2 Measures

measure space

Let $(X,\mathcal{A})$ be a measurable space, where $X$ is a set and $\mathcal{A}$ is a $\sigma$-algebra on $X$. A measure on $(X,\mathcal{A})$ is a function $\mu :\mathcal{A}\to [0,\infty ]$ such that

• $\mu (\varnothing )=0$;
• if $\{A_i{\}}_{i=1}^{\infty }\subseteq \mathcal{A}$ are pairwise disjoint, then

$$\mu \left({\cup }_{i=1}^{\infty }{A}_i\right)=\sum _{i=1}^{\infty }\mu \left(A_i\right)$$

The triple $(X,\mathcal{A},\mu )$ is called a measure space.

Examples.

• Let $X$ be any set, $\mathcal{A}$ the collection of all subsets of $X$ and let $\mu (A)=$ the number of elements in $A$ if $A$ is finite and $\mu (A)=\infty$ if $A$ is infinite. Then $\mu$ is called counting measure.
• Let ${\delta }_x(A)=1$ if $x\in A$ and $0$ otherwise. This measure is called Dirac measure at $x$.
• Let $X$ be an uncountable set and $\mathcal{C}$ the collection of those subsets that are either countable or the complement of a countable set. Define $\mu :\mathcal{C}\to \{0,1\}$ by $\mu (A)=0$ if $A$ is countable and $\mu (A)=1$ if $A$ is uncountable. Then $(X,\mathcal{C},\mu )$ is a measure space.
• Let $\mathcal{L}$ be the collection of Lebesgue measurable sets of ${\mathbb{R}}^n$ and $m$ be the Lebesgue measure; the $({\mathbb{R}}^n,\mathcal{L},m)$ is a measure space. Also, $({\mathbb{R}}^n,\mathcal{B}({\mathbb{R}}^n),m)$ is a measure space.
• Let $X$ be a set that has at least two members, and let $\mathcal{A}=\mathcal{P}(X)$. Define a function $\mu :\mathcal{A}\to [0,\infty ]$ by letting $\mu (A)$ be $1$ if $A\ne \varnothing$ and letting $\mu (A)$ be $0$ if $A=\varnothing$. Then $\mu$ is not a measure.

Lemma

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

• (Monotonicity) If $A,B\in \mathcal{A}$ with $A\subset B$, then $\mu (A)⩽\mu (B)$;
• (Subadditivity) If $A_i\in \mathcal{A}$ and $A={\cup }_{i=1}^{\infty }{A}_i$, then $\mu (A)⩽{\sum }_{i=1}^{\infty }\mu \left(A_i\right)$;
• (Continuity from below) If $A_i\in \mathcal{A}$ and $A_i\uparrow A$, then $\mu (A)={\lim }_{i\to \infty }\mu \left(A_i\right)$;
• (Continuity from above) Let $A_i\in \mathcal{A}$ and $A_i\downarrow A$. If $\mu \left(A_1\right)<\infty$, then

$$\mu (A)=\underset{i\to \infty }{\lim }\mu \left(A_i\right)$$

\begin{proof} i) $B=A\cup (B-A)$. Then $\mu (B)=\mu (A)+\mu (B-A)⩾\mu (A)$.

ii) Let $B_1=A_1$, and let $B_i=A_i-{\cup }_{j=1}^{i-1}{A}_j$. Then $\{B_i{\}}_{i=1}^{\infty }$ are disjoint, and so $\mu ({\cup }_{i=1}^{\infty }{A}_i)=\mu ({\cup }_{i=1}^{\infty }{B}_i)={\sum }_{i=1}^{\infty }\mu (B_i)⩽{\sum }_{i=1}^{\infty }\mu (A_i)$.

iii) Let $A_0=\varnothing$. Then $\mu (A)={\sum }_{i=1}^{\infty }\mu (A_i-A_{i-1})={\lim }_{n\to \infty }\mu (A_n)-\mu (A_0)={\lim }_{n\to \infty }\mu (A_n)$.

iv) Let $B_n=A_1-A_n$. Since $B_n\uparrow A_1-A$, we have $\mu (A_1)-\mu (A)=\mu (A_1-A)={\lim }_{n\to \infty }\mu (B_n)=\mu (A_1)-{\lim }_{n\to \infty }\mu (A_n)$. Therefore, $\mu (A)={\lim }_{i\to \infty }\mu \left(A_i\right)$. \end{proof}

Remark. In iv), $\mu (A_1)<\infty$ is necessary. Consider

• $[1,\infty )\supset [2,\infty )\supset \cdots \supset [n,\infty )\cdots$ and
• $A_i=\{i,i+1,\cdots \}$.

Definition

A measure $\mu$ is a finite measure if $\mu (X)<\infty$.

A measure $\mu$ is $\sigma$-finite if there exist sets $E_i\in \mathcal{A}$ for $i=1,2,\dots$ such that $\mu \left(E_i\right)<\infty$ for each $i$ and $X={\cup }_{i=1}^{\infty }{E}_i$.

If $\mu$ is a finite measure, then $(X,\mathcal{A},\mu )$ is called a finite measure space, and similarly, if $\mu$ is a $\sigma$-finite measure, then $(X,\mathcal{A},\mu )$ is called a $\sigma$-finite measure space.

almost everywhere holds

For a measure space $(X,\mathcal{A},\mu )$ and a measurable subset $E$ of $X$, we say that a property holds almost everywhere on $E$, or it holds for almost all $x$ in $E$, provided it holds on $E-E_0$, where $E_0$ is a measurable subset of $E$ for which $\mu \left(E_0\right)=0$.

Borel-Cantelli

Let $(X,\mathcal{A},\mu )$ be a measure space and ${\left\{E_k\right\}}_{k=1}^{\infty }$ a countable collection of measurable sets for which ${\sum }_{i=1}^{\infty }\mu \left(E_k\right)<\infty$. Then almost all $x$ in $X$ belong to at most a finite number of the $E_k$ ‘s.

\begin{proof} Let $D=\{x:x\text{mbox}belongstoatmostfinitenumberofthe{E}_k\text{mbox}{}^{\prime }s\}$. It suffices to show that $\mu (D^c)=0$. Note that $D^c=\{x:x\text{mbox}belongstoinfinitelymany{E}_k\text{mbox}{}^{\prime }s\}=\lim {\sup }_{n\to \infty }{E}_n$. So we have $\mu (D^c)=\mu ({\cap }_{n=1}^{\infty }{\cup }_{k=n}^{\infty }{E}_k)⩽\mu ({\cap }_{k=n}^{\infty }{E}_k)⩽{\sum }_{k=n}^{\infty }\mu (E_k)=0$. \end{proof}

complete

A measure space $(X,\mathcal{A},\mu )$ is said to be complete if

$$N\in \mathcal{A},\mu (N)=0,E\subset N\Rightarrow E\in \mathcal{A}.$$

Example. Let $X=\{a,b,c\}$. Then $\mathcal{A}=\{\varnothing ,\{a\},\{b,c\},X\}$ is a $\sigma$-algebra of subsets of $X$. Define function $\mu$ on $\mathcal{A}$ by $\mu (\varnothing )=0,\mu (\{a\})=$ $1,\mu (\{b,c\})=0$, and $\mu (X)=1$, then $\mu$ is a measure on $\mathcal{A}$. The set $\{b,c\}$ is a null set in the measure space $(X,\mathcal{A},\mu )$, but its subset $\{b\}$ is not a member of $\mathcal{A}$. Therefore $(X,\mathcal{A},\mu )$ is not a complete measure space.

Theorem

Let $(X,\mathcal{A},\mu )$ be a measure space and define

$${\mathcal{A}}^{\ast }=\{E\subseteq X:\exists A,B\in \mathcal{A}\text{ such that }A\subseteq E\subseteq B\text{ and }\mu (B-A)=0\}.$$

Then the following holds.

• ${\mathcal{A}}^{\ast }$ is a $\sigma$-algebra and $\mathcal{A}\subset {\mathcal{A}}^{\ast }$.
• There exists a unique measure ${\mu }^{\ast }$ on ${\mathcal{A}}^{\ast }$ such that ${{\mu }^{\ast }|}_{\mathcal{A}}=\mu$.
• The triple $\left(X,{\mathcal{A}}^{\ast },{\mu }^{\ast }\right)$ is a complete measure space. It is called the completion of $(X,\mathcal{A},\mu )$.

\begin{proof} i) Note that $\varnothing ,X\in {\mathcal{A}}^{\ast }$. Let $E\in {\mathcal{A}}^{\ast }$. Then there exists $A\subseteq E\subseteq B$ such that $\mu (B-A)=0$. Then $A^c\supseteq E^c\supseteq B^c$ and $\mu (A^c-B^c)=\mu (A^c\cap B)=\mu (B-A)=0$. So $E^c\in {\mathcal{A}}^{\ast }$. Let $\{E_n\}\subseteq {\mathcal{A}}^{\ast }$, and let $A_n\subseteq E_n\subseteq B_n$ with $\mu (B_n-A_n)=0$. Let $A={\cup }_{n=1}^{\infty }{A}_n$ and $B={\cup }_{n=1}^{\infty }{B}_n$. Then $\mu (B-A)⩽{\sum }_{i=1}^{\infty }\mu (B_n-A_n)=0$ and $A\subseteq E={\cup }_{n=1}^{\infty }{E}_n\subseteq B$. Therefore, $E\in {\mathcal{A}}^{\ast }$.

ii) For any $E\in {\mathcal{A}}^{\ast }$, we have $A\subset E\subset B$ and $\mu (B-A)=0$. If ${\mu }^{\ast }$ is defined, then ${\mu }^{\ast }(E)=\mu (A)$. Define ${\mu }^{\ast }(E)=\mu (A)$ if $A\subset E\subset B$. Claim that ${\mu }^{\ast }$ is well defined. Otherwise, assume that $A_1\subset E\subset B_1$ and $A_2\subset E\subset B_2$ with $\mu (B_i-A_i)=0$. Then $\mu (A_1)⩽\mu (B_2)=\mu (A_2)⩽\mu (B_1)=\mu (A_1)$, i.e., $\mu (A_1)=\mu (A_2)$. If $\{E_i\}\subseteq {\mathcal{A}}^{\ast }$ with $E_i\cap E_j=\varnothing$, then there exist $A_i\subseteq E_i\subseteq B_i$ such that $\mu (B_i-A_i)=\varnothing$. Let $A={\cup }_{i=1}^{\infty }{A}_i$, and let $B={\cup }_{i=1}^{\infty }{B}_i$. Then $B-A\subseteq {\cup }_{i=1}^{\infty }(B_i-A_i)$ and so $\mu (B-A)=0$. It follows that ${\mu }^{\ast }({\cup }_{i=1}^{\infty }{E}_i)={\mu }^{\ast }(A)=\mu (A)={\sum }_{i=1}^{\infty }{\mu }^{\ast }(E_i)$. Therefore, ${\mu }^{\ast }$ is a measure.

iii) If $N\in {\mathcal{A}}^{\ast }$ and ${\mu }^{\ast }(N)=0$, then there exists $A,B$ such that $A\subseteq N\subseteq B$ such that $\mu (B-A)=0$. Since $\mu (A)=0$, we have $\mu (B)=0$. For any $M\subseteq N$, there is $\varnothing \subseteq M\subseteq B$ and $\mu (B-\varnothing )=0$. Therefore, $M\in {\mathcal{A}}^{\ast }$ and we finish the proof. \end{proof}

Definition

A probability or probability measure is a measure $\mu$ such that $\mu (X)=1$.