Why Densities Multiply
A Radon–Nikodym derivative describes one measure relative to another. If a measure \(\nu\) is absolutely continuous with respect to \(\lambda\), and \(\lambda\) is absolutely continuous with respect to \(\mu\), it is natural to ask whether the two corresponding densities combine into a density of \(\nu\) relative to \(\mu\). They do: the densities multiply, in the order dictated by their reference measures. The key is to verify the proposed product by integrating it over every measurable set.
Throughout, \((X,\mathcal{M})\) is a measurable space, and \(\mu,\lambda,\nu\) are sigma-finite positive measures on it. A measure is sigma-finite if \(X\) is a countable union of measurable sets of finite measure. We write \(\nu\ll\lambda\) when \(\lambda(E)=0\) implies \(\nu(E)=0\) for every \(E\in\mathcal{M}\). By the Radon–Nikodym theorem, if both measures are sigma-finite and \(\nu\ll\lambda\), there is a nonnegative measurable function \(r\) such that \(\nu(E)=\int_E r\,d\lambda\) for all \(E\in\mathcal{M}\); this function is unique \(\lambda\)-almost everywhere. The corresponding statement applies to \(\lambda\ll\mu\). These are equalities of measures on every measurable set, while the densities themselves are determined only almost everywhere relative to their reference measures.
The Integral Identity Behind the Rule
The intermediate step is an identity for integrating a nonnegative function against \(\lambda\). It follows directly from the defining relation \(\lambda(E)=\int_E g\,d\mu\); no interchange of two integrals is needed.
Proof. First let \(q=\sum_{j=1}^k a_j\chi_{A_j}\) be a nonnegative simple function, where the \(A_j\) are disjoint measurable sets and \(a_j\ge0\). The definition of the integral of a simple function and the assumed identity for \(\lambda\) give $$ \int_E q\,d\lambda =\sum_{j=1}^k a_j\lambda(E\cap A_j) =\sum_{j=1}^k a_j\int_{E\cap A_j}g\,d\mu =\int_E qg\,d\mu. $$ For general nonnegative measurable \(q\), choose nonnegative simple functions \(q_n\uparrow q\), as in the definition of the nonnegative Lebesgue integral. Then \(q_ng\uparrow qg\). The Monotone Convergence Theorem, applied once with measure \(\lambda\) and once with measure \(\mu\), gives $$ \int_E q\,d\lambda =\lim_{n\to\infty}\int_E q_n\,d\lambda =\lim_{n\to\infty}\int_E q_ng\,d\mu =\int_E qg\,d\mu. $$ All functions in both applications are nonnegative and measurable, and the sequences increase pointwise. This proves the identity, including when the common value is infinite. \(\square\)
Proof of the Chain Rule. Let \(r=d\nu/d\lambda\) and \(g=d\lambda/d\mu\). The Radon–Nikodym identities give \(\nu(E)=\int_E r\,d\lambda\) for every \(E\in\mathcal{M}\). Apply the lemma with \(q=r\). For every such \(E\), $$ \nu(E)=\int_E r\,d\lambda=\int_E rg\,d\mu. $$ Thus \(rg\) is a nonnegative measurable density of \(\nu\) with respect to \(\mu\). Since \(\nu\ll\mu\) follows from \(\nu\ll\lambda\ll\mu\), the Radon–Nikodym theorem gives a density \(d\nu/d\mu\). Uniqueness in that theorem implies \(rg=d\nu/d\mu\) \(\mu\)-almost everywhere. The product can be understood using finite-valued versions of the densities: for sigma-finite measures their derivatives are finite almost everywhere relative to their respective reference measures, and changing a derivative on a null set does not change its defining integrals. \(\square\)
The measure identity in this proof holds for every \(E\in\mathcal{M}\); the conclusion that the two density functions agree is only an almost-everywhere statement. Also note the order: \(d\nu/d\lambda\) is multiplied by \(d\lambda/d\mu\), so the intermediate reference measure \(\lambda\) cancels.
Worked Examples
Worked Example: Weighted Measures on the Unit Interval
Let \(X=(0,1)\), \(\mathcal{M}\) be the Lebesgue sigma-algebra, and let \(\mu=m\) be Lebesgue measure. Define \(\lambda(E)=\int_E x\,dm\) and \(\nu(E)=\int_E 2x^2\,dm\). Find the two intermediate densities and check the chain rule on \(E=(1/2,1)\).
By their definitions, \(g=d\lambda/dm=x\). Since \(x>0\) throughout \(X\), the density of \(\nu\) relative to \(\lambda\) is \(r=2x\): for every measurable \(E\), $$ \int_E 2x\,d\lambda=\int_E 2x\cdot x\,dm=\int_E 2x^2\,dm=\nu(E). $$ The product is \(rg=2x^2\), which is the density defining \(\nu\) relative to \(m\). For the specified set, direct integration gives $$ \nu((1/2,1))=\int_{1/2}^1 2x^2\,dx =\left[\frac{2x^3}{3}\right]_{1/2}^1 =\frac{2}{3}-\frac{1}{12} =\frac{7}{12}. $$ Using the chain-rule product gives the same calculation: $$ \int_{1/2}^1 rg\,dm =\int_{1/2}^1 2x^2\,dx =\frac{7}{12}. $$ Check: the intermediate integral is \(\int_{1/2}^1 r\,d\lambda=\int_{1/2}^1 2x\cdot x\,dx=7/12\), agreeing with \(\nu(E)\).
Worked Example: A Zero Intermediate Density on an Atom
Let \(X=\{a,b,c\}\), let \(\mathcal{M}\) be all subsets of \(X\), and let \(\mu\) be counting measure. Define \(\lambda(\{a\})=2\), \(\lambda(\{b\})=0\), \(\lambda(\{c\})=3\), and \(\nu(\{a\})=6\), \(\nu(\{b\})=0\), \(\nu(\{c\})=3\). Determine the densities and check them on \(E=\{a,c\}\).
Relative to counting measure, the density \(g=d\lambda/d\mu\) takes the values \(g(a)=2\), \(g(b)=0\), and \(g(c)=3\). On the atoms with positive \(\lambda\)-mass, the ratios defining \(r=d\nu/d\lambda\) are \(r(a)=6/2=3\) and \(r(c)=3/3=1\). At \(b\), \(\lambda(\{b\})=0\), so the value of \(r(b)\) does not affect its \(\lambda\)-almost-everywhere class; choose \(r(b)=0\). The product has values \(rg(a)=3\cdot2=6\), \(rg(b)=0\cdot0=0\), and \(rg(c)=1\cdot3=3\), exactly the masses of \(\nu\).
On \(E=\{a,c\}\), the direct measure is \(\nu(E)=6+3=9\). Integrating the product against counting measure gives \(\int_E rg\,d\mu=6+3=9\). The intermediate computation is \(\int_E r\,d\lambda=3\cdot2+1\cdot3=9\). Check: on the remaining atom \(b\), both \(\nu(\{b\})\) and \(\int_{\{b\}}rg\,d\mu\) are zero, so the equality also holds for \(E=X\).
Worked Example: Densities on the Natural Numbers
Let \(X=\mathbb{N}=\{1,2,\ldots\}\) with all subsets measurable. Let \(\mu\) be counting measure, set \(\lambda(\{n\})=2^{-n}\), and set \(\nu(\{n\})=n2^{-n}\). Find the chain-rule product and verify the measures of \(A=\{1,3\}\).
At each point \(n\), \(g(n)=d\lambda/d\mu(n)=2^{-n}\), since integration against counting measure is summation. The ratio of the atomic masses gives \(r(n)=d\nu/d\lambda(n)=(n2^{-n})/2^{-n}=n\). Hence \(r(n)g(n)=n2^{-n}\), the mass of \(\nu\) at \(n\). For \(A=\{1,3\}\), the direct value is $$ \nu(A)=1\cdot2^{-1}+3\cdot2^{-3} =\frac12+\frac38 =\frac78. $$ The product integral is \(\int_A rg\,d\mu=1\cdot2^{-1}+3\cdot2^{-3}=7/8\); the intermediate integral is \(\int_A r\,d\lambda=1\cdot2^{-1}+3\cdot2^{-3}=7/8\). Check: the calculation uses the same atom-by-atom contributions in all three expressions. The measures are finite as well as sigma-finite: \(\lambda(X)=\sum_{n\ge1}2^{-n}=1\) and \(\nu(X)=\sum_{n\ge1}n2^{-n}=2\).
Common Pitfall: Reversing the Density Order
The chain rule is not a rule for multiplying any two density expressions in either order. Each derivative records a specific reference measure, and the middle measure must match: \(d\nu/d\lambda\) goes with \(d\lambda/d\mu\). In the interval example, reversing the operation by dividing \(2x\) by \(x\) would give the incorrect proposed density \(2\) for \(\nu\) relative to \(m\). Its integral over \((0,1)\) is \(\int_0^1 2\,dx=2\), whereas the actual total mass is \(\nu((0,1))=\int_0^1 2x^2\,dx=2/3\). Multiplication in the correct order gives \((2x)x=2x^2\), whose integral is \(2/3\).
A second point is that densities need not be specified at every point. In the finite example, \(r(b)\) could be changed without changing \(d\nu/d\lambda\), because \(\lambda(\{b\})=0\). The theorem asserts equality of densities \(\mu\)-almost everywhere, not equality for every chosen pair of representatives. Its proof avoids any ambiguity by checking the defining integral on every measurable set, then using uniqueness of the Radon–Nikodym derivative.
Check Your Understanding
For each question, keep track of the reference measure in every derivative and state the measure relative to which any density equality holds.
- Suppose \(d\lambda/d\mu=g\) and \(d\nu/d\lambda=r\) for sigma-finite positive measures with \(\nu\ll\lambda\ll\mu\). What integral identity verifies that \(rg\) is a density of \(\nu\) relative to \(\mu\)?
- In the interval example, compute \(\lambda((1/2,1))\) and \(\int_{1/2}^1 r\,d\lambda\). Verify that the second quantity equals \(\nu((1/2,1))\).
- In the three-point example, why can the value of \(r(b)\) be changed without changing its Radon–Nikodym derivative class?
- For counting measure on \(\mathbb{N}\), use the atomic masses \(\lambda(\{n\})=3^{-n}\) and \(\nu(\{n\})=2\cdot3^{-n}\) to identify \(d\nu/d\lambda\), \(d\lambda/d\mu\), and their product.
- Why does the chain-rule conclusion assert equality almost everywhere rather than equality at every point?