Equivalence between Density and Radon-Nikodym Derivative

Theorem (Radon-Nikodym Derivative is the Density)

Let I=[a,b]I=[a,b]. Let F:IRF:I\to \mathbb{R} be Continuous and non-decreasing. Then the following are equivalent:

  1. FF is absolutely continuous on [a,b][a,b]
  2. FF maps sets of Lebesgue Measure zero to sets of Lebesgue measure zero
  3. FF is differentiable mm-a.e. with FL1(m)F'\in L^{1}(m) and x[a,b]:F(x)F(a)=[a,x]Fdm\forall x \in [a,b]:F(x)-F(a)=\int\limits _{[a,x]}F' \, dm

Definition (Lebesgue point)

Let fL1(R)f\in L^{1}(\mathbb{R}). We say that xRx \in\mathbb{R} is a Lebesgue point for ff if and only if limr012r(xr,x+r)(ff(x))dm=0\lim_{ r \to 0 } \frac{1}{2r}\int\limits _{(x-r,x+r)}(f-f(x) )\, dm =0

Remark

ff continuous at xx     \implies xx is a Lebesgue point for ff. Also, intuitively, xx is a Lebesgue point for ff if ff does not oscillate too much near xx.

Theorem (Almost every point is a Lebesgue point)

Let fL1(R)f\in L^{1}(\mathbb{R}). Then almost every xRx \in\mathbb{R} is a for ff.

Definition (Shrink nicely)

Let xRx \in\mathbb{R}. A sequence of Borel subsets (Ej)j1(E_{j})_{j\ge 1} is said to shrink nicely to xx if and only if α>0\exists\alpha>0 with the following property: There is a sequence of Open intervals JjJ_{j} with limjm(Jj)=0\lim_{ j \to \infty }m(J_{j})=0 such that EjJj and m(Ej)αm(Jj)j1E_{j}\subseteq J_{j}\text{ and }m(E_{j})\ge \alpha \cdot m(J_{j})\quad\forall j\ge 1

Theorem

Suppose that for every xRx \in\mathbb{R} we have a sequence (Ej(x))j1(E_{j}(x))_{j\ge 1} that to xx. Let fL1(R)f\in L^{1}(\mathbb{R}). Then for every xx of ff we have that f(x)=limj1m(Ej(x))Ej(x)fdmf(x)=\lim_{ j \to \infty } \frac{1}{m(E_{j}(x))}\int\limits _{E_{j}(x)}f \, dm

Theorem (Density is the Radon-Nikodym Derivative)

Suppose that fL1(R)f\in L^{1}(\mathbb{R}). Define, xR\forall x \in\mathbb{R}: F(x)=[,x]fdmF(x)=\int\limits _{[-\infty,x]}f \, dm Then for every xx of ff we have F(x)=f(x)F'(x)=f(x)

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