The Lebesgue Decomposition from a Common Dominating Measure

Why the Sum of the Measures Helps

Suppose \(\mu\) and \(\nu\) are finite positive measures on the same measurable space. The part of \(\nu\) that has a density with respect to \(\mu\) may be difficult to identify directly, especially when \(\nu\) also assigns mass to a set that \(\mu\) regards as null. The measure \(\lambda=\mu+\nu\) provides a reference that dominates both measures. Their Radon–Nikodym derivatives with respect to \(\lambda\) reveal where \(\mu\) has mass and where it does not; taking the ratio of those derivatives on the first region gives the desired density.

Throughout, \((X,\mathcal{M})\) is a measurable space, and \(\mu,\nu\) are finite positive measures on it. No completeness assumption is needed. In this tutorial, \(\alpha\ll\beta\) means that \(\beta(A)=0\) implies \(\alpha(A)=0\) for every \(A\in\mathcal{M}\). The measures \(\alpha\) and \(\beta\) are singular, written \(\alpha\perp\beta\), if there is an \(S\in\mathcal{M}\) with \(\beta(S)=0\) and \(\alpha(X\setminus S)=0\).

Construction from the Common Dominating Measure

Set \(\lambda=\mu+\nu\). This is a finite measure, since \(\lambda(X)=\mu(X)+\nu(X)<\infty\). Also \(\mu\ll\lambda\) and \(\nu\ll\lambda\): if \(\lambda(A)=0\), then \(\mu(A)+\nu(A)=0\), and both summands are nonnegative. By the Radon–Nikodym theorem for finite measures, there are nonnegative \(\mathcal{M}\)-measurable functions \(f\) and \(g\) such that \(\mu(A)=\int_A f\,d\lambda\) and \(\nu(A)=\int_A g\,d\lambda\) for every \(A\in\mathcal{M}\).

Proof. For every \(A\in\mathcal{M}\), the definitions of \(\lambda,f,g\) give \(\int_A(f+g)\,d\lambda=\mu(A)+\nu(A)=\lambda(A)=\int_A1\,d\lambda\). Thus \(f+g\) and the constant function \(1\) are both Radon–Nikodym derivatives of \(\lambda\) with respect to \(\lambda\). Almost-everywhere uniqueness of Radon–Nikodym derivatives gives \(f+g=1\) \(\lambda\)-almost everywhere. \(\square\)

Choose representatives of \(f\) and \(g\) that are finite and nonnegative everywhere and satisfy \(f+g=1\) everywhere. This is possible by changing their values on a \(\lambda\)-null set; such a set is also \(\mu\)-null and \(\nu\)-null. Define \(E=\{x:f(x)>0\}\) and \(F=\{x:f(x)=0\}\). On \(E\), define \(h=g/f\); on \(F\), set \(h=0\). The function \(h\) is measurable and nonnegative.

Proof. We first record the integral identity used in the construction: for every nonnegative measurable \(q\), \(\int q\,d\mu=\int qf\,d\lambda\). For an indicator \(q=\chi_A\), this is the defining identity \(\mu(A)=\int_A f\,d\lambda\). Additivity and nonnegative scalar multiplication give the identity for nonnegative simple functions. Approximating a general nonnegative measurable \(q\) increasingly by simple functions and applying the Monotone Convergence Theorem on both sides gives the stated identity.

The formula \(\nu_{\mathrm{ac}}(A)=\int_A h\,d\mu\) defines a positive measure. Applying the identity just established, and using \(hf=g\) on \(E\) and \(hf=0\) on \(F\), yields \(\nu_{\mathrm{ac}}(A)=\int_A hf\,d\lambda=\int_{A\cap E}g\,d\lambda=\nu(A\cap E)\). In particular, this measure is finite, since \(\nu_{\mathrm{ac}}(X)=\nu(E)\leq\nu(X)<\infty\). If \(\mu(A)=0\), then \(\int_A f\,d\lambda=0\), so \(f=0\) \(\lambda\)-almost everywhere on \(A\). Hence \(\lambda(A\cap E)=0\), and \(\nu_{\mathrm{ac}}(A)=\nu(A\cap E)=0\). Thus \(\nu_{\mathrm{ac}}\ll\mu\).

The formula \(\nu_{\mathrm{s}}(A)=\nu(A\cap F)\) defines a finite positive measure by countable additivity of \(\nu\). Also \(\mu(F)=\int_F f\,d\lambda=0\), while \(\nu_{\mathrm{s}}(X\setminus F)=0\). Therefore \(\nu_{\mathrm{s}}\perp\mu\), with \(F\) as a witnessing set. Finally, \(E\) and \(F\) partition \(X\), so for every \(A\in\mathcal{M}\), \(\nu_{\mathrm{ac}}(A)+\nu_{\mathrm{s}}(A)=\nu(A\cap E)+\nu(A\cap F)=\nu(A)\). This proves all the claims. \(\square\)

The construction separates the two regions using \(f\), the density of \(\mu\) relative to \(\lambda\). Where \(f>0\), the ratio \(g/f\) converts the density of \(\nu\) relative to \(\lambda\) into a density relative to \(\mu\). Where \(f=0\), \(\mu\) assigns no mass, so the \(\nu\)-mass there belongs to the singular part.

A Common Density-Conversion Error

The ratio \(g/f\), not \(g\) alone, is the density of the absolutely continuous part relative to \(\mu\). The function \(g\) is a density relative to \(\lambda\), so using it directly with \(\mu\) changes the reference measure without justification.

For a concrete check, let \(X=[0,1]\), \(\mu=2m\), and \(\nu=m\). Then \(\lambda=3m\), so \(f=2/3\) and \(g=1/3\). The correct density relative to \(\mu\) is \(h=g/f=1/2\), and \(\int h\,d\mu=\int_0^1(1/2)2\,dm=1=\nu(X)\). The incorrect calculation \(\int g\,d\mu=\int_0^1(1/3)2\,dm=2/3\) fails to recover \(\nu(X)=1\). The conversion is required precisely because \(g\) measures mass relative to \(\lambda\), whereas the desired density is relative to \(\mu\).

There is a related issue at points where \(f=0\): the quotient \(g/f\) is not defined there. Assigning it an arbitrary value on that set and then treating it as a density for all of \(\nu\) can incorrectly absorb singular mass. The construction instead sets \(h=0\) on \(F=\{f=0\}\) and places \(\nu\)'s mass on \(F\) in the singular measure. The two formulas together account for all of \(\nu\).