The Radon-Nikodym Theorem

Yongheng Zhang

Theorem 1 (Johann Radon-Otton Nikodym). Let (X,B, µ) be a σ-finite measure space and let ν bea measure defined on B such that ν � µ. Then there is a unique nonnegative measurable function fup to sets of µ-measure zero such that

ν(E) =

∫E

fdµ,

for every E ∈ B. f is called the Radon-Nikodym derivative of ν with respect to µ and it is oftendenoted by [ dνdµ ].

The assumption that the measure µ is σ-finite is crucial in the theorem (but we don’t have thisrequirement directly for ν). Consider the following two examples in which the µ’s are not σ-finite.

Example 1. Let (R,M, ν) be the Lebesgue measure space. Let µ be the counting measure onM. So µ is not σ-finite. For any E ∈ M, if µ(E) = 0, then E = ∅ and hence ν(E) = 0. This showsν � µ. Do we have the associated Radon-Nikodym derivative? Never. Suppose we have it and call itf , then for each x ∈ R, 0 = ν({x}) =

∫{x} fdµ =

∫{x} fχ{x}dµ = f(x)µ({x}) = f(x). Hence, f = 0.

This means for every E ∈M, ν(E) =∫E

0dµ = 0, which contradicts that ν is the Lebesgue measure.

Example 2. ν is still the same measure above but we let µ to be defined by µ(∅) = 0 and µ(A) =∞if A 6= ∅. Clearly, µ is not σ-finite and ν � µ. The Radon-Nikodym derivative does not exist, neither.Suppose f is one. Then for any x ∈ R, 0 = ν({x}) =

∫{x} fdµ = f(x)µ({x}) = f(x)∞. Thus, f = 0.

Hence for any E ∈M, ν(E) = infE 0dµ = 0.

[Also notice that if µ(X) = 0, then the measure of every set in M with respect to µ and ν is zero,which is not interesting at all. So we ignore this case.]

Now we prove the Radon-Nikodym Theorem.

Proof(Wikipedia): We assume that (X,B, µ) is a finite measure space and show the existence anduniqueness of the Radon-Nikodym derivative. Then the case when µ is σ-finite follows by the “past-ing” of the derivatives on each set of finite measure.

Let F = {f is measurable|ν(E) ≥∫Efdµ, for all E ∈ M}. F is nonempty because the zero func-

tion is in it. Let s = supf∈F∫Xfdµ. Then there is a sequence 〈hn〉 in F such that limn→∞

∫Xfndµ =

s.

Let f1, f2 ∈ F , then for any E ∈M,∫Ef1∨f2dµ =

∫{x∈E|f1(x)≥f2(x)} f1∨f2dµ+

∫{x∈E|f1(x)<f2(x)} f1∨

f2dµ =∫{x∈E|f1(x)≥f2(x)} f1dµ +

∫{x∈E|f1(x)<f2(x)} f2dµ ≤ ν({x ∈ E|f1(x) ≥ f2(x)}) + ν({x ∈

E|f1(x) < f2(x)}) = ν(E). Therefore, f1 ∨ f2 ∈ F .

Let fn =∨nk=1 hk. Then 〈fn〉 is a nonnegative increasing sequence in F and limn→∞

∫Xfndµ = s.

Define g by g(x) = limn→∞ fn(x) for x ∈ X. Then by the motonone convergence theorem, for anyE ∈M,

∫Egdµ = limn→∞

∫Efndµ ≤ ν(E). This shows g ∈ F and

∫Xgdµ = limn→∞

∫Xfndµ = s.

1

2

Therefore, the function ν0 defined on M by ν0(E) = ν(E) −∫Egdµ is a measure. We want to

show that ν0 = 0 and then g is the desired function. Suppose ν0 is not zero. Since ν0(X) > 0and µ(X) < ∞, there is ε > 0 such that ν0(X) − εµ(X) > 0. Let {A,B} be a Hahn decompo-sition for the signed measure ν0 − εµ. Then for every E ∈ M, ν0(A ∩ E) − εµ(A ∩ E) ≥ 0. So,ν(E) = ν0(E)+

∫Egdµ ≥ ν0(E∩A)+

∫Egdµ ≥ εµ(A∩E)+

∫Egdµ =

∫E

(g+εχA). Therefore, g+εχAis also in F . However, if µ(A) > 0, then

∫X

(g+εχA)dµ =∫Xgdµ+εµ(A) >

∫Xgdµ = s, which is a con-

tradiction. In fact, if µ(A) = 0, since ν � µ, ν(A) = 0. So ν0(A) = ν(A)−∫Agdµ ≤ ν(A) = 0. Hence

ν0(A) = 0. Consequently, ν0(X)−εµ(X) = ν0(B)−εµ(B) ≤ 0, contradicting that ν0(X)−εµ(X) > 0.

Therefore, ν0 = 0, which means ν(E) =∫Egµ for every E ∈M.

To show uniqueness, let ν(E) =∫Efdµ =

∫Egdµ. Then

∫E

(f − g)dµ = 0. Since E is arbitrary,∫{f−g≥0} f − gdµ = 0. This shows f = g a.e. on {x ∈ X|f(x) ≥ g(x)}. Similarly, f = g a.e. on

{x ∈ X|f(x) < g(x)}. Hence f = g a.e. on X. �

Remark: If we add the condition that ν is finite, then the function g in our proof isintegrable.