# 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 $\mu $ and $\nu $ are real, nonnegative, and finite.
The extension to the $\sigma $-finite case is a standard exercise,
as is $\mu $-a.e. uniqueness of Radon-Nikodym derivative^{}.
Having done this, the thesis also holds for signed and complex-valued measures^{}.

Let $(X,\mathcal{F})$ be a measurable space^{}
and let $\mu ,\nu :\mathcal{F}\to [0,R]$ two finite measures on $X$
such that $\nu (A)=0$ for every $A\in \mathcal{F}$ such that $\mu (A)=0$.
Then $\sigma =\mu +\nu $ is a finite measure on $X$
such that $\sigma (A)=0$ if and only if $\mu (A)=0$.

Consider the linear functional $T:{L}^{2}(X,\mathcal{F},\sigma )\to \mathbb{R}$ defined by

$$Tu={\int}_{X}u\mathit{d}\mu \forall u\in {L}^{2}(X,\mathcal{F},\sigma ).$$ | (1) |

$T$ is well-defined
because $\mu $ is finite and dominated by $\sigma $, so that
${L}^{2}(X,\mathcal{F},\sigma )\subseteq {L}^{2}(X,\mathcal{F},\mu )\subseteq {L}^{1}(X,\mathcal{F},\mu );$
it is also linear and bounded^{} because
$|Tu|\le {\parallel u\parallel}_{{L}^{2}(X,\mathcal{F},\sigma )}\cdot \sqrt{\sigma (X)}.$
By Riesz representation theorem^{}, there exists $g\in {L}^{2}(X,\mathcal{F},\sigma )$ such that

$$Tu={\int}_{X}u\mathit{d}\mu ={\int}_{X}u\cdot g\mathit{d}\sigma $$ | (2) |

for every $u\in {L}^{2}(X,\mathcal{F},\sigma )$.
Then
$\mu (A)={\int}_{A}g\mathit{d}\sigma $
for every $A\in \mathcal{F}$,
so that $$ $\mu $- and $\sigma $-a.e.
(Consider the former with $A=\{x\mid g(x)\le 0\}$ or $A=\{x\mid g(x)>1\}$.)
Moreover, the second equality in (LABEL:eq:q)
holds when $u={\chi}_{A}$ for $A\in \mathcal{F}$,
thus also when $u$ is a simple measurable function^{}
by linearity of integral,
and finally when $u$ is a ($\mu $- and $\sigma $-a.e.)
nonnegative $\mathcal{F}$-measurable function
because of the monotone convergence theorem^{}.

Now, $1/g$ is $\mathcal{F}$-measurable and nonnegative $\mu $- and $\sigma $-a.e.; moreover, $\frac{1}{g}}\cdot g=1$ $\sigma $- and $\mu $-a.e. Thus, for every $A\in \mathcal{F}$,

$${\int}_{A}\frac{1}{g}\mathit{d}\mu ={\int}_{A}\mathit{d}\sigma =\sigma (A)$$ | (3) |

Since $\sigma $ is finite, $1/g\in {L}^{1}(X,\mathcal{F},\mu )$, and so is $f={\displaystyle \frac{1}{g}}-1$. Then for every $A\in \mathcal{F}$

$$\nu (A)=\sigma (A)-\mu (A)={\int}_{A}\left(\frac{1}{g}-1\right)\mathit{d}\mu ={\int}_{A}f\mathit{d}\mu .$$ |

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^{} proof of Radon-Nikodym’s theorem |

Synonym | measure- theoretic proof of Radon-Nikodym theorem |