Chaining Densities Across Three Measures

Composing Derivatives Through an Intermediate Measure

A Radon–Nikodym derivative depends on its reference measure. If $\mu$, $\lambda$, and $\nu$ are related by absolute continuity, one can sometimes calculate a derivative relative to $\nu$ by passing through $\lambda$. The order matters: the derivative of $\mu$ relative to $\lambda$ must be multiplied by the derivative of $\lambda$ relative to $\nu$. The null sets also matter, because each density is determined only almost everywhere with respect to its own reference measure.

Throughout, $(X,\mathcal M)$ is a measurable space, and $\mu,\lambda,\nu$ are positive measures on it. A measure is sigma-finite if $X$ is a countable union of measurable sets of finite measure. We assume sigma-finiteness when invoking the Radon–Nikodym theorem. The finite examples below are also sigma-finite. No completeness assumption is needed. Densities are nonnegative, extended-real-valued measurable functions, chosen finite almost everywhere with respect to their reference measures.

Definition: For measures $\alpha,\beta$ on $(X,\mathcal M)$, $\alpha\ll\beta$ means that $\beta(E)=0$ implies $\alpha(E)=0$ for every $E\in\mathcal M$. When $\alpha$ and $\beta$ are sigma-finite and $\alpha\ll\beta$, the Radon–Nikodym theorem gives a measurable function $r\ge0$ such that $\alpha(E)=\int_E r\,d\beta$ for every $E\in\mathcal M$. This function, unique $\beta$-almost everywhere, is denoted $d\alpha/d\beta$.

The chain rule below uses the Radon–Nikodym theorem and its uniqueness, as established in the earlier material on the theorem. One integration identity will be useful: if $d\lambda/d\nu=g$, then $\int h\,d\lambda=\int hg\,d\nu$ for every nonnegative measurable $h$. For an indicator $h=\chi_E$, this is the defining identity $\lambda(E)=\int_Eg\,d\nu$; finite nonnegative simple functions follow by addition and scaling. For general $h$, choose an increasing sequence of nonnegative simple functions converging to $h$. The Monotone Convergence Theorem on each measure space gives the identity by taking limits. Thus this use of integration against a density does not require either measure to be finite.

The Three-Measure Chain Rule

Theorem (Chain rule): Let $\mu,\lambda,\nu$ be sigma-finite positive measures on the same measurable space $(X,\mathcal M)$. If $\mu\ll\lambda$ and $\lambda\ll\nu$, then $\mu\ll\nu$, and $$ \frac{d\mu}{d\nu} = \frac{d\mu}{d\lambda}\frac{d\lambda}{d\nu} \qquad \nu\text{-almost everywhere}. $$

Proof. If $\nu(E)=0$, then $\lambda(E)=0$ because $\lambda\ll\nu$, and then $\mu(E)=0$ because $\mu\ll\lambda$. Thus $\mu\ll\nu$. Write $f=d\mu/d\lambda$ and $g=d\lambda/d\nu$. Choose a measurable version of $f$ that is finite outside a $\lambda$-null set $N$, and redefine it as zero on $N$. This does not change its $\lambda$-almost-everywhere class or its derivative identity. Since $\lambda(N)=0$, the identity for $g$ gives $\int_Ng\,d\nu=0$. As $g\ge0$, this implies $g=0$ $\nu$-almost everywhere on $N$. Hence changing the version of $f$ on $N$ cannot change the product $fg$ except on a $\nu$-null set; in particular, the product can be taken to be measurable and finite where $g>0$, up to a $\nu$-null set.

For every $E\in\mathcal M$, apply the integration identity above with $h=f\chi_E$. It gives $$ \int_E fg\,d\nu=\int_E f\,d\lambda=\mu(E). $$ Therefore $fg$ is a Radon–Nikodym derivative of $\mu$ with respect to $\nu$. Uniqueness of that derivative gives $d\mu/d\nu=fg$ $\nu$-almost everywhere. $\square$

The null-set check explains why the product is well-defined as a derivative even though $f$ is specified only $\lambda$-almost everywhere. Any $\lambda$-null set on which its version might be changed is a set where $g=0$, except for a $\nu$-null subset. This is not the same as saying that all $\lambda$-null sets are $\nu$-null.

Worked Example: Chaining Densities on a Finite Space

Let $X=\{a,b,c\}$, let $\mathcal M$ be all subsets of $X$, and give the points the masses $\nu=(2,3,5)$, $\lambda=(1,0,4)$, and $\mu=(3,0,2)$, in that order. These are finite measures. Compute the three densities and check the chain rule at every point.

On a finite space, the density at a point of positive reference mass is the ratio of the two point masses. Since $\lambda$ has mass zero at $b$, $\mu\ll\lambda$ still holds: both $\mu$ and $\lambda$ assign mass zero to $\{b\}$. We may choose $f=d\mu/d\lambda$ to be $3$ at $a$, $1/2$ at $c$, and $0$ at $b$. For $g=d\lambda/d\nu$, all $\nu$-masses are positive, so the ratios are defined at each point:

f=(3,0,\tfrac12),\qquad g=(\tfrac12,0,\tfrac45),\qquad fg=(\tfrac32,0,\tfrac25).

Directly, $d\mu/d\nu$ has values $3/2$ at $a$, $0/3=0$ at $b$, and $2/5$ at $c$. These agree point by point with $fg$. Check: integrating the product against $\nu$ gives $(3/2)\cdot2+0\cdot3+(2/5)\cdot5=3+0+2=5$, the total mass of $\mu$. The middle point also displays the null-set issue: $\lambda(\{b\})=0$ but $\nu(\{b\})=3$; here $g(b)=0$, so the chosen value of $f(b)$ does not affect the product.

A Sigma-Finite Example on the Real Line

Worked Example: Chaining Densities on an Infinite Space

Take $X=\mathbb R$, $\mathcal M=\mathcal B_{\mathbb R}$, and let $\nu=m$, Lebesgue measure restricted to the Borel sets. Define $\lambda(E)=\int_E e^{-x}\,dm(x)$ and $\mu(E)=\int_E(1+x^2)e^{-x}\,dm(x)$. All three measures are sigma-finite: each has finite measure on every bounded interval, and $\mathbb R$ is a countable union of bounded intervals. None is finite on all of $\mathbb R$; for example, $\nu(\mathbb R)=\infty$ and $\int_{-\infty}^0e^{-x}\,dm(x)=\infty$.

The defining density identities give $d\lambda/d\nu=e^{-x}$ and $d\mu/d\lambda=1+x^2$. Both absolute-continuity hypotheses hold because each measure is defined by integrating a nonnegative density against the preceding measure. The chain rule therefore gives $\frac{d\mu}{d\nu}=(1+x^2)e^{-x}$, $\nu$-almost everywhere.

To check the calculation on the finite interval $[0,1]$, integrate first through $\lambda$. Direct integration yields $\int_0^1 e^{-x}\,dx=1-e^{-1}$. For the second term, an antiderivative of $x^2e^{-x}$ is $-(x^2+2x+2)e^{-x}$, so $\int_0^1x^2e^{-x}\,dx=2-5e^{-1}$. Consequently, \[ \mu([0,1])=\int_0^1(1+x^2)\,d\lambda =(1-e^{-1})+(2-5e^{-1})=3-6e^{-1}. \] Using the direct derivative gives $\int_0^1(1+x^2)e^{-x}\,dm(x)$, the same two integrals and the same value $3-6e^{-1}$. Check: this is positive, since $e>2$ implies $3-6/e>0$; the two routes compute exactly the same mass on $[0,1]$.

When the Chain Cannot Be Used

The absolute-continuity hypotheses are not bookkeeping: they are what make the two derivatives exist. In particular, $\lambda\ll\nu$ gives the second derivative, and $\mu\ll\lambda$ gives the first. The direct derivative $d\mu/d\nu$ may exist even when the intermediate derivative $d\mu/d\lambda$ does not.

Worked Example: A Missing Middle Absolute-Continuity Condition

Let $X=\{a,b,c\}$ with all subsets measurable, and define finite measures by $\nu=(1,1,1)$, $\lambda=(1,1,0)$, and $\mu=(1,0,1)$. We have $\lambda\ll\nu$, since every $\nu$-null set is empty. But $\mu\not\ll\lambda$: for $E=\{c\}$, $\lambda(E)=0$ while $\mu(E)=1$. Thus $d\lambda/d\nu=(1,1,0)$ exists, and $d\mu/d\nu=(1,0,1)$ exists, but $d\mu/d\lambda$ does not exist.

A tempting but invalid pointwise calculation at $c$ would try to use the ratio $\mu(\{c\})/\lambda(\{c\})=1/0$ as a density and then multiply by $d\lambda/d\nu=0$. This is not a Radon–Nikodym derivative: no $\lambda$-integral can assign positive mass to $\{c\}$, since $\lambda(\{c\})=0$. The undefined expression also cannot be repaired by treating it as an ordinary product. The failure is exactly the missing hypothesis $\mu\ll\lambda$.

Common Pitfall: Reversing the Derivative

The chain rule is directional. Under $\mu\ll\lambda\ll\nu$, it multiplies $d\mu/d\lambda$ by $d\lambda/d\nu$; it does not assert that either density can be inverted. In the first finite example, $\lambda(\{b\})=0$ while $\nu(\{b\})=3$, so $\nu\not\ll\lambda$. There can be no derivative $d\nu/d\lambda$. The mistaken calculation $d\nu/d\lambda=1/(d\lambda/d\nu)$ would require division by $g(b)=0$. Reciprocal densities are available under equivalence of measures, not from one-way absolute continuity alone.

For a practical computation, first write the intended chain of absolute-continuity relations. Then identify the reference measure for each derivative, multiply in the order specified by the theorem, and check which measure controls the almost-everywhere equality. If a density vanishes on a set of positive measure for the final reference measure, do not divide by it without an additional equivalence hypothesis.

Check Your Understanding

For each question, identify the reference measure governing any almost-everywhere statement.

If $\mu\ll\lambda\ll\nu$ are sigma-finite, which two derivatives multiply to give $d\mu/d\nu$, and with respect to which measure does the resulting equality hold almost everywhere?
In the finite example with masses $\nu=(2,3,5)$, $\lambda=(1,0,4)$, and $\mu=(3,0,2)$, why does the value chosen for $d\mu/d\lambda$ at $b$ not change the chained derivative $\nu$-almost everywhere?
For the real-line example, compute $\mu([0,1])$ by integrating the direct density $(1+x^2)e^{-x}$ with respect to Lebesgue measure.
Give the precise absolute-continuity condition that fails in the example with $\nu=(1,1,1)$, $\lambda=(1,1,0)$, and $\mu=(1,0,1)$, and state which derivative therefore does not exist.
Why does the chain rule not, by itself, justify taking the reciprocal of $d\lambda/d\nu$?