Differentiation

The Lebesgue-Radon-Nikodym theorem

Let ${\displaystyle (\Omega ,F)}$ be a measurable space and let ${\displaystyle \mu }$ and ${\displaystyle \nu }$ be two measures on ${\displaystyle (\Omega ,F)}$. ${\displaystyle \mu }$ is dominated by ${\displaystyle \nu }$, ${\displaystyle \mu \langle \nu }$ if ${\displaystyle \nu \left(A\right)=0\Rightarrow \mu \left(A\right)=0\forall A\in F}$.

Let ${\displaystyle f}$ be a non-negative measurable function, ${\displaystyle \mu \left(A\right)={\int }_{A}fd\nu \forall A\in F}$, then ${\displaystyle \mu }$ is also a measure and ${\displaystyle \mu \langle \nu }$.

${\displaystyle \mu }$ is singular wrt to ${\displaystyle \nu }$, ${\displaystyle \mu \perp \nu }$ if ${\displaystyle \exists B\in F\text{st}\mu \left(B\right)=0\text{and}v\left(B^c\right)=0}$. If ${\displaystyle \mu }$ and ${\displaystyle \nu }$ are mutally singular, then for ${\displaystyle B}$ as above ${\displaystyle \mu \left(A\right)=\mu (A\cap B^c)}$ and ${\displaystyle \nu \left(A\right)=\nu (A\cap B)}$.

• Lebesgue decomposition theorem: ${\displaystyle \mu }$ can be uniquely decomposed as ${\displaystyle \mu ={\mu }_a+{\mu }_s}$ where ${\displaystyle {\mu }_a\langle \nu }$ and ${\displaystyle {\mu }_s\perp \nu }$ where ${\displaystyle {\mu }_a}$ and ${\displaystyle {\mu }_s}$ a ${\displaystyle \sigma }$-finite.
• Radon Nikodym theorem: there exists a non-negative measurable function ${\displaystyle h}$ on ${\displaystyle (\Omega ,F)}$ st. ${\displaystyle {\mu }_s\left(A\right)={\int }_{A}hd\nu \forall A\in F}$

Let ${\displaystyle h}$ be a non-negative measurable function st ${\displaystyle \mu \left(A\right)={\int }_{A}hd\nu }$, then ${\displaystyle h}$ is called the Radon-Nikodym derivative${\displaystyle of}$mu${\displaystyle wrt}$nu${\displaystyle \text{and}iswrien}$dmu/dnu${\displaystyle .}$

Let ${\displaystyle \nu ,{\mu }_1,{\mu }_2,...}$ be ${\displaystyle \sigma }$-finite measures, then:

• ${\displaystyle {\mu }_1\langle {\mu }_2,{\mu }_2\langle {\mu }_3}$, then ${\displaystyle {\mu }_1\langle {\mu }_3}$ and ${\displaystyle d{\mu }_1/d{\mu }_2=d{\mu }_1/d{\mu }_{2}d{\mu }_2/d{\mu }_{3}ae\left({\mu }_3\right)}$
• ${\displaystyle {\mu }_1\langle {\mu }_3}$, ${\displaystyle {\mu }_2\langle {\mu }_3}$, then ${\displaystyle \forall a,b\in ℝ}$ ${\displaystyle a{\mu }_1+\beta {\mu }_2\langle {\mu }_3}$ and ${\displaystyle d\frac{\alpha {\mu }_1+\beta {\mu }_2}{d}{\mu }_3=\alpha d{\mu }_1/d{\mu }_3+\beta d{\mu }_2/d{\mu }_{3}ae\left({\mu }_3\right)}$
• ${\displaystyle \mu \langle \nu }$ and ${\displaystyle d\frac{\mu }{d}\nu >0ae\left(\nu \right)}$ then ${\displaystyle \nu \langle \mu }$ and ${\displaystyle d\frac{\nu }{d}\mu ={\left(d\frac{\mu }{d}\nu \right)}^{‐1}}$
• Let ${\displaystyle {\left\{{\mu }_n\right\}}_{n\ge 1}}$ be a sequence of measures and ${\displaystyle {\left\{a_n\right\}}_{n\ge 1}\subset {ℝ}^{+}}$, define ${\displaystyle \mu =\sum a_n{\mu }_n}$ then ${\displaystyle \mu \langle \nu \iff {\mu }_n\langle \nu \forall n}$ and ${\displaystyle d\frac{\mu }{d}\nu =\sum a_{n}d{\mu }_n/d\nu ae\left(\nu \right)}$, ${\displaystyle \mu \perp \nu \iff {\mu }_n{|}_n\forall n\ge 1}$

Signed measures

Let ${\displaystyle {\mu }_1,{\mu }_2}$ be finite measures on MS ${\displaystyle (\Omega ,F)}$. Let ${\displaystyle \nu \left(A\right)={\mu }_1\left(A\right)‐{\mu }_2\left(A\right)\forall A\in F}$. A finite signed measure satisfies:

• ${\displaystyle \nu \left(o\right)=0}$
• ${\displaystyle v\left(A\right)=\sum v\left(A_i\right)}$, ${\displaystyle A_i}$ a partition of ${\displaystyle A}$
• ${\displaystyle \left|\right|\nu \left|\right|=\underset{\text{all partitions}}{sup}\{\sum |v\left(A_i\right)\left|\right\}<\infty }$.

Let ${\displaystyle \nu }$ be a finite signed measure and ${\displaystyle \left|\nu \right|\left(A\right)=\underset{\text{all partitions}}{sup}\{\sum |v\left(A_i\right)\left|\right\}}$, then ${\displaystyle \left|\nu \right|}$ is a finite measure.

A set function ${\displaystyle \nu }$ is a finite signed measure iff ${\displaystyle \exists {\mu }_1,{\mu }_2\text{st}\nu ={\mu }_1‐{\mu }_2}$ OR ${\displaystyle \exists \mu }$ and ${\displaystyle f\in L^1\left(\mu \right)}$ st ${\displaystyle \forall A\in Fv\left(A\right)={\int }_{A}fd\mu }$.

${\displaystyle A\in F}$ is called a *negative set for ${\displaystyle \nu }$ if for any ${\displaystyle B\subset A}$ ${\displaystyle \nu \left(B\right)\le 0}$.

Hahn decomposition theorem: ${\displaystyle \nu }$ a finite signed measure, then ${\displaystyle \exists }$ a positive set ${\displaystyle {\Omega }^{+}}$ and negative set ${\displaystyle {\Omega }^{‐}}$ st ${\displaystyle \Omega ={\Omega }^{+}\cup {\Omega }^{‐}}$ and ${\displaystyle {\Omega }^{+}\cap {\Omega }^{‐}=\varnothing }$. Jordan decomposition theorem: ${\displaystyle \nu ={\nu }^{+}‐{\nu }^{\_}}$.

${\displaystyle S=\{\nu :\nu \text{is a finite signed measure}\}}$ is a vector space, with total variation norm ${\displaystyle \left|\right|\nu \left|\right|=\left|\nu \right|(\Omega )}$.

Functions of bounded variation

${\displaystyle f:[a,b]\to ℝ}$, then for some partition ${\displaystyle Q}$:

• positive variation of f ${\displaystyle P(f,Q)=\sum (f\left(x_i\right)‐f{\left(x_{i+1}\right)}_{+}}$
• negative variation of f ${\displaystyle N(f,Q)=\sum (f\left(x_i\right)‐f{\left(x_{i+1}\right)}_{‐}}$
• total variation of f ${\displaystyle P(f,Q)=\sum |f\left(x_i\right)‐f\left(x_{i+1}\right|}$

And for without partition take supremum over all possible partitions. ${\displaystyle f}$ is said to be of bounded variation if ${\displaystyle T(f,[a,b])<\infty }$. If ${\displaystyle f\in BV[a,b]}$ and ${\displaystyle f_1=N(f,[a,b]),f_2=P(f,[a,b])}$ then ${\displaystyle f_1,f_2}$ are non-decreasing and ${\displaystyle f\left(x\right)=f_1\left(x\right)‐f_2\left(x\right)}$. ${\displaystyle f\in BV[a,b]}$ iff ${\displaystyle \exists }$ a finite signed measure ${\displaystyle \mu }$ on ${\displaystyle (ℝ,B(ℝ)}$ st ${\displaystyle f\left(x\right)=\mu [a,x]}$.

Absolutely continuous functions

A function is absolutely continuous if ${\displaystyle \forall \epsilon >0\exists \delta >0\text{st}{I}_j[a_j,b_j]\text{disjoint}\text{and}\sum m(b_j‐a_j)<\delta \Rightarrow \sum |F\left(b_j\right)‐F\left(a_j\right)|<\epsilon }$. By mean value theorem, F differentiable and ${\displaystyle F\text{'}}$ bounded, then F is ac.

A function ${\displaystyle F:[a,b]\to R}$ is absolutely continuous if ${\displaystyle \overline{F}=F\left(a\right)I(x<a)+F\left(x\right)I(a\le x<b)+F\left(b\right)I(x\ge b)}$ is ac.

Fundamental theorem of Lebesgue integral calculus: ${\displaystyle F:[a,b]\to ℝ}$ is ac iff there is a function ${\displaystyle f:[a,b]\to ℝ}$ st ${\displaystyle f}$ is Lebesgue measurable and integrable and ${\displaystyle F\left(x\right)=F\left(a\right)+{\int }_{[a,x]}f{d}_{{\mu }_L}\forall a\le x\le b}$.