Dominating Two Measures with Their Sum

A Common Reference Measure

To compare two finite measures, it is useful to place them under one measure that controls both. Their sum does exactly this: each measure is bounded above by the sum on every measurable set, and the sum has the same null sets as the two measures taken together. This elementary construction is also the starting point for the \(L^2\) argument used in von Neumann’s approach to the Radon–Nikodym theorem.

Throughout, let \((X,\mathcal{M})\) be a measurable space and let \(\mu\) and \(\nu\) be finite positive measures on it. Thus \(\mu(X)<\infty\) and \(\nu(X)<\infty\). No completeness assumption is made. Define a set function \(\lambda\) on \(\mathcal{M}\) by \(\lambda(E)=\mu(E)+\nu(E)\). All functions used in the \(L^2\) setup below are real-valued and measurable; equalities between elements of \(L^2(\lambda)\) are understood \(\lambda\)-almost everywhere.

Why the Sum Dominates Both Measures

Proof. Both summands are nonnegative, and \(\lambda(\varnothing)=\mu(\varnothing)+\nu(\varnothing)=0\). If \((E_j)_{j\ge1}\) are pairwise disjoint sets in \(\mathcal{M}\), countable additivity of \(\mu\) and \(\nu\) gives \(\lambda(\bigcup_j E_j)=\mu(\bigcup_jE_j)+\nu(\bigcup_jE_j)=\sum_j\mu(E_j)+\sum_j\nu(E_j)=\sum_j\lambda(E_j)\). Thus \(\lambda\) is a measure. Its total mass is \(\lambda(X)=\mu(X)+\nu(X)<\infty\), so it is finite. For each \(E\), nonnegativity gives \(\mu(E)\le\mu(E)+\nu(E)=\lambda(E)\) and \(\nu(E)\le\lambda(E)\). If \(\lambda(E)=0\), then the sum of the two nonnegative numbers \(\mu(E)\) and \(\nu(E)\) is zero, so each is zero. Conversely, if both are zero, then \(\lambda(E)=0\). The null-set equivalence implies \(\mu\ll\lambda\) and \(\nu\ll\lambda\). \(\square\)

In particular, domination gives absolute continuity, but it is stronger: \(\mu\le\lambda\) is an inequality for every set, whereas \(\mu\ll\lambda\) only asserts that \(\lambda\)-null sets are also \(\mu\)-null. The sum construction requires neither a density nor a prior relationship between \(\mu\) and \(\nu\).

The Integral Functionals on \(L^2(\lambda)\)

The next step is to turn each measure into a linear functional. For a measurable real-valued simple function \(\varphi=\sum_{j=1}^k a_j\chi_{E_j}\), with disjoint \(E_j\in\mathcal{M}\), define \(T_\mu(\varphi)=\int\varphi\,d\mu\). The same definition gives \(T_\nu\). We will show that these functionals satisfy a bound measured by the \(L^2(\lambda)\) norm.

Proof. First, the triangle inequality for the integral gives \(\left|\int\varphi\,d\mu\right|\le\int|\varphi|\,d\mu\). For a simple function on a disjoint partition, the finite-sum Cauchy–Schwarz inequality, applied to the numbers \(|a_j|\sqrt{\mu(E_j)}\) and \(\sqrt{\mu(E_j)}\), yields \(\int|\varphi|\,d\mu\le\left(\sum_j |a_j|^2\mu(E_j)\right)^{1/2}\left(\sum_j\mu(E_j)\right)^{1/2}\). The two factors are \((\int|\varphi|^2\,d\mu)^{1/2}\) and \(\sqrt{\mu(X)}\). Since \(\mu\le\lambda\), integration of a nonnegative simple function preserves the inequality, so \(\int|\varphi|^2\,d\mu\le\int|\varphi|^2\,d\lambda\). Also \(\mu(X)\le\lambda(X)\). Combining these inequalities proves the asserted bound for \(\mu\). Repeating the same argument with \(\nu\le\lambda\) and \(\nu(X)\le\lambda(X)\) proves it for \(\nu\). \(\square\)

If two simple functions agree \(\lambda\)-almost everywhere, they agree outside some \(N\in\mathcal{M}\) with \(\lambda(N)=0\). The null-set equivalence proved above gives \(\mu(N)=\nu(N)=0\), so their integrals against either measure agree. Thus these functionals are well-defined on simple-function equivalence classes, not merely on particular pointwise representatives. The bound is the key reason the sum measure is useful: it controls both functionals using the same \(L^2(\lambda)\) norm.

Here \(L^2(\lambda)\) denotes the space of measurable real-valued functions \(u\) with \(\int|u|^2\,d\lambda<\infty\), where functions equal \(\lambda\)-almost everywhere are identified, and \(\|u\|_2=(\int|u|^2\,d\lambda)^{1/2}\). The estimate above is the initial boundedness step in von Neumann’s strategy. Passing from simple functions to all of \(L^2(\lambda)\), and representing a bounded functional by an inner product, require additional Hilbert-space results; those steps are not being assumed or proved here.

What a Representation Would Give

The preceding bound alone does not produce densities. The point of the setup is that if a later Hilbert-space representation supplies functions \(h_\mu,h_\nu\in L^2(\lambda)\) with \(T_\mu(\varphi)=\int\varphi h_\mu\,d\lambda\) and \(T_\nu(\varphi)=\int\varphi h_\nu\,d\lambda\) for every simple \(\varphi\), then testing with indicators gives \(\mu(E)=\int_E h_\mu\,d\lambda\) and \(\nu(E)=\int_E h_\nu\,d\lambda\) for every \(E\in\mathcal{M}\). Because \(\lambda(X)<\infty\), these \(L^2\) functions are integrable: Cauchy–Schwarz gives \(\int|h|\,d\lambda\le\sqrt{\lambda(X)}\|h\|_2\). Thus all the integrals in these identities are well-defined. The identities make the functions Radon–Nikodym densities, if such a representation has been established.

Proof. To prove \(h_\mu\ge0\) almost everywhere, suppose instead that \(\{h_\mu<0\}\) has positive \(\lambda\)-measure. Since \(\{h_\mu<0\}=\bigcup_{n\ge1}\{h_\mu\le-1/n\}\), some \(E=\{h_\mu\le-1/n\}\) must have positive measure. Then \(\int_Eh_\mu\,d\lambda\le-\lambda(E)/n<0\), contradicting \(\int_Eh_\mu\,d\lambda=\mu(E)\ge0\). If \(\{h_\mu>1\}\) has positive measure, some \(E=\{h_\mu\ge1+1/n\}\) has positive measure. It follows that \(\int_Eh_\mu\,d\lambda\ge(1+1/n)\lambda(E)>\lambda(E)\), contradicting \(\int_Eh_\mu\,d\lambda=\mu(E)\le\lambda(E)\). Hence \(0\le h_\mu\le1\) almost everywhere. The same argument applies to \(\nu\).

For every \(E\in\mathcal{M}\), the representation identities and the definition of \(\lambda\) give \(\int_E(h_\mu+h_\nu-1)\,d\lambda=\mu(E)+\nu(E)-\lambda(E)=0\). Put \(q=h_\mu+h_\nu-1\), which is integrable. If \(\{q>0\}\) had positive measure, then some \(E=\{q\ge1/n\}\) would have positive measure and \(\int_Eq\,d\lambda\ge\lambda(E)/n>0\), a contradiction. Applying the same argument to \(-q\) shows that \(\{q<0\}\) is null as well. Therefore \(q=0\) almost everywhere. \(\square\)

A Common Pitfall: Densities Are Only Almost-Everywhere Objects

Once densities are constructed, their defining measure identities do not fix their values on \(\lambda\)-null sets. For instance, in the unit-interval example, \(\lambda(\{0\})=\int_{\{0\}}(3+x)\,dm(x)=0\). Replace \(f(0)=3/(3+0)=1\) by \(f(0)=7\), leaving \(f(x)=3/(3+x)\) for \(x\ne0\). The modified function is still a density of \(\mu\), because its integral over every set is unchanged on the null singleton. But at \(x=0\), the modified \(f(0)\) plus \(g(0)=0\) equals \(7\), not \(1\).

The incorrect conclusion is that the chosen density representatives must satisfy \(d\mu/d\lambda+d\nu/d\lambda=1\) at every point. The correct conclusion is equality \(\lambda\)-almost everywhere. In the same example the densities \(3/(3+x)\) and \(x/(3+x)\) happen to sum to \(1\) everywhere, but this is a convenient feature of those representatives, not a general pointwise requirement. The robust content is the measure identity and its almost-everywhere consequence.