Lebesgue Integral Note

Almost Everywhere

Definition

The Propositional Function is satisfied for every point except where Lebesgue Measure is 0

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Lebesgue Integral

Definition

Riemann integral and Lebesgue integral

Consider a measure spaces $(\Omega ,\mathcal{F},\mu )$ and a Measure Space $(\overline{\mathbb{R}},\mathcal{R})$, where $\mathcal{R}$ is the Borel Sigma-Field on $\overline{\mathbb{R}}$, and a Measurable Function $f:(\Omega ,\mathcal{F})\to (\overline{\mathbb{R}},\mathcal{R})$. Then the Lebesgue integral of the function $f$ with respect to $\mu$ is denoted as:

${\int }_{\Omega }fd\mu ={\int }_{\Omega }f(x)d\mu (x)={\int }_{\Omega }f(x)\mu (dx)$

Approximation of Lebesgue Integral with Finite Sum

$\int fd\lambda \approx \sum _{i=1}^N{a}_n\lambda (A_n)$ where $A_n:=\{x|f(x)=a_n\}$ or $A_n=f^{-1}(\{a_n\})$.

To compute the Riemann Integral of $f$, one partitions the domain $[a,b]$ into sub-intervals, while in the Lebesgue integral, one partitions the image of $f$.

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Simple Function

Definition

$f(\omega )=\sum _{i=1}^n{\alpha }_i{1}_{A_i}(\omega )$ where $A_i\in \mathcal{F}$ are disjoint measurable sets, $a_i$ are the sequence of real or complex numbers, and $1_A$ is a Indicator Function

Finite linear combination of indicator functions of measurable sets

Facts

Simple function $f$ is $\mathcal{F}-\mathcal{R}$ measurable map

A Measurable Function $f$ with a finite Image is a Simple Function

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Lebesgue Integral of Simple Function

Definition

Consider a measure spaces $(\Omega ,\mathcal{F},\mu )$ and a Measure Space $(\overline{\mathbb{R}},\mathcal{R})$, and a positive $\mathcal{F}-\mathcal{R}$ measurable Simple Function, $f:(\Omega ,\mathcal{F})\to (\overline{\mathbb{R}},\mathcal{R})$, with Codomain $R:=\{r_1,r_2,\dots ,r_n\}$ that have corresponding preimages $f^{-1}(r_i)=A_i\in \mathcal{F}$. Then the Lebesgue integral of the function $f$ with respect to $\mu$ is defined as ${\int }_{\Omega }fd\mu ={\sum }_{i=1}^n{r}_i\mu (A_i)$

Facts

Consider a measure spaces $(\Omega ,\mathcal{F},\mu )$ and a Measure Space $(\overline{\mathbb{R}},\mathcal{R})$, and a positive $\mathcal{F}-\mathcal{R}$ measurable Simple Function, $f:(\Omega ,\mathcal{F})\to (\overline{\mathbb{R}},\mathcal{R})$, with Codomain $R:=\{r_1,r_2,\dots ,r_n\}$ that have corresponding preimages $f^{-1}(r_i)=A_i\in \mathcal{F}$. Then, $f={\sum }_{i=1}^m{\alpha }_i{1}_{A_i}={\sum }_{j=1}^n{\beta }_j{1}_{B_j}⟹\int fd\mu ={\sum }_{i=1}^n{\alpha }_i\mu (A_i)={\sum }_{i=1}^n{\beta }_i\mu (B_i)$

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Lebesgue Integral of Non-Negative Function

Definition

Consider a measure spaces $(\Omega ,\mathcal{F},\mu )$ and a Measure Space $(\overline{\mathbb{R}},\mathcal{R})$ and a positive Measurable Function $f:(\Omega ,\mathcal{F})\to (\overline{\mathbb{R}},\mathcal{R})$. Then the Lebesgue integral of the function $f$ with respect to $\mu$ is defined as ${\int }_{\Omega }fd\mu :=\sup \left\{{\int }_{\Omega }\phi d\mu \right|0\le \phi \le f,\text{a.e. w.r.t.}\mu \}$ where each $\phi :(\Omega ,\mathcal{F})\to (\mathbb{R},\mathcal{R})$ is a Simple Function.

Facts

$\int fd\mu$ can be infinity

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Lebesgue Integral of Measurable Function

Definition

Consider a measure spaces $(\Omega ,\mathcal{F},\mu )$ and a Measure Space $(\overline{\mathbb{R}},\mathcal{R})$ and a Measurable Function $f:(\Omega ,\mathcal{F})\to (\overline{\mathbb{R}},\mathcal{R})$. Then the Lebesgue integral of the function $f$ with respect to $\mu$ is defined as ${\int }_{\Omega }fd\mu :={\int }_{\Omega }{f}^{+}d\mu -{\int }_{\Omega }{f}^{-}d\mu$ where $f^{+}:=\max (0,f)$ and $f^{-}:=\max (0,-f)$ for the function $f$.

Existence of Integral

$\int fd\mu$ exists

• $\int f^{+}d\mu <\infty$ and $\int f^{-}d\mu <\infty$ ($f$ is integrable w.r.t $\mu$) $\int fd\mu =\int f^{+}d\mu -\int f^{-}d\mu$
• $\int f^{+}d\mu =\infty$ and $\int f^{-}d\mu <\infty$ $\int fd\mu =\infty$
• $\int f^{+}d\mu <\infty$ and $\int f^{-}d\mu =\infty$ $\int fd\mu =-\infty$

$\int fd\mu$ not exists

• $\int f^{+}d\mu =\infty$ and $\int f^{-}d\mu =\infty$ $\int fd\mu$ is not defined

Notations

For a Density Function $F_X={\mu }_X((-\infty ,x])$ on the Measurable Space $(\mathbb{R},\mathcal{R},{\mu }_X)$ $\int gd{\mu }_X=\int gd{F}_X=\int g(x)d{F}_X(x)$

For a Density Function $F_X={\mu }_X((-\infty ,x])$ that has Density Function $f_X(x)$ on the Measurable Space $(\mathbb{R},\mathcal{R},{\mu }_X)$ $\int g(x)d{F}_X(x)=\int g(x)f_X(x)dx$

Consider a Measure Space $(\Omega ,\mathcal{F},\#)$ where $\Omega$ is a countable set, $\mathcal{F}=2^{\Omega }$ is a Sigma-Field defined on the set $\Omega$, and $\#$ is a Counting measure on $\mathcal{F}$. Then $\int fd\#=\sum _{i\in \Omega }f(i)$

Facts

For the functions $f^{+}:=\max (0,f)$ and $f^{-}:=\max (0,-f)$ of $f$,

• $f^{+},f^{-}$ are non-negative
• $|f|=f^{+}+f^{-}$
• $f=f^{+}-f^{-}$
• If $f$ is measurable, then $f^{+}$ and $f^{-}$ are also measurable.

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