Measure Theory/Integration

Let ${\displaystyle (X,\sigma ,\mu )}$ be a ${\displaystyle \sigma }$-finite measure space. Suppose ${\displaystyle s}$ is a positive simple measurable function, with ${\displaystyle s=\displaystyle \sum _{i=1}^{3}y_{i}\chi _{A_{i}}}$; ${\displaystyle A_{i}\in \sigma }$ are disjoint.

Define ${\displaystyle \displaystyle \int _{X}s~d\mu =\sum y_{i}\mu (A_{i})}$

Let ${\displaystyle f:X\to {\overline {\mathbb {R} }}}$ be measurable, and let ${\displaystyle f\geq 0}$.

Define ${\displaystyle \displaystyle \int _{X}f~d\mu =\sup\{\int _{X}s~d\mu =\sum y_{i}\mu (A_{i})|s{\text{ simple }},s\geq 0,s\leq f\}}$

Now let ${\displaystyle f}$ be any measurable function. We say that ${\displaystyle f}$ is integrable if ${\displaystyle f^{+}}$ and ${\displaystyle f^{-}}$ are integrable and if ${\displaystyle \displaystyle \int _{X}f^{+}~d\mu ,\int _{X}f^{-}~d\mu <\infty }$. Then, we write

${\displaystyle \displaystyle \int _{X}f~d\mu =\int _{X}f^{+}~d\mu -\int _{X}f^{-}~d\mu }$

The class of measurable functions on ${\displaystyle X}$ is denoted by ${\displaystyle {\mathcal {L}}^{1}(X)}$

For ${\displaystyle 0, we define ${\displaystyle {\mathcal {L}}^{p}}$ to be the collection of all measurable functions ${\displaystyle f}$ such that ${\displaystyle |f|^{p}\in {\mathcal {L}}^{1}}$

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 ${\displaystyle (X,\sigma ,\mu )}$ be a measure space and let ${\displaystyle f,g}$ be measurable on ${\displaystyle X}$. Then

1. If ${\displaystyle f\leq g}$, then ${\displaystyle \displaystyle \int _{X}fd\mu \leq \int _{X}gd\mu }$
2. If ${\displaystyle A,B\in \sigma }$, ${\displaystyle A\subset B}$, then ${\displaystyle \displaystyle \int _{A}fd\mu \leq \int _{B}fd\mu }$
3. If ${\displaystyle f\geq 0}$ and ${\displaystyle c\geq 0}$ then ${\displaystyle \displaystyle \int _{X}cfd\mu =c\int _{X}fd\mu }$
4. If ${\displaystyle E\in \sigma }$, ${\displaystyle \mu (E)=0}$, then ${\displaystyle \displaystyle \int _{E}fd\mu =0}$, even if ${\displaystyle f(E)=\{\infty \}}$
5. If ${\displaystyle E\in \sigma }$, ${\displaystyle f(E)=\{0\}}$, then ${\displaystyle \displaystyle \int _{E}fd\mu =0}$, even if ${\displaystyle \mu (E)=\infty }$

Proof

Monotone Convergence Theorem

Suppose ${\displaystyle f_{n}\geq 0}$ and ${\displaystyle f_{n}}$ are measurable for all ${\displaystyle n}$ such that

1. ${\displaystyle f_{1}(x)\leq f_{2}(x)\leq \ldots }$ for every ${\displaystyle x\in X}$
2. ${\displaystyle f_{n}(x)\to f(x)}$ almost everywhere on ${\displaystyle X}$

Then, ${\displaystyle \displaystyle \int _{X}f_{n}d\mu \to \int _{X}fd\mu }$

Proof

${\displaystyle \displaystyle \int _{X}f_{n}d\mu }$ is an increasing sequence in ${\displaystyle \mathbb {R} }$, and hence, ${\displaystyle \displaystyle \int _{X}f_{n}d\mu \to \alpha \in {\overline {\mathbb {R} }}}$ (say). We know that ${\displaystyle f}$ is measurable and that ${\displaystyle f\geq f_{n}\forall n}$. That is,

${\displaystyle \displaystyle \int _{X}f_{1}d\mu \leq \int _{X}f_{2}d\mu \leq \ldots \int _{X}f_{n}d\mu \leq \ldots \int _{X}fd\mu }$

Hence, ${\displaystyle \displaystyle \alpha \leq \int _{X}fd\mu =\sup\{\int _{X}sd\mu :s{\text{ is simple }},0\leq s\leq 1\}}$

Let ${\displaystyle c\in [0,1]}$

Define ${\displaystyle E_{n}=\{x|f_{n}(x)\geq cs(x)\}}$; ${\displaystyle n=1,2\ldots }$. Observe that ${\displaystyle E_{1}\subset E_{2}\subset \ldots }$ and ${\displaystyle \bigcup E_{n}=X}$

Suppose ${\displaystyle x\in X}$. If ${\displaystyle f(x)=0}$ then ${\displaystyle s(x)=0}$ implying that ${\displaystyle x\in E_{1}}$. If ${\displaystyle f(x)>0}$, then there exists ${\displaystyle n}$ such that ${\displaystyle f_{n}(x)>cs(x)}$ and hence, ${\displaystyle x\in E_{n}}$.

Thus, ${\displaystyle \bigcup E_{n}=X}$, therefore ${\displaystyle \displaystyle \int _{X}f_{n}(x)d\mu \geq \int _{E_{n}}f_{n}d\mu \geq c\int _{E_{n}}sd\mu }$. As this is true if ${\displaystyle c\in [0,1]}$, we have that ${\displaystyle \alpha \geq \displaystyle \int _{X}fd\mu }$. Thus, ${\displaystyle \displaystyle \int _{X}f_{n}d\mu \to \int _{X}fd\mu }$.

Fatou's Lemma

Let ${\displaystyle f_{n}\geq 0}$ be measurable functions. Then,

${\displaystyle \displaystyle \int _{X}\liminf f_{n}d\mu \leq \liminf \int _{X}f_{n}d\mu }$

Proof

For ${\displaystyle k=1,2,\ldots }$ define ${\displaystyle g_{k}(x)=\displaystyle \inf _{i\geq k}f_{i}(x)}$. Observe that ${\displaystyle g_{k}}$ are measurable and increasing for all ${\displaystyle x}$.

As ${\displaystyle k\to \infty }$, ${\displaystyle g_{k}(x)\to \displaystyle \liminf _{n\geq k}f_{n}(x)}$. By monotone convergence theorem,

${\displaystyle \displaystyle \int _{X}g_{k}d\mu \to \int _{X}\liminf f_{n}(x)d\mu }$ and as ${\displaystyle \displaystyle \int _{X}g_{k}(x)d\mu \leq \liminf \int g_{k}(x)d\mu }$, we have the result.

Dominated convergence theorem

Let ${\displaystyle (X,\sigma ,\mu )}$ be a complex measure space. Let ${\displaystyle \{f_{n}\}}$ be a sequence of complex measurable functions that converge pointwise to ${\displaystyle f}$; ${\displaystyle f(x)=\displaystyle \lim _{n\to \infty }f_{n}(x)}$, with ${\displaystyle x\in X}$

Suppose ${\displaystyle |f_{n}(x)|\leq g(x)}$ for some ${\displaystyle g\in {\mathcal {L}}^{1}(X)}$ then

${\displaystyle f\in {\mathcal {L}}^{1}}$ and ${\displaystyle \displaystyle \int _{X}|f_{n}-f|d\mu \to 0}$ as ${\displaystyle n\to \infty }$

Proof

We know that ${\displaystyle |f|\leq g}$ and hence ${\displaystyle |f_{n}-f|\leq 2g}$, that is, ${\displaystyle 0\leq 2g-|f_{n}-f|}$

Therefore, by Fatou's lemma, ${\displaystyle \displaystyle \int _{X}2gd\mu \leq \liminf \int _{X}(2g-|f_{n}-f|)d\mu \leq \displaystyle \int _{X}2gd\mu +\liminf \int _{X}(-|f_{n}-f|)d\mu }$

${\displaystyle =\displaystyle \int _{X}2gd\mu -\limsup \int _{X}|f_{n}-f|d\mu }$

As ${\displaystyle g\in {\mathcal {L}}^{1}}$, ${\displaystyle \displaystyle \limsup _{n\to \infty }\int _{X}|f_{n}-f|d\mu \leq }$ implying that ${\displaystyle \displaystyle \limsup \int _{X}|f_{n}-f|d\mu }$

Theorem

1. Suppose ${\displaystyle f:X\to [0,\infty ]}$ is measurable, ${\displaystyle E\in \sigma }$ with ${\displaystyle \mu (E)>0}$ such that ${\displaystyle \displaystyle \int _{E}fd\mu =0}$. Then ${\displaystyle f(x)=0}$ almost everywhere ${\displaystyle E}$
2. Let ${\displaystyle f\in {\mathcal {L}}^{1}(X)}$ and let ${\displaystyle \displaystyle \int _{E}fd\mu =0}$ for every ${\displaystyle E\in \sigma }$. Then, ${\displaystyle f=0}$ almost everywhere on ${\displaystyle X}$
3. Let ${\displaystyle f\in {\mathcal {L}}^{1}(X)}$ and ${\displaystyle \displaystyle \left|\int _{X}fd\mu \right|=\int _{X}|f|d\mu }$ then there exists constant ${\displaystyle \alpha }$ such that ${\displaystyle |f|=\alpha f}$ almost everywhere on ${\displaystyle E}$

Proof

1. For each ${\displaystyle n\in \mathbb {N} }$ define ${\displaystyle A_{n}=\{x\in E|f(x)>{\frac {1}{n}}\}}$. Observe that ${\displaystyle A_{n}\uparrow E}$
but ${\displaystyle {\frac {1}{n}}\mu (A_{n})\leq \displaystyle \int _{A_{n}}fd\mu \leq \int _{E}fd\mu =0}$ Thus ${\displaystyle \mu (A_{n})=0}$ for all ${\displaystyle n}$, by continuity, ${\displaystyle f=0}$ almost everywhere on ${\displaystyle E}$
2. Write ${\displaystyle \displaystyle \int _{E}fd\mu =\int _{E}u^{+}d\mu -\int _{E}u^{-}d\mu +i\left(\int _{E}v^{+}d\mu -\int _{E}v^{-}d\mu \right)}$, where ${\displaystyle u^{+},u^{-},v^{+},v^{-}}$ are non-negative real measurable.
Further as ${\displaystyle \displaystyle \int _{E}u^{+}d\mu ,\int _{E}(-u^{-})d\mu }$ are both non-negative, each of them is zero. Thus, by applying part I, we have that ${\displaystyle u^{+},u^{-}}$ vanish almost everywhere on ${\displaystyle E}$. We can similarly show that ${\displaystyle v^{+},v^{-}}$ vanish almost everywhere on ${\displaystyle E}$.