The Finite Radon-Nikodym Theorem by von Neumann

From Hilbert Space Representation to a Density

The Radon–Nikodym theorem asks when one measure can be represented by integrating a function against another. In the finite case, the key move is to place both measures under a single finite measure, their sum. Integration against either measure then defines a bounded linear functional on an \(L^2\) space. Hilbert space representation supplies a function, and positivity of the two measures forces that function to lie between zero and one. Absolute continuity is exactly what rules out the endpoint one on a set of positive measure, allowing a change of denominator to produce the desired density.

Throughout this tutorial, \((X,\mathcal{M})\) is a measurable space, and \(\mu\) and \(\nu\) are finite positive measures on it. Finite means \(\mu(X)<\infty\) and \(\nu(X)<\infty\). The measures need not be complete or sigma-finite beyond finiteness. The functions in the Hilbert space argument are real-valued measurable functions, identified when equal \(\lambda\)-almost everywhere. The resulting density is nonnegative and finite \(\mu\)-almost everywhere; its integral identity holds for every \(E\in\mathcal{M}\).

We use the Hilbert representation theorem established in the earlier material on Hilbert representation for measure functionals: if \(H\) is a real Hilbert space and \(T:H\to\mathbb{R}\) is a bounded linear functional, then there is a unique \(g\in H\) such that \(T(h)=\langle h,g\rangle\) for every \(h\in H\). Here \(L^2(\lambda)\), with inner product \(\langle h,k\rangle=\int hk\,d\lambda\), is a real Hilbert space when \(\lambda\) is a finite measure. The \(L^2\) completeness and bounded-functional facts needed for this application are part of the earlier material on \(L^2\) and measure functionals.

The Finite Radon–Nikodym Theorem

Proof. Set \(\lambda=\mu+\nu\), so \(\lambda\) is a finite positive measure. If \(\lambda(X)=0\), then \(0\le\mu(E)\le\lambda(E)=0\) and \(0\le\nu(E)\le\lambda(E)=0\) for every \(E\). Thus both measures are zero, and \(q=0\) is a density. We may now suppose \(\lambda(X)>0\).

For \(h\in L^2(\lambda)\), define \(T(h)=\int h\,d\nu\). This integral exists: by Cauchy–Schwarz for the finite measure \(\nu\),

The last inequality follows from \(\nu(E)\le\lambda(E)\) for measurable sets, first for nonnegative simple functions and then for their increasing approximations. If two representatives of \(h\) agree \(\lambda\)-almost everywhere, they agree \(\nu\)-almost everywhere because \(\nu\le\lambda\); hence \(T\) is well-defined on \(L^2(\lambda)\). It is linear, and the displayed bound shows it is bounded. Hilbert space representation therefore gives \(g\in L^2(\lambda)\) such that \(T(h)=\int hg\,d\lambda\) for every \(h\in L^2(\lambda)\). In particular, indicators belong to \(L^2(\lambda)\), since \(\lambda\) is finite. Taking \(h=\chi_E\) gives, for every \(E\in\mathcal{M}\),

We next prove the bounds on \(g\). For each positive integer \(n\), the set \(A_n=\{g\le-1/n\}\) is measurable. If \(\lambda(A_n)>0\), then \(\int_{A_n}g\,d\lambda\le-\lambda(A_n)/n<0\), contradicting \(\nu(A_n)\ge0\). Thus every \(A_n\) is \(\lambda\)-null. Since \(\{g<0\}=\bigcup_{n\ge1}A_n\), we have \(g\ge0\) \(\lambda\)-almost everywhere. Likewise, for \(B_n=\{g\ge1+1/n\}\), positivity of \(\mu\) gives \(0\le\int_{B_n}(1-g)\,d\lambda\le-\lambda(B_n)/n\). Hence every \(B_n\) is null, and \(\{g>1\}=\bigcup_{n\ge1}B_n\) is null. Changing \(g\) on a \(\lambda\)-null set if needed, take \(0\le g\le1\) everywhere.

Let \(N=\{g=1\}\). The identity for \(\mu(E)\), applied to \(N\), gives \(\mu(N)=\int_N(1-g)\,d\lambda=0\). The assumption \(\nu\ll\mu\) then gives \(\nu(N)=0\), so \(\lambda(N)=\mu(N)+\nu(N)=0\). Define \(q=g/(1-g)\) on \(X\setminus N\), and set \(q=0\) on \(N\). This is a nonnegative measurable function, finite outside \(N\), and \(N\) is \(\mu\)-null.

For completeness, the measure identity \(\mu(A)=\int_A(1-g)\,d\lambda\) implies \(\int u\,d\mu=\int u(1-g)\,d\lambda\) for every nonnegative measurable \(u\): the identity holds first for indicators, then by finite linearity for nonnegative simple functions, and finally by increasing simple approximation and the Monotone Convergence Theorem. Apply it to \(u=q\chi_E\). Off \(N\), \(q(1-g)=g\), and on \(N\) both sides of the integral identity are unaffected because \(\lambda(N)=0\). Therefore, for every \(E\in\mathcal{M}\),

This proves the asserted density identity and completes the proof. \(\square\)

Examples of the Construction

What the Representing Function Does—and Does Not—Mean

The function \(g\) first represents \(\nu\) relative to the sum measure \(\lambda=\mu+\nu\); it is not yet the density relative to \(\mu\). In the two-point example, \(g(a)=1/3\), but the density relative to \(\mu\) at \(a\) is \(q(a)=1/2\). Indeed, the incorrect choice \(q(a)=g(a)\) would give \(\int_{\{a\}}g\,d\mu=(1/3)\cdot2=2/3\), not \(\nu(\{a\})=1\). The correction is the change of denominator: \(q=g/(1-g)\), which gives \((1/2)\cdot2=1\).

The endpoint \(g=1\) has a precise measure-theoretic meaning. On \(N=\{g=1\}\), the identity \(\mu(N)=\int_N(1-g)\,d\lambda\) forces \(\mu(N)=0\). Absolute continuity then forces \(\nu(N)=0\) as well. This is why the quotient is usable almost everywhere under \(\nu\ll\mu\), even though it need not be defined at every point. No claim of pointwise uniqueness is needed: the density identity is a statement about integrals over measurable sets, and values on a \(\mu\)-null set do not affect those integrals.

The proof’s finite-measure assumptions are used to make \(\lambda\) finite and the functional \(h\mapsto\int h\,d\nu\) bounded on \(L^2(\lambda)\). Extending the theorem when the measures are only sigma-finite requires an additional reduction and belongs to the separate sigma-finite case.