Radon-Nikodym Theorem

Motivation

Let $(X,\mathcal{F})$ be a Measurable Space. Let $\mu$ be a Measure on $\mathcal{F}$ . Let $f\in L^{1}(X,\mathcal{F},\mu)$ , define a new measure $\nu$ as $\nu(E)=\int\limits _{E}f \, d\mu$ we see that $\nu$ is a finite measure. Now we would like to do the reverse, i.e. given two measures $\mu,\nu$ where $\nu\ll \mu$ can we find a $f$ s.t. the above holds?

Theorem (Radon-Nikodym)

Let $(X,\mathcal{F},\mu)$ be a measure space, with $\mu$ as a finite signed measure. Let $\nu$ be a measure on $(X,\mathcal{F})$ and assume $\nu\ll\mu$ (i.e. $\nu$ absolutely continuous w.r.t $\mu$ ). Then $\exists!f\in L^{1}(X,\mathcal{F},\mu)$ such that $\nu(A)=\int\limits _{A}f \, d\mu$ $\forall A\in\mathcal{F}$ .

Theorem (Continuous and Compactly supported functions dense in Lp)

Let $1\le p < \infty$ , and let $C_{c}(\mathbb{R})$ be the space of Continuous and Compactly Supported functions on $\mathbb{R}$ . Then $C_{c}(\mathbb{R})$ is Dense in $L^{p}(\mathbb{R})$ .

Theorem (S is dense in Lp)

Let $1\le p <\infty$ and let $(X,\mathscr{M},\mu)$ be a Measure Space. Let $S:=\{ f:X\to \mathbb{C}:f\text{ measurable, }f(X)\text{ is a finite set, and }\mu(\{ x \in X:f(x)\not= 0 \})<\infty\}$ then $S$ is a Dense subspace in $L^{p}(X,\mathscr{M},\mu)$ .

Theorem (Radon-Nikodym (Lebesgue Decomposition))

Let $(X,\mathscr{M})$ be a Measurable Space and $\mu$ a σ-finite Measure on $\mathscr{M}$ . Let $\nu$ be a finite measure on $\mathscr{M}$ .

There exists a unique pair of measures $\nu_{a},\nu_{s}$ on $\mathscr{M}$ such that:
$\nu=\nu_{a}+\nu_{s}$
$\nu_{a}\ll \mu$
$\nu_{s}\perp \mu$ , (i.e. $\nu_{s},\mu$ are Mutually Singular)
$\exists h\in L^{1}(X,\mathscr{M},\mu)$ such that $\nu_{a}(E)=\int\limits _{E}h \, d\mu\quad \forall E\in\mathscr{M}$

Linked from

A Summary of MATH 891

Radon-Nikodym Derivative

Lebesgue Integral