5.1 Measure Space
Definition 5.1.1 (Measurable Space).label Let $X$ be a set and $\Si\suf 2^{X}$ be a $\s$-algebra, then the pair $(X,\Si)$ is a measurable space
Definition 5.1.2 (Measure Space).label Let $(X,\Si)$ be a measurable space and $\mu:\Si\to[0,\infty]$ then $\mu$ is a (countably additive) measure if
- (M1)
$\mu(\emp)=0$
- (M2)
For any $\curl{E_n}_{n\in\N}\suf\Si$ pairwise disjoint, $\mu(\dcups{}{n\in\N}E_{n})=\sums{}{n\in\N}\mu(E_{n})$
In which case, $(X,\Si,\mu)$ is a measure space. If $\mu$ instead satisfies (M1) and
- (M2’)
For any $\curl{E_n}_{1}^{n}\suf\Si$ pairwise disjoint, $\mu(\dcups{n}{i=1}E_{i})=\sums{n}{i=1}\mu(E_{i})$
then $\mu$ is a finitely-additive measure.
Definition 5.1.3 (Null Set).label Let $(X,\Si,\mu)$ be a measure space, then $A\suf X$ is $\mu$-null/null/negligible if $\exists E\in\Si$ such that $A\suf E$ and $\mu(E)=0$. A set $A\suf X$ is co-null/co-negligible if $X\del A$ is null.
Definition 5.1.4 (Almost Everywhere).label Let $(X,\Si,\mu)$ be a measure space. A statement $P$ holds $\mu$-almost everywhere/$\mu$-a.e./a.e. if the set on which $\neg P$ holds is $\mu$-null.
Proposition 5.1.5.label Let $(X,\Si,\mu)$ be a measure space then
- (1)
For any $E,F\in\Si$ such that $E\suf F$ then $\mu(E)\leq \mu(F)$
- (2)
For any $\curl{E_n }_{n\in\N}\suf\Si$ we have $\mu \parens{\cups{}{n\in\N}E_n }\leq \sums{}{n\in\N}\mu(E_{n})$
- (3)
For any $\curl{E_n }_{n\in\N}\suf\Si$ with $E_{n}\suf E_{n+1}$ for all $n\in\N$ we have $\mu \parens{\cups{}{n\in\N}E_n }= \limit{n\to\infty}\mu(E_{n})=\SUP{n\in\N}\mu(E_{n})$
- (4)
For any $\curl{E_n }_{n\in\N}\suf\Si$ with $E_{n}\supf E_{n+1}$ for all $n\in\N$ and $\mu(E_{k})<\infty$ for some $k\in\N$ then $\mu \parens{\caps{}{n\in\N}E_n}=\limit{n\to\infty}\mu(E_{n})=\INF{n\in\N}\mu(E_{n})$
- (5)
For any $\curl{E_n}_{n\in\N}\suf\Si$
\begin{align*}\mu \parens{\limitinf{n\to\infty }E_n}\leq \limitinf{n\to\infty}\mu(E_{n})\end{align*} - (6)
For any $\curl{E_n}_{n\in\N}\suf\Si$ with $\mu\parens{\cups{}{n\in\N}E_n}<\infty$
\begin{align*}\mu \parens{\limitsup{n\to\infty }E_n}\leq \limitsup{n\to\infty}\mu(E_{n})\end{align*}
Proof.
- (1)
$\mu(F)=\mu(E)+\mu(F\del E)\geq \mu(E)$
- (2)
For each $n\in\N^{+}$ let $F_{n}\define E_{n}\del \cups{n-1}{i=1}E_{i}$ then $\curl{F_n}_{n\in\N}$ is pairwise disjoint and $\dcups{}{n\in\N}F_{n}=\cups{}{n\in\N}E_{n}$ hence
\begin{align*}\mu \parens{\cups{}{n\in\N}E_n}=\mu \parens{\dcups{}{n\in\N}F_n }=\sums{}{n\in\N}\mu(F_{n})\leq \sums{}{n\in\N}\mu(E_{n})\end{align*} - (3)
For each $n\in\N^{+}$ let $F_{n}\define E_{n}\del E_{n-1}$ then $\curl{F_n}_{n\in\N}$ is pairwise disjoint and $\dcups{}{n\in\N}F_{n}=\cups{}{n\in\N}E_{n}$ hence
\begin{align*}\mu \parens{\cups{}{n\in\N}E_n}&=\mu \parens{\dcups{}{n\in\N}F_n }=\sums{}{n\in\N}\mu(F_{n})\\&=\limit{n\to\infty}\sums{}{m\leq n}\mu(F_{m})=\limit{n\to\infty}\mu(E_{n})=\SUP{n\in\N}\mu(E_{n})\end{align*} - (4)
Since $\mu(E_{k})<\infty$
\begin{align*}\limit{n\to\infty}\mu(E_{n})&=\mu(E_{k})-\limit{n\to\infty}\mu(E_{k}\del E_{k+n})\\&=\mu(E_{k})-\mu\parens{E_k\del\caps{}{n\in\N^+}E_{k+n}}=\mu\parens{\caps{}{n\in\N^+}E_{k+n}}=\mu\parens{\caps{}{n\in\N^+}E_n}\end{align*} - (5)
By (1) and (3)
\begin{align*}\mu\parens{\limitinf{n\to\infty}E_n}&=\mu\parens{\cups{}{n\in\N^+}\caps{}{m\geq n}E_m}=\limit{n\to\infty}\mu\parens{\caps{}{m\geq n}E_m}\\&\leq\limit{n\to\infty}\INF{m\geq n}\mu(E_{m})=\limitinf{n\to\infty}\mu(E_{n})\end{align*} - (6)
By (1) and (4)
\begin{align*}\mu\parens{\limitsup{n\to\infty}E_n}&=\mu\parens{\caps{}{n\in\N^+}\cups{}{m\geq n}E_m}=\limit{n\to\infty}\mu\parens{\cups{}{m\geq n}E_m}\\&\geq\limit{n\to\infty}\SUP{m\geq n}\mu(E_{m})=\limitsup{n\to\infty}\mu(E_{n})\end{align*}
$\square$
Definition 5.1.6 (Borel Measure).label Let $(X,\cali{T})$ be a topological space and $(X,\Si,\mu)$ be a measure space then $\mu$ is a Borel measure if $\Si=\cali{B}_{X}$. Denote by $\cali{M}_{lf}(X)$ the space of Borel measures on $X$ that is finite on compact sets.
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