check Your Understanding
Use the hypotheses and the almost-everywhere qualification carefully in each question.
- State the meaning of \(\mu\sim\nu\) and explain why it ensures that a \(\mu\)-null exceptional set is also \(\nu\)-null.
- In the finite-set example, compute \(d\nu/d\mu\) and \(d\mu/d\nu\) at each of \(a,b,c\), then verify their products.
- For \(\nu(E)=\int_E 3x^2\,dm(x)\) on \([0,1]\), identify the reciprocal density on \((0,1]\) and give a finite convention at \(0\).
- Explain why the chain rule can be applied with \(\alpha=\gamma=\mu\) and \(\beta=\nu\), and identify the derivative \(d\mu/d\mu\).
- For the countable example, compute \(d\nu/d\mu\) at \(n=4\) and its reciprocal, and verify that their product is \(1\).