Measures

Measures provide a method for mapping sets to a values in the interval $[0,+\infty ]$. The resulting value can interpreted as the size of the subset. From a geometric perspective, the measure of a set can be viewed as the generalization of length, area, and volume.

Definition

Let $X$ be a set and let ${\mathcal {M}}\subseteq 2^{X}$ be a $\sigma$-algebra. Tbe structure $\left(X,{\mathcal {M}}\right)$ is called a measurable space and each set in ${\mathcal {M}}$ is called a measurable set. A measure on $(X,{\mathcal {M}})$ (also referred to simply as a measure on $X$ if ${\mathcal {M}}$ is understood) is a function $\mu :{\mathcal {M}}\rightarrow [0,+\infty ]$ that satisfies the following criteria:

1. $\mu \left(\emptyset \right)=0$,
2. Let $\left\{E_{k}\right\}_{k=1}^{\infty }$ be a disjoint sequence of sets such that each $E_{k}\in {\mathcal {M}}$. Then, $\mu \left(\cup _{k=1}^{\infty }E_{k}\right)=\sum _{k=1}^{\infty }\mu \left(E_{k}\right)$.

If the previous conditions are satisfied, the structure $\left(X,{\mathcal {M}},\mu \right)$ is called a measure space.

Additional Terminology

Let $\left(X,{\mathcal {M}},\mu \right)$ be a measure space.

• The measure $\mu$ is called finite if $\mu \left(X\right)<+\infty$.
• Let $E\in {\mathcal {M}}$. If there exist $\left\{E_{k}\right\}_{k=1}^{\infty }\subseteq {\mathcal {M}}$ such that $E=\cup _{k=1}^{\infty }E_{k}$ and $\mu \left(E_{k}\right)<+\infty$ (for all $k\in \mathbb {N}$), then $E$ is $\sigma$-finite for $\mu$.
• If $X$ is $\sigma$-finite for $\mu$, then $\mu$ is called $\sigma$-finite.
• Let $S$ be the collection of all the sets in ${\mathcal {M}}$ with infinite $\mu$-measure. The measure $\mu$ is called semifinite if there exists $F\in {\mathcal {M}}$ such that $F\subseteq E$ and $0<\mu (F)<+\infty$, for all $E\in S$.

Properties

Let $\left(X,{\mathcal {M}},\mu \right)$ be a measure space.

1. Finite Additivity: Let $\left\{E_{k}\right\}_{k=1}^{n}$ be a finite disjoint sequence of sets such that each $E_{k}\in {\mathcal {M}}$. Then, $\mu \left(\cup _{k=1}^{n}E_{k}\right)=\sum _{k=1}^{n}\mu \left(E_{k}\right)$. This follows directly from the defintion of measures by taking $E_{n+1}=E_{n+2}=...=\emptyset$.
2. Monotonicity: Let $E,F\in {\mathcal {M}}$ such that $E\subseteq F$. Then, $\mu \left(E\right)\leq \mu \left(F\right)$.
3. Subadditivity: Let $\left\{E_{k}\right\}_{k=1}^{\infty }\subseteq {\mathcal {M}}$. Then, $\mu \left(\cup _{k=1}^{\infty }E_{k}\right)\leq \sum _{k=1}^{\infty }\mu \left(E_{k}\right)$.
4. Continuity from Below: Let $\left\{E_{k}\right\}_{k=1}^{\infty }\subseteq {\mathcal {M}}$ such that $E_{1}\subseteq E_{2}\subseteq ...$. Then, $\mu \left(\cup _{k=1}^{\infty }E_{k}\right)=\lim _{k\rightarrow +\infty }\mu \left(E_{k}\right)$.
5. Continuity from Above: Let $\left\{E_{k}\right\}_{k=1}^{\infty }\subseteq {\mathcal {M}}$ such that $E_{1}\supseteq E_{2}\supseteq ...$ and $\mu \left(E'\right)<+\infty$ for some $E'\in \left\{E_{k}\right\}_{k=1}^{\infty }$. Then, $\mu \left(\cap _{k=1}^{\infty }E_{k}\right)=\lim _{k\rightarrow +\infty }\mu \left(E_{k}\right)$.

Examples

• Let $X$ be a non-empty set and ${\mathcal {M}}=2^{X}$. Let $f$ be any function from $X$ to $[0,+\infty ]$. Given $E\in {\mathcal {M}}$, define $A_{E}=\left\{x\in E:f(x)>0\right\}$. Then, the function $\mu :{\mathcal {M}}\rightarrow [0,+\infty ]$ defined by $\mu (E)={\begin{cases}\sum _{x\in E}f(x),A_{E}{\text{ is countable}}\\+\infty ,A_{E}{\text{ is uncountable}}\end{cases}}$ is a measure. This measure has the following properties:

1. The measure $\mu$ is semifinite if and only if $f(x)<+\infty$ for every $x\in X$.
2. The measure $\mu$ is $\sigma$-finite if and only if $\mu$ is semifinite and $A_{E}$ is countable for every $E\in {\mathcal {M}}$.

There are special cases of this measure that are frequently used:

1. When fixing $f(x)=1$, the resulting measure is referred to as the counting measure.
2. Let $x_{0}\in X$ be fixed. By defining $f(x)={\begin{cases}1,x=x_{0}\\0,x\neq x_{0}\end{cases}}$, the resulting measure is referred to as the point mass measure or the Dirac measure.

• Let $X$ be an uncountable set. Let ${\mathcal {M}}$ be the $\sigma$-algebra of countable or co-cocountable sets of $X$. The function $\mu :0\rightarrow [0,+\infty ]$ defined as $\mu (E)={\begin{cases}0,E{\text{ is countable}}\\1,E{\text{ is co-countable}}\end{cases}}$ is a measure.

• Let $X$ be an infinite set. Let ${\mathcal {M}}=2^{X}$. The function $\mu :0\rightarrow [0,+\infty ]$ defined as $\mu (E)={\begin{cases}0,E{\text{ is finite}}\\+\infty ,E{\text{ is infinite}}\end{cases}}$ is not a measure. To verify that it is not a measure, it is sufficient to take $X=\mathbb {N}$, and note that $\sum _{k=1}^{\infty }\mu \left(\{k\}\right)=0\neq +\infty =\mu \left(\mathbb {N} \right)=\mu \left(\cup _{k=1}^{\infty }\{k\}\right)$. In other words. the countable additivity property is not satisfied. However, $\mu$ does satisfy the finite additivity property.

Complete Measures

Consider a measure space $(X,{\mathcal {M}},\mu )$. A set $E\in {\mathcal {M}}$ is called a $\mu$-null set (or simply null set) if $\mu (E)=0$. A property $P(x)$ holds $\mu$-almost everywhere (or simply almost everywhere) if $N=\left\{x\in X:P(x){\text{ does not hold}}\right\}$ satisfies $N\in {\mathcal {M}}$ and $\mu (N)=0$.

A measure space $(X,{\mathcal {M}},\mu )$ is called complete if ${\mathcal {M}}$ contains all subsets of its null sets. An incomplete measure space can be constructed by taking $X=\{a.b,c\}$ and ${\mathcal {M}}=\{\emptyset ,\{a\},\{b,c\},X\}$ with $\mu (E)={\begin{cases}0,E\neq \{a\}\\1,E=\{a\}\end{cases}}$. The set $\{b,c\}$ is a null set in this case, but $\{b\}\notin {\mathcal {M}}$.

Given an incomplete measure $(X,{\mathcal {M}},\mu )$, the following theorem guarantees that a complete measure space this measure space can be extended to a complete measure space $(X,{\overline {\mathcal {M}}},{\overline {\mu }})$. The measure ${\overline {\mu }}$ is called the completion of $\mu$, and ${\overline {\mathcal {M}}}$ is called the completion of ${\mathcal {M}}$ with respect to $\mu$.

Theorem Suppose that $(X,{\mathcal {M}},\mu )$ is a measure space. Let ${\mathcal {N}}=\left\{N\in {\mathcal {M}}:\mu (N)=0\right\}$ and ${\overline {\mathcal {M}}}=\left\{E\cup F:E\in {\mathcal {M}}{\text{ and }}F\subseteq N{\text{ for some }}N\in {\mathcal {N}}\right\}$. Then, ${\overline {\mathcal {M}}}$ is a $\sigma$-algebra, and there is a unique extension ${\overline {\mu }}$ of $\mu$ to a complate measure on ${\overline {\mathcal {M}}}$.

Borel Measures and Lebesgue Measures

A measure whose domain is the Borel $\sigma$-algebra ${\mathcal {B}}_{\mathbb {R} }$ is called a Borel measure on $\mathbb {R}$. The following theorem provides a method for constructing Borel measures.

Theorem If $F:\mathbb {R} \rightarrow \mathbb {R}$ is any increasing, right continuous function, there is a unique Borel measure $\mu _{F}$ on $\mathbb {R}$ such that $\mu _{F}((a,b])=F(b)-F(a)$, for all $a,b\in \mathbb {R}$. If $G:\mathbb {R} \rightarrow \mathbb {R}$ is another such function, we have $\mu _{F}=\mu _{G}$ if and only if $F-G$ is constant. Conversely, if $\mu$ is a Borel measure on $\mathbb {R}$ that is finite on all bounded Borel sets and we define $F(x)={\begin{cases}\mu ((0,x]),x>0\\0,x=0\\-\mu ((x,0]),x<0\end{cases}}$, then $F$ is increasing and right continuous, and $\mu =\mu _{F}$.

A few things should be noted regarding the previous theorem. The $(a,b]$ intervals can be replaced by intervals of the form $[a,b)$; in this case, the function $F$ would have to be left continuous. Additionally, the completion of $\mu _{F}$, ${\overline {\mu _{F}}}$, is known as the Lebesgue-Stieljes measure; this complete measure has a domain that is strictly greater than the ${\mathcal {B}}_{\mathbb {R} }$. Finally, taking $F(x)=x$ gives rise to the Lebesgue measure.

References

1. Folland, Gerald B. (1999). Real Analysis: Modern Techniques and Their Applications, John Wiley and Sons, ISBN 0471317160, Second edition.
2. Craig, Katy. MATH 201A Lectures 4-5, 7-8. UC Santa Barbara, Fall 2020.