Changing the Reference Measure by a Positive Weight

Changing the Measure, Not the Underlying Sets

A Radon–Nikodym derivative describes one measure relative to another. If the reference measure changes, the derivative changes too. In this tutorial the change is made by weighting a measure: sets with larger weight receive more measure, and sets with smaller weight receive less. When the weight is positive almost everywhere, no set of positive measure is discarded, so the two reference measures have the same null sets. The new derivative is therefore obtained by dividing the old one by the weight.

Throughout, \((X,\mathcal{M},\mu)\) is a sigma-finite measure space. Let \(w:X\to[0,\infty)\) be a finite-valued measurable function that is strictly positive \(\mu\)-almost everywhere. Define a measure \(\rho\) on \((X,\mathcal{M})\) by \(\rho(E)=\int_E w\,d\mu\) for \(E\in\mathcal{M}\). The measure \(\rho\) need not be finite, but it is sigma-finite, as shown below. Let \(\nu\) be a finite signed measure on \((X,\mathcal{M})\) with \(\nu\ll\mu\), and let \(q=d\nu/d\mu\). Thus \(q\) is real-valued and \(\mu\)-integrable, and \(\nu(E)=\int_Eq\,d\mu\) for every \(E\in\mathcal{M}\).

Why the Weighted Measure Is a Suitable Reference

First, \(\rho\) is a measure: it is the integral of the nonnegative measurable function \(w\) over each set, and countable additivity follows from the additivity of the nonnegative integral over disjoint sets. It is absolutely continuous with respect to \(\mu\), meaning that every \(\mu\)-null set is a \(\rho\)-null set. Indeed, if \(\mu(E)=0\), then \(\rho(E)=\int_Ew\,d\mu=0\).

For the other direction, suppose \(\rho(E)=0\). Let \(N=\{x:w(x)=0\}\), which is \(\mu\)-null by the assumption on \(w\). For every positive integer \(k\), on \(E\cap\{w\ge 1/k\}\) the weight is at least \(1/k\), so \(\rho(E)\ge \frac1k\mu(E\cap\{w\ge1/k\})\). Since \(\rho(E)=0\), each set \(E\cap\{w\ge1/k\}\) is \(\mu\)-null. Every point of \(E\setminus N\) belongs to one of these sets, because \(w(x)>0\) implies \(w(x)\ge1/k\) for some \(k\). Thus \(E\) is contained in the union of \(N\) and countably many \(\mu\)-null sets, and \(\mu(E)=0\). This proves that \(\mu\) and \(\rho\) have the same null sets.

The sigma-finiteness of \(\rho\) is also useful. Cover \(X\) by countably many measurable sets \(A_j\) with \(\mu(A_j)<\infty\). For each \(j,k\), the set \(A_j\cap\{w\le k\}\) satisfies \(\rho(A_j\cap\{w\le k\})\le k\mu(A_j)<\infty\). These sets, together with the \(\rho\)-null set \(N\), cover \(X\). Hence \(\rho\) is sigma-finite. Since \(\nu\ll\mu\) and \(\mu\ll\rho\), we also have \(\nu\ll\rho\).

Proof of the Quotient Formula

We will use a direct consequence of the definition of \(\rho\): for every nonnegative measurable \(g\), \(\int g\,d\rho=\int gw\,d\mu\). To justify it, first take \(g=\chi_E\), when the identity is precisely \(\rho(E)=\int_Ew\,d\mu\). Additivity gives the identity for nonnegative simple functions. For general nonnegative measurable \(g\), choose nonnegative simple functions \(g_n\uparrow g\). The Monotone Convergence Theorem, applied on the sigma-finite measure spaces \((X,\mathcal M,\rho)\) and \((X,\mathcal M,\mu)\), gives \(\int g\,d\rho=\lim_n\int g_n\,d\rho=\lim_n\int g_nw\,d\mu=\int gw\,d\mu\). No finiteness of these nonnegative integrals is required.

Proof. Choose a \(\mu\)-measurable representative of \(q\), and define \(h=q/w\) where \(w>0\), setting \(h=0\) on \(N=\{w=0\}\). The function \(h\) is measurable. The identity just proved, applied to \(|h|\), gives \(\int |h|\,d\rho=\int |h|w\,d\mu=\int |q|\,d\mu<\infty\); the equality in the middle holds off \(N\), and \(N\) is \(\mu\)-null. Thus \(h\) is \(\rho\)-integrable. Applying the same identity to \(\chi_Eh^+\) and \(\chi_Eh^-\), which are nonnegative and measurable, yields \(\int_Eh\,d\rho=\int_Ehw\,d\mu=\int_Eq\,d\mu=\nu(E)\) for every \(E\in\mathcal M\). Here the signed integral is legitimate because \(\int|h|\,d\rho<\infty\), and \(hw=q\) \(\mu\)-almost everywhere. Therefore \(h\) is a Radon–Nikodym derivative of \(\nu\) with respect to \(\rho\). By the uniqueness of Radon–Nikodym derivatives up to almost-everywhere equality, \(d\nu/d\rho=h=q/w\) \(\rho\)-almost everywhere. \(\square\)

The proof uses the Radon–Nikodym theorem for the existence and almost-everywhere uniqueness of densities, established in the earlier material on the Radon–Nikodym theorem. The key calculation here is that multiplying the candidate quotient by the weight recovers the original density. In particular, the formula does not require \(w\) to be bounded above or bounded away from zero.

When the Formula Fails

Positivity almost everywhere is essential. For a concrete counterexample, let \(X=[0,1]\) with Lebesgue measure \(\mu\), set \(w=\chi_{[0,1/2)}\), and define \(\rho(E)=\int_Ew\,dm\). Let \(\nu=\mu\), so \(d\nu/d\mu=1\). The set \(E=[1/2,1]\) has \(\rho(E)=0\) but \(\nu(E)=1/2\). Consequently \(\nu\not\ll\rho\), and \(d\nu/d\rho\) does not exist. Dividing the old density \(1\) by \(w\) on \(E\) is not a valid calculation: the denominator vanishes on a set of positive \(\mu\)-measure, and the weighted measure has lost information about that set.

In contrast, when \(w>0\) \(\mu\)-almost everywhere, the two reference measures have exactly the same null sets. The quotient \(q/w\) can still be unbounded, as in the preceding example, but it is integrable with respect to \(\rho\) whenever \(q\) is integrable with respect to \(\mu\). The weighted-integral calculation shows why: the factor \(w\) in the new measure cancels the denominator in the quotient.