proof of Radon-Nikodym theorem
The following proof of Radon-Nikodym theorem
is based on the original argument by John von Neumann.
We suppose that μ and ν are real, nonnegative, and finite.
The extension to the σ-finite case is a standard exercise,
as is μ-a.e. uniqueness of Radon-Nikodym derivative.
Having done this, the thesis also holds for signed and complex-valued measures
.
Let (X,ℱ) be a measurable space
and let μ,ν:ℱ→[0,R] two finite measures on X
such that ν(A)=0 for every A∈ℱ such that μ(A)=0.
Then σ=μ+ν is a finite measure on X
such that σ(A)=0 if and only if μ(A)=0.
Consider the linear functional T:L2(X,ℱ,σ)→ℝ defined by
Tu=∫Xu𝑑μ∀u∈L2(X,ℱ,σ). | (1) |
T is well-defined
because μ is finite and dominated by σ, so that
L2(X,ℱ,σ)⊆L2(X,ℱ,μ)⊆L1(X,ℱ,μ);
it is also linear and bounded because
|Tu|≤∥u∥L2(X,ℱ,σ)⋅√σ(X).
By Riesz representation theorem
, there exists g∈L2(X,ℱ,σ) such that
Tu=∫Xu𝑑μ=∫Xu⋅g𝑑σ | (2) |
for every u∈L2(X,ℱ,σ).
Then
μ(A)=∫Ag𝑑σ
for every A∈ℱ,
so that 0<g≤1 μ- and σ-a.e.
(Consider the former with A={x∣g(x)≤0} or A={x∣g(x)>1}.)
Moreover, the second equality in (LABEL:eq:q)
holds when u=χA for A∈ℱ,
thus also when u is a simple measurable function
by linearity of integral,
and finally when u is a (μ- and σ-a.e.)
nonnegative ℱ-measurable function
because of the monotone convergence theorem
.
Now, 1/g is ℱ-measurable and nonnegative μ- and σ-a.e.; moreover, 1g⋅g=1 σ- and μ-a.e. Thus, for every A∈ℱ,
∫A1g𝑑μ=∫A𝑑σ=σ(A) | (3) |
Since σ is finite, 1/g∈L1(X,ℱ,μ), and so is f=1g-1. Then for every A∈ℱ
ν(A)=σ(A)-μ(A)=∫A(1g-1)𝑑μ=∫Af𝑑μ. |
Title | proof of Radon-Nikodym theorem |
---|---|
Canonical name | ProofOfRadonNikodymTheorem |
Date of creation | 2013-03-22 18:58:03 |
Last modified on | 2013-03-22 18:58:03 |
Owner | Ziosilvio (18733) |
Last modified by | Ziosilvio (18733) |
Numerical id | 5 |
Author | Ziosilvio (18733) |
Entry type | Proof |
Classification | msc 28A15 |
Synonym | Hilbert spaces![]() |
Synonym | measure- theoretic proof of Radon-Nikodym theorem |