proof of Radon-Nikodym theorem

The following proof of Radon-Nikodym theoremis 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 $(X,ℱ)$ be a measurable spaceand let $\mu ,\nu :ℱ\to [0,R]$ two finite measures on $X$such that $\nu (A)=0$ for every $A\in ℱ$ 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$T:L^2(X,ℱ,\sigma )\to ℝ$defined by

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

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

$$Tu={\int }_{X}u𝑑\mu ={\int }_{X}u\cdot g𝑑\sigma$$(2)

for every $u\in L^2(X,ℱ,\sigma )$.Then$\mu (A)={\int }_{A}g𝑑\sigma$for every $A\in ℱ$,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 $A\in ℱ$,thus also when $u$ is a simple measurable functionby linearity of integral,and finally when $u$ is a ($\mu$- and $\sigma$-a.e.)nonnegative $ℱ$-measurable functionbecause of the monotone convergence theorem.

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

$${\int }_A\frac{1}{g}𝑑\mu ={\int }_A𝑑\sigma =\sigma (A)$$(3)

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

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