Measures

Subjects: Measure Theory
Links: Rings and Algebras of Sets

Def: A set function is a function whose domain is a family of sets. An extended eral values set $\mu$ defined on a family $\mathcal{E}$ of sets is additive if, whenever $E,F,E\cup F\in \mathcal{E}$ and $E\cap F=\varnothing$, then $$\mu(E \cup F) = \mu(E) + \mu(F).$$An extended real valued set function $\mu$ defined on a set $\mathcal{E}$ is finitely additive, if for every disjoint family $\{E_1,\dots ,E_n\}$ of sets in $\mathcal{E}$ whose union is also in $\mathcal{E}$, we have $$\mu\left(\bigcup_{i = 1}^n E_i\right) = \sum_{i = 1}^n\mu(E_i).$$An extended real valued set function $\mu$ defined on a class $\mathcal{E}$ is $\sigma$-additive if, for every disjoint sequence $\{E_n:n<\omega \}$ of sets in $\mathcal{E}$ whose union is also $\mathcal{E}$, we have $$\mu\left(\bigcup_{n < \omega} E_n\right) = \sum_{n = 1}^\infty\mu(E_n).$$
Def: A measure is an extended real valued, non negative, and $\sigma$-additive set function $\mu$, defined on a ring $\mathcal{R}$, and such that $\mu (\varnothing )=0$.

Prop: If $\mu$ is an extended real valued, non negative, and additive set function defined on a ring $\mathcal{R}$, and such that exists a $E\in \mathcal{R}$ such that $\mu (E)<\mathrm{\infty }$, then $\mu (\varnothing )=0$.

Def: If $\mu$ is a measure on a ring $\mathcal{R}$, a set $E\in \mathcal{R}$ is said to have finite measure if $\mu (E)<\mathrm{\infty }$; the measure of $E$ is $\sigma$-finite if there exists a sequence $\{E_n:n<\omega \}$ of sets in $R$ such that $E\subseteq \bigcup _{n<\omega }{E}_n$ and $\mu (E_n)<\mathrm{\infty }$ for all $n<\omega$.

Prop: If $\mathcal{E}$ is a non empty family of sets and $\mu$ a measure on $\mathcal{R}\mathcal{(}\mathcal{E}\mathcal{)}$ such that if $E\in \mathcal{E}$, then $\mu (E)<\mathrm{\infty }$, then $\mu$ is finite on $\mathcal{R}\mathcal{(}\mathcal{E}\mathcal{)}$.

Def: If the measure of every set $E$ in $\mathcal{R}$ is finite or $\sigma$-finite, the measure $\mu$ is called finite of $\sigma$-finite on $\mathcal{R}$. If $X\in \mathcal{R}$ (i.e. if $\mathcal{R}$ is an algebra) and $\mu (X)$ is finite or $\sigma$-finite, then $\mu$ is called totally finite or totally $\sigma$-finite, respectively.

Prop: If $\mu$ is a measure on a $\sigma$-ring, then the family of all sets of finite measure is a ring and the family of all sets of $\sigma$-finite measure is a $\sigma$-ring.

Prop: Let $\mu$ be a $\sigma$-finite measure on a $\sigma$-ring. Then the family of all sets of finite measure a $\sigma$-ring iff $\mu$ is finite.

Def: The measure $\mu$ is called complete if the conditions $E\in \mathcal{R}$, $F\subseteq E$ and $\mu (E)=0$ imply that $F\in \mathcal{R}$.

Def: Let $\mathcal{P}$ be a semiring, and $\{E_1,\dots ,E_n\}$ a finite pairwise disjoint family of elements of $\mathcal{P}$ whose union, $E$ is also in $\mathcal{P}$ is called a $\mathcal{P}$-partition of $E$. Let $\mu$ be an extended real valued, non negative and additive set function. The $\mathcal{P}$-partition $\{E_1,\dots ,E_n\}$ is called a $\mu$ partition, if for every $F$ in $\mathcal{P}$, $$\mu(E \cap F) = \sum_{i = 1}^n \mu(E_i \cap F).$$If $\{E_1,\dots ,E_n\}$ and $\{F_1,\dots ,F_m\}$ are $\mathcal{P}$-partitions of $E$, then $\{E_1,\dots ,E_n\}$ is called a subpartition of $\{F_1,\dots ,F_m\}$ if each set $E_i$ is contained in one of the sets $F_j$.

Lemma: If $\{E_1,\dots ,E_n\}$ and $\{F_1,\dots ,F_m\}$ are partitions of $E$, then so is their product, consisting of all sets of the form $E_i\cap F_j$.

Lemma: If a subpartition of a partition $\{E_1,\dots ,E_n\}$ is a $\mu$-partition, then $\{E_1,\dots ,E_n\}$ is a $\mu$-partition.

Lemma: The product of two $\mu$-partitions is a $\mu$-partition.

Lemma: If $E=C_0\subseteq C_1\subseteq \cdots \subseteq C_n=F$, where $C_i\in \mathcal{P}$ for $i\in \{0,\dots ,n\}$ and if $D_i=C_i-C_{i-1}\in \mathcal{P}$, for $i\in \{1,\dots ,n\}$, then $\{E,D_1,\dots ,D_n\}$ is a $\mu$-partition of $F$.

Prop: Every partition of a set $E$ in $\mathcal{P}$ is a $\mu$-partition. Equivalently, we get that If $\mu$ is a extended real valued, non negative and additive set function defined on a Halmos semiring $\mathcal{P}$ such that $\mu (\varnothing )=0$, then $\mu$ is finitely additive.

Prop: If $\mu$ is a countably additive and non negative set function on a Halmos semiring $\mathcal{P}$, such that $\mu (\varnothing )=0$, then there is a unique measure $\bar{\mu }$ on the ring $\mathcal{R}\mathcal{(}\mathcal{P}\mathcal{)}$ such that $E\in \mathcal{P}$, $\bar{\mu }(E)=\mu (E)$. If $\mu$ is (totally) finte or $\sigma$-finite, then so is $\bar{\mu }$.

Def: An extended real value set function $\mu$ on a family $\mathcal{E}$ is monotone if, whenever $E,F\in \mathcal{E}$, and $E\subseteq F$, then $\mu (E)\le \mu (F)$.

Def: An extended real value set function $\mu$ on a family $\mathcal{E}$ is subtractive if, whenever $E,F,E\setminus F\in \mathcal{E}$, $E\subseteq F$, and $|\mu (E)|<\mathrm{\infty }$, then $\mu (F\setminus E)=\mu (F)-\mu (E)$.

Th: If $\mu$ is a measure on a ring $\mathcal{R}$, then $\mu$ is monotone and subtractive.

Th: If $\mu$ is a measure on a ring $\mathcal{R}$, if $E\in \mathcal{R}$, and if $\{E_n:n<\omega \}$ of sets in $\mathcal{R}$ such that $E\subseteq \bigcup _{n<\omega }{E}_n$, then $$\mu(E) \le \sum_{n < \omega} \mu(E_n).$$
Th: If $\mu$ is a measure on a ring $\mathcal{R}$, if $E\in \mathcal{R}$, and if $\{E_n:n<\omega \}$ inifinite disjoint sequence of sets in $\mathcal{R}$ such that $\bigcup _{n<\omega }{E}_n\subseteq E$, then $$\sum_{n <\omega} \mu(E_n) \le \mu(E).$$
Th: If $\mu$ is a measure on a ring $\mathcal{R}$ and $\{E_n:n<\omega \}$ is an increasing sequence of sets in $\mathcal{R}$ for for which $\underset{n\to \mathrm{\infty }}{lim}{E}_n\in \mathcal{R}$, then $$\mu\left(\lim_{n \to \infty} E_n\right) = \lim_{n \to \infty} \mu(E_n).$$
Th: If $\mu$ is a measure on a ring $\mathcal{R}$ and $\{E_n:n<\omega \}$ is an decreasing sequence of sets in $\mathcal{R}$ of which one has finite measure and for which $\underset{n\to \mathrm{\infty }}{lim}{E}_n\in \mathcal{R}$, then $$\mu\left(\lim_{n \to \infty} E_n\right) = \lim_{n \to \infty} \mu(E_n).$$
Def: An extended real valued set function $\mu$ defined on a family $\mathcal{E}$, is continuous from below at a set $E$ if for every increasing sequence $\{E_n:n<\omega \}$ of sets in $\mathcal{E}$ for which $\underset{n\to \mathrm{\infty }}{lim}{E}_n=E$, we have $\underset{n\to \mathrm{\infty }}{lim}\mu (E_n)=\mu (E)$. Similarly, $\mu$ is continuous from above at $E$ if, for every decreasing sequence $\{E_n:n<\omega \}$ of sets in $\mathcal{E}$ for which $|\mu (E_m)|<\mathrm{\infty }$ for at least one value of $m$ and for which $\underset{n\to \mathrm{\infty }}{lim}\mu (E_n)=\mu (E)$.

Obs: The two theorems above, we see that if $\mu$ is a measure, then $\mu$ is continuous from above and from below.

Cor: If $\mu$ is a measure on a ring $\mathcal{R}$ and if $\{E_n:n<\omega \}$ is a sequence of sets in $\mathcal{R}$ for which $\bigcap _{n\le m<\omega }{E}_n\in \mathcal{R}$, for $m<\omega$ and $\underset{n\to \mathrm{\infty }}{lim inf}{E}_n\in \mathcal{R}$, then $$\mu\left(\liminf_{n \to \infty} E_n\right) \le \liminf_{n \to \infty} \mu(E_n).$$
Cor: If $\mu$ is a measure on a ring $\mathcal{R}$ and if $\{E_n:n<\omega \}$ is a sequence of sets in $\mathcal{R}$ for which $\bigcup _{n\le m<\omega }{E}_n\in \mathcal{R}$, for $m<\omega$, $\underset{n\to \mathrm{\infty }}{lim inf}{E}_n\in \mathcal{R}$, and if $\mu \left(\bigcup _{n\le m<\omega }{E}_m\right)<\mathrm{\infty }$ then $$ \limsup_{n \to \infty} \mu(E_n)\le \mu\left(\limsup_{n \to \infty} E_n\right).$$
Cor: If $\mu$ is a measure on a ring $\mathcal{R}$ and if $\{E_n:n<\omega \}$ is a sequence of sets in $\mathcal{R}$ for which $\bigcup _{n\le m<\omega }{E}_n\in \mathcal{R}$, for $m<\omega$, $\underset{n\to \mathrm{\infty }}{lim inf}{E}_n\in \mathcal{R}$, if $\mu \left(\bigcup _{n\le m<\omega }{E}_m\right)<\mathrm{\infty }$, and $\sum _{n<\omega }\mu (E_n)<\mathrm{\infty }$, then $\mu (lim sup{E}_n)=0$.

Th: Let $\mu$ a finite, non negative, and additive set function on a ring $\mathcal{R}$. If $\mu$ is either continuous from below at every $E\in \mathcal{R}$, or continuous from below at $\varnothing$, then $\mu$ is a measure on $\mathcal{R}$.

Prop: If $\mu$ is a measure on a ring $\mathcal{R}$, then if $E,F\in \mathcal{R}$, then $\mu (E)+\mu (F)=\mu (E\cup F)+\mu (E\cap F)$.