This section gives a measure-theoretic proof of the theorem. There is also a functional-analytic proof, using Hilbert space methods, that was first given by von Neumann.
For finite measures
Constructing an extended-valued candidate First, suppose μ and ν are both finite-valued nonnegative measures. Let F be the set of those extended-value measurable functions f : X → [0, ∞] such that:
- :\qquad \int _{A}f\,d\mu \leq \nu (A)}
F ≠ ∅, since it contains at least the zero function. Now let f1, f2 ∈ F, and suppose A is an arbitrary measurable set, and define:
Then one has
and therefore, max{ f 1, f 2} ∈ F.
Now, let { fn } be a sequence of functions in F such that
By replacing fn with the maximum of the first n functions, one can assume that the sequence { fn } is increasing. Let g be an extended-valued function defined as
By Lebesgue's monotone convergence theorem, one has
for each A ∈ Σ, and hence, g ∈ F. Also, by the construction of g,
Proving equality Now, since g ∈ F,
defines a nonnegative measure on Σ. To prove equality, we show that ν0 = 0.
Suppose ν0 ≠ 0; then, since μ is finite, there is an ε > 0 such that ν0(X) > ε μ(X). To derive a contradiction from ν0 ≠ 0, we look for a positive set P ∈ Σ for the signed measure ν0 − ε μ (i.e. a measurable set P, all of whose measurable subsets have non-negative ν0 − εμ measure), where also P has positive μ-measure. Conceptually, we're looking for a set P, where ν0 ≥ ε μ in every part of P. A convenient approach is to use the Hahn decomposition (P, N) for the signed measure ν0 − ε μ.
Note then that for every A ∈ Σ one has ν0(A ∩ P) ≥ ε μ(A ∩ P), and hence,
where 1P is the indicator function of P. Also, note that μ(P) > 0 as desired; for if μ(P) = 0, then (since ν is absolutely continuous in relation to μ) ν0(P) ≤ ν(P) = 0, so ν0(P) = 0 and
contradicting the fact that ν0(X) > εμ(X).
Then, since also
g + ε 1P ∈ F and satisfies
This is impossible because it violates the definition of a supremum; therefore, the initial assumption that ν0 ≠ 0 must be false. Hence, ν0 = 0, as desired.
Restricting to finite values Now, since g is μ-integrable, the set {x ∈ X : g(x) = ∞} is μ-null. Therefore, if a f is defined as
then f has the desired properties.
Uniqueness As for the uniqueness, let f, g : X → [0, ∞) be measurable functions satisfying
for every measurable set A. Then, g − f is μ-integrable, and
In particular, for A = {x ∈ X : f(x) > g(x)}, or {x ∈ X : f(x) < g(x)}. It follows that
and so, that (g − f )+ = 0 μ-almost everywhere; the same is true for (g − f )−, and thus, f = g μ-almost everywhere, as desired.
For signed and complex measures
If ν is a σ-finite signed measure, then it can be Hahn–Jordan decomposed as ν = ν+ − ν− where one of the measures is finite. Applying the previous result to those two measures, one obtains two functions, g, h : X → [0, ∞), satisfying the Radon–Nikodym theorem for ν+ and ν− respectively, at least one of which is μ-integrable (i.e., its integral with respect to μ is finite). It is clear then that f = g − h satisfies the required properties, including uniqueness, since both g and h are unique up to μ-almost everywhere equality.
If ν is a complex measure, it can be decomposed as ν = ν1 + iν2, where both ν1 and ν2 are finite-valued signed measures. Applying the above argument, one obtains two functions, g, h : X → [0, ∞), satisfying the required properties for ν1 and ν2, respectively. Clearly, f = g + ih is the required function.