# Measure Theory/Integration

Let $(X,\sigma ,\mu )$ be a $\sigma$-finite measure space. Suppose $s$ is a positive simple measurable function, with $s=\displaystyle \sum _{i=1}^{3}y_{i}\chi _{A_{i}}$; $A_{i}\in \sigma$ are disjoint.
Define $\displaystyle \int _{X}s~d\mu =\sum y_{i}\mu (A_{i})$
Let $f:X\to {\overline {\mathbb {R} }}$ be measurable, and let $f\geq 0$.
Define $\displaystyle \int _{X}f~d\mu =\sup\{\int _{X}s~d\mu =\sum y_{i}\mu (A_{i})|s{\text{ simple }},s\geq 0,s\leq f\}$
Now let $f$ be any measurable function. We say that $f$ is integrable if $f^{+}$ and $f^{-}$ are integrable and if $\displaystyle \int _{X}f^{+}~d\mu ,\int _{X}f^{-}~d\mu <\infty$. Then, we write
$\displaystyle \int _{X}f~d\mu =\int _{X}f^{+}~d\mu -\int _{X}f^{-}~d\mu$
The class of measurable functions on $X$ is denoted by ${\mathcal {L}}^{1}(X)$
For $0<p<\infty$, we define ${\mathcal {L}}^{p}$ to be the collection of all measurable functions $f$ such that $|f|^{p}\in {\mathcal {L}}^{1}$
A property is said to hold almost everywhere if the set of all points where the property does not hold has measure zero.

## Properties

Let $(X,\sigma ,\mu )$ be a measure space and let $f,g$ be measurable on $X$. Then
* If $f\leq g$, then $\displaystyle \int _{X}fd\mu \leq \int _{X}gd\mu$
* If $A,B\in \sigma$, $A\subset B$, then $\displaystyle \int _{A}fd\mu \leq \int _{B}fd\mu$
* If $f\geq 0$ and $c\geq 0$ then $\displaystyle \int _{X}cfd\mu =c\int _{X}fd\mu$
* If $E\in \sigma$, $\mu (E)=0$, then $\displaystyle \int _{E}fd\mu =0$, even if $f(E)=\{\infty \}$
* If $E\in \sigma$, $f(E)=\{0\}$, then $\displaystyle \int _{E}fd\mu =0$, even if $\mu (E)=\infty$
Proof

## Monotone Convergence Theorem

Suppose $f_{n}\geq 0$ and $f_{n}$ are measurable for all $n$ such that
* $f_{1}(x)\leq f_{2}(x)\leq \ldots$ for every $x\in X$
* $f_{n}(x)\to f(x)$ almost everywhere on $X$
Then, $\displaystyle \int _{X}f_{n}d\mu \to \int _{X}fd\mu$
Proof
$\displaystyle \int _{X}f_{n}d\mu$ is an increasing sequence in $\mathbb {R}$, and hence, $\displaystyle \int _{X}f_{n}d\mu \to \alpha \in {\overline {\mathbb {R} }}$ (say). We know that $f$ is measurable and that $f\geq f_{n}\forall n$. That is,
$\displaystyle \int _{X}f_{1}d\mu \leq \int _{X}f_{2}d\mu \leq \ldots \int _{X}f_{n}d\mu \leq \ldots \int _{X}fd\mu$
Hence, $\displaystyle \alpha \leq \int _{X}fd\mu =\sup\{\int _{X}sd\mu :s{\text{ is simple }},0\leq s\leq 1\}$
Let $c\in [0,1]$
Define $E_{n}=\{x|f_{n}(x)\geq cs(x)\}$; $n=1,2\ldots$. Observe that $E_{1}\subset E_{2}\subset \ldots$ and $\bigcup E_{n}=X$
Suppose $x\in X$. If $f(x)=0$ then $s(x)=0$ implying that $x\in E_{1}$. If $f(x)>0$, then there exists $n$ such that $f_{n}(x)>cs(x)$ and hence, $x\in E_{n}$.
Thus, $\bigcup E_{n}=X$, therefore $\displaystyle \int _{X}f_{n}(x)d\mu \geq \int _{E_{n}}f_{n}d\mu \geq c\int _{E_{n}}sd\mu$. As this is true if $c\in [0,1]$, we have that $\alpha \geq \displaystyle \int _{X}fd\mu$. Thus, $\displaystyle \int _{X}f_{n}d\mu \to \int _{X}fd\mu$.

## Fatou's Lemma

Let $f_{n}\geq 0$ be measurable functions. Then,
$\displaystyle \int _{X}\liminf f_{n}d\mu \leq \liminf \int _{X}f_{n}d\mu$
Proof
For $k=1,2,\ldots$ define $g_{k}(x)=\displaystyle \inf _{i\geq k}f_{i}(x)$. Observe that $g_{k}$ are measurable and increasing for all $x$.
As $k\to \infty$, $g_{k}(x)\to \displaystyle \liminf _{n\geq k}f_{n}(x)$. By monotone convergence theorem,
$\displaystyle \int _{X}g_{k}d\mu \to \int _{X}\liminf f_{n}(x)d\mu$ and as $\displaystyle \int _{X}g_{k}(x)d\mu \leq \liminf \int g_{k}(x)d\mu$, we have the result.

## Dominated convergence theorem

Let $(X,\sigma ,\mu )$ be a complex measure space. Let $\{f_{n}\}$ be a sequence of complex measurable functions that converge pointwise to $f$; $f(x)=\displaystyle \lim _{n\to \infty }f_{n}(x)$, with $x\in X$
Suppose $|f_{n}(x)|\leq g(x)$ for some $g\in {\mathcal {L}}^{1}(X)$ then
$f\in {\mathcal {L}}^{1}$ and $\displaystyle \int _{X}|f_{n}-f|d\mu \to 0$ as $n\to \infty$
Proof
We know that $|f|\leq g$ and hence $|f_{n}-f|\leq 2g$, that is, $0\leq 2g-|f_{n}-f|$
Therefore, by Fatou's lemma, $\displaystyle \int _{X}2gd\mu \leq \liminf \int _{X}(2g-|f_{n}-f|)d\mu \leq \displaystyle \int _{X}2gd\mu +\liminf \int _{X}(-|f_{n}-f|)d\mu$
$=\displaystyle \int _{X}2gd\mu -\limsup \int _{X}|f_{n}-f|d\mu$
As $g\in {\mathcal {L}}^{1}$, $\displaystyle \limsup _{n\to \infty }\int _{X}|f_{n}-f|d\mu \leq$ implying that $\displaystyle \limsup \int _{X}|f_{n}-f|d\mu$

## Theorem

* Suppose $f:X\to [0,\infty ]$ is measurable, $E\in \sigma$ with $\mu (E)>0$ such that $\displaystyle \int _{E}fd\mu =0$. Then $f(x)=0$ almost everywhere $E$
* Let $f\in {\mathcal {L}}^{1}(X)$ and let $\displaystyle \int _{E}fd\mu =0$ for every $E\in \sigma$. Then, $f=0$ almost everywhere on $X$
* Let $f\in {\mathcal {L}}^{1}(X)$ and $\displaystyle \left|\int _{X}fd\mu \right|=\int _{X}|f|d\mu$ then there exists constant $\alpha$ such that $|f|=\alpha f$ almost everywhere on $E$
Proof
* For each $n\in \mathbb {N}$ define $A_{n}=\{x\in E|f(x)>{\frac {1}{n}}\}$. Observe that $A_{n}\uparrow E$ but ${\frac {1}{n}}\mu (A_{n})\leq \displaystyle \int _{A_{n}}fd\mu \leq \int _{E}fd\mu =0$ Thus $\mu (A_{n})=0$ for all $n$, by continuity, $f=0$ almost everywhere on $E$
* Write $\displaystyle \int _{E}fd\mu =\int _{E}u^{+}d\mu -\int _{E}u^{-}d\mu +i\left(\int _{E}v^{+}d\mu -\int _{E}v^{-}d\mu \right)$, where $u^{+},u^{-},v^{+},v^{-}$ are non-negative real measurable. Further as $\displaystyle \int _{E}u^{+}d\mu ,\int _{E}(-u^{-})d\mu$ are both non-negative, each of them is zero. Thus, by applying part I, we have that $u^{+},u^{-}$ vanish almost everywhere on $E$. We can similarly show that $v^{+},v^{-}$ vanish almost everywhere on $E$.