Positive Measures

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Definition (Positive measure)

Let [math](E,\A)[/math] be a measurable space. A positive measure [math]\mu[/math] on [math](E,\mathcal{A})[/math] is an application [math]\mu:\mathcal{A}\longrightarrow[0,\infty]\subset\overline{\mathbb{R}}[/math], which satisfies the following.

• (measure of the empty set is zero) [math]\mu(\varnothing)=0[/math],
• ([math]\sigma[/math]-additivity) For all sequences [math](A_n)_{n\in\mathbb{N}}\in\mathcal{A}[/math] of disjoint measurable sets, that is [math]A_i\cap A_j=\varnothing[/math] for [math]i\not=j[/math], we have
  [[math]] \mu\left(\bigcup_{n\in\N}A_n \right)=\sum_{n\in\N}\mu(A_n). [[/math]]
  Moreover, we call a triple [math](E,\A,\mu)[/math], that is a measurable space endowed with a specific measure, a measure space.

A nice observation of [math]\bar\R_+[/math] is that all sums are convergent, that is, for any sequence [math](x_n)_{n\in\N}\subset[0,\infty][/math], we get [math]\sum_{n\in\N}x_n\in[0,\infty][/math]. We can formulate an equivalent definition of this sum as

[[math]] \sum_{n\in\N}x_n=\sup_{I\subset\N\atop \text{$I$ finite}}\sum_{n\in I}x_n. [[/math]]

More general, for any sequence [math](x_\alpha)_{\alpha\in A}[/math] of real numbers [math]x_\alpha\in[0,\infty][/math] with an arbitrary index set [math]A[/math], either countable or uncountable, we can define

[[math]] \sum_{\alpha\in A}x_\alpha:=\sup_{I\subset A\atop\text{$A$ finite}}\sum_{\alpha\in I}x_\alpha. [[/math]]

Moreover, for any two index sets [math]A[/math] and [math]B[/math] and for any bijection [math]\phi:B\xrightarrow{\sim}A[/math], we get

[[math]] \sum_{\alpha\in A}x_\alpha=\sum_{\beta\in B}x_{\phi(\beta)}. [[/math]]

Proposition

Let [math](E,\mathcal{A},\mu)[/math] be a measure space. Then the following hold.

• Let [math]A,B\in\A[/math] be two measurable sets such that [math]A\subset B[/math]. Then [math]\mu(A)\leq \mu(B)[/math]. Moreover, if [math]\mu(A) \lt \infty[/math], then [math]\mu(B\setminus A)=\mu(B)-\mu(A)[/math].
• (Inclusion-exclusion) Let [math]A,B\in \A[/math] be two measurable sets. Then [math]\mu(A)+\mu(B)=\mu(A\cup B)+\mu(A\cap B)[/math].
• Let [math](A_n)_{n\in\N}\subset\A[/math] be an increasing sequence of measurable sets. Then
  [[math]] \mu\left(\bigcup_{n\in\mathbb{N}}A_n\right)=\lim_{n\to\infty}\uparrow\mu(A_n). [[/math]]
• Let [math](B_n)_{n\in\N}\subset\A[/math] be a decreasing sequence of measurable sets. Moreover, let [math]\mu(B_0) \lt \infty[/math]. Then
  [[math]] \mu\left(\bigcap_{n\in\N}B_n\right)=\lim_{n\to\infty}\downarrow\mu(B_n). [[/math]]
• ([math]\sigma[/math]-subadditivity) Let [math](A_n)_{n\in\N}\subset\A[/math] be a sequence of measurable sets. Then
  [[math]] \mu\left(\bigcup_{n\in\mathbb{N}}A_n\right)\leq \sum_{n\in\mathbb{N}}\mu(A_n). [[/math]]

Show Proof

For [math](i)[/math], observe that [math]B=(B\setminus A)\sqcup A[/math] and thus [math]\mu(B)=\mu(B\setminus A)+\mu(A)[/math]. Moreover, if [math]\mu(A) \lt \infty[/math], then [math]\mu(B\setminus A)=\mu(B)-\mu(A)[/math]. For [math](ii)[/math], observe that [math]A\cup B= (A\setminus B)\sqcup (B\setminus A)\sqcup(A\cap B)[/math] and thus [math]\mu(A\cup B)=\mu(A\setminus B)+\mu(B\setminus A)+\mu(A\cap B)[/math]. Assume [math]\mu(A\cap B)=\infty[/math], then we get [math](ii)[/math], since [math]\mu(A\cup B),\mu(A),\mu(B)\geq \mu(A\cap B)=\infty[/math]. On the other hand, assume [math]\mu(A\cap B) \lt \infty[/math], then [math]\mu(A\setminus B)=\mu(A\cap B^C)[/math] and [math]\mu(B\setminus A)=\mu(B\cap A^C)[/math], which implies that [math]\mu(A\cup B)=\mu(A)+\mu(B)-2\mu(A\cap B)+\mu(A\cap B)[/math]. Rearranging things, we get the claim. For [math](iii)[/math], let [math]C_0=A_0[/math] and [math]C_n=A_n\setminus A_{n-1}[/math] for all [math]n\geq 1[/math]. Then we get

[[math]] \bigcup_{n\in\N}C_n=\bigcup_{n\in\N}A_n, [[/math]]

where the [math]C_n[/math]'s are disjoint and moreover, [math]\mu(C_n)=\mu(A_n)-\mu(A_{n-1})[/math]. Therefore we get

[[math]] \mu\left(\bigcup_{n\in\N}A_n\right)=\mu\left(\bigcup_{n\in\N}C_n\right)=\sum_{n\in\N}\mu(C_n)=\lim_{N\to\infty}\sum_{n=0}^N\mu(C_n)=\lim_{N\to\infty}\uparrow\mu(A_N). [[/math]]

For [math](iv)[/math], let [math]A_n=B_0\setminus B_1[/math], which implies that [math]A_n\subset A_{n+1}[/math] for all [math]n\in \N[/math]. Thus we get

[[math]] \begin{multline*} \mu(A_n)=\mu(B_0)-\mu\left(\bigcup_{n\in\N}B_n\right)=\mu\left(b_0\setminus \bigcup_{n\in\N}B_n\right)\\=\mu\left(\bigcup_{n\in\N}A_n\right)=\lim_{n\to\infty}\uparrow \mu(A_n)=\lim_{n\to\infty}(\mu(B_0)-\mu(B_n)). \end{multline*} [[/math]]

Now, since [math]\mu(B_0) \lt \infty[/math], we get that

[[math]] \mu\left(\bigcap_{n\in\N}B_n\right)=\lim_{n\to\infty}\downarrow\mu(B_n), [[/math]]

where we have used the fact that

[[math]] \bigcup_{n\in\N}A_n=\bigcup_{n\in\N}(B_0\setminus B_n)=\bigcup_{n\in\N}(B_0\cap B_n^C)=B_0\cap\left(\bigcup_{n\in\N}B_n^C\right)=B_0\cap\left(\bigcap_{n\in\N}B_n\right)^C=B_0\setminus\bigcap_{n\in\N}B_n, [[/math]]

and that [math]\bigcap_{n\in\N}\subseteq B_0[/math]. For [math](v)[/math], let [math]C_0=A_0[/math] and for [math]n\geq 1[/math] let

[[math]] C_n=A_n\setminus\bigcup_{k=0}^{n-1}A_k, [[/math]]

where the [math]C_n[/math]'s are disjoint and moreover, [math]\bigcup_{n\in\N}A_n=\bigcup_{n\in\N}C_n\subset A_n[/math]. Therefore we get

[[math]] \mu\left(\bigcup_{n\in\N}A_n\right)=\mu\left(\bigcup_{n\in\N}C_n\right)=\sum_{n\in\N}\mu(C_n)\leq \sum_{n\in\N}\mu(A_n). [[/math]]

This can also been proved by induction, which we leave as an exercise for the reader.

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Example

[Dirac measure] Let [math](E,\A)[/math] be a measurable space such that for any [math]x\in E[/math], we get that [math]\{ x\}\in\A[/math]. we can define a measure [math]\delta:E\times \A\longrightarrow \{1,0\}[/math] by

[[math]] \delta_x(A)=\one_A(x)=\begin{cases}1,&x\in A\\ 0,& x\not\in A\end{cases} [[/math]]

This measure is called the Dirac measure or the Dirac mass at [math]x[/math]. More generally, if we consider sequences [math](x_n)_{n\in\N}\subset E[/math] and [math](\alpha_n)_{n\in\N}\subset [0,\infty][/math], we can define a measure [math]\mathscr{D}_{(x_n)}^{(\alpha_n)}:\A\longrightarrow \bar\R_+[/math], which is defined by

[[math]] \mathscr{D}_{(x_n)}^{(\alpha_n)}(A)=\left(\sum_{n\in\N}\alpha_n\delta_{x_n}\right)=\sum_{n\in\N}\alpha_n\delta_{x_n}(A). [[/math]]

Example

[Lebesgue measure]

There exists a unique measure on the measurable space [math](\mathbb{R},\mathcal{B}(\mathbb{R}))[/math], which is denoted by [math]\lambda[/math], such that for all open intervals [math](a,b)\in \B(\R)[/math] it is given by

[[math]] \lambda((a,b))=b-a. [[/math]]

Definition (Finite-, [math]\sigma[/math]-finite- and probability measures)

Let [math](E,\A)[/math] be a measurable space. We say that a measure [math]\mu[/math] is

• finite if [math]\mu(E) \lt \infty[/math].
• a probability measure if [math]\mu(E)=1[/math].
• [math]\sigma[/math]-finite if there exists an increasing sequence (partition of the total space) [math](E_n)_{n\in\N}\subset\mathcal{A}[/math], such that [math]E=\bigcup_{n\in\N}E_n[/math] and with [math]\mu(E_n) \lt \infty[/math] for all [math]n\in\N[/math].

Definition (Atom)

Let [math](E,\A,\mu)[/math] be a measure space. An element [math]x\in E[/math] is called an atom for [math]\mu[/math] if the set [math]\{x\}\in\A[/math] and

[[math]] \mu(\{x\}) \gt 0. [[/math]]

Definition (Product [math]\sigma[/math]-Algebra)

Let [math](E_1,\mathcal{A}_1)[/math] and [math](E_2,\mathcal{A}_2)[/math] be two measurable spaces. Then we can define the product [math]\sigma[/math]-Algebra [math]\A_1\otimes\A_2[/math] on the product space [math]E_1\times E_2[/math] by

[[math]] \A_1\otimes\A_2=\sigma(A_1\times A_2), [[/math]]

where [math]A_1\in\A_1[/math] and [math]A_2\in\A_2[/math]. This is actually the [math]\sigma[/math]-Algebra which contains all sets of the form [math](A_1\times A_2)[/math].

Let [math](E,\A)[/math] and [math](F,\B)[/math] be two measure spaces. Consider a map [math]f:E\longrightarrow F[/math]. Moreover, let [math]I[/math] be an arbitrary index set and for [math]i\in I[/math], let [math]A_i\subset E[/math] and [math]B_i\subset F[/math]. We can write, for [math]A\subset E[/math]

[[math]] f(A):=\{ f(x)\mid x\in A\}, [[/math]]

and similarly, for [math]B\subset F[/math], we can write

[[math]] f^{-1}(B)=\{ x\in E\mid f(x)\in B\}. [[/math]]

Moreover, it is easy to observe the following relations.

• [math]f\left(\bigcup_{i\in I}A_i\right)=\bigcup_{i\in I}f(A_i).[/math]
• [math]f\left(\bigcap_{i\in I}A_i\right)\subseteq \bigcap_{i\in I}f(A_i). \text{(where equality holds if $f$ is injective)}[/math]
• [math]f^{-1}\left(\bigcup_{i\in I}B_i\right)=\bigcup_{i\in I}f^{-1}(B_i).[/math]
• [math]f^{-1}\left(\bigcap_{i\in I}B_i\right)=\bigcap_{i\in I}f^{-1}(B_i).[/math]
• [math]f^{-1}(B)^C=f^{-1}(B)^C.[/math]
• If [math]\mathcal{C}\subset \mathcal{P}(F)[/math], then [math]f^{-1}(\mathcal{C}):=f^{-1}(C)\mid C\in\mathcal{C}\}[/math].

Proposition

Let [math]E[/math] and [math]F[/math] be two measurable spaces, where [math]\mathcal{B}[/math] is a [math]\sigma[/math]-Algebra on [math]F[/math]. Then

[[math]] \A:=f^{-1}(\B)=\{f^{-1}(B)\mid B\in \B\} [[/math]]

is a [math]\sigma[/math]-Algebra on [math]E[/math].

Show Proof

First, it is obvious that [math]f^{-1}(F)=E[/math], which implies that if [math]F\in\B[/math] then [math]E\in \A[/math]. Moreover, it holds that

[[math]] f^{-1}(B)^C=f^{-1}(B^C)\in\A [[/math]]

for all [math]B\in\B[/math] since arbitrary unions of elements in [math]\B[/math] are again in [math]\B[/math].

■

It is sometimes usual to write [math]\sigma(f)[/math] instead of [math]f^{-1}(B)[/math].

Example

Let [math](E,\A)[/math] be a measurable space and let [math]F\subset E[/math] be a subset of [math]E[/math]. Moreover, let [math]\iota:F\hookrightarrow (E,\A)[/math] be the canonical injection. Then we get

[[math]] \iota^{-1}(\A)=\{\iota^{-1}(A)\mid A\in\A\}=\{F\cap A\mid A\in\A\}. [[/math]]

Example

Let [math](E,\A)[/math] be a measurable space and let [math]F\subset E[/math] be a subset of [math]E[/math]. Moreover, let [math]\pi_E:E\times F\longrightarrow (E,\A)[/math] be the canonical projection. Then we get

[[math]] \pi_E^{-1}(\A)=\{\pi_E^{-1}(A)\mid A\in\A\}=\{A\times F\mid A\in\A\}. [[/math]]

Definition (Image [math]\sigma[/math]-Algebra)

Let [math](E.\A)[/math] and [math](F,\B)[/math] be measurable spaces and let [math]f:E\longrightarrow F[/math] be a map. The image [math]\sigma[/math]-Algebra of [math]\A[/math] by [math]f[/math] is defined by

[[math]] \mathcal{I}=\{I\in\mathcal{P}(F)\mid f^{-1}(I)\in\A\}. [[/math]]

Proposition

Let [math](X,d)[/math] be a metric space and let [math]Y\subset X[/math]. Then the Borel [math]\sigma[/math]-Algebra of [math]Y[/math] is given by

[[math]] \B(Y)=\{A\cap Y\mid A\in\B(X)\}. [[/math]]

Moreover, if [math]Y\in\B(X)[/math] then [math]\B(Y)\subset \B(X)[/math] and [math]\B(Y)=\{A\in\B(X)\mid A\subset Y\}[/math].

Show Proof

Let [math]\iota:Y\hookrightarrow X[/math] be the canonical injection of [math]Y[/math] into [math]X[/math]. Then

[[math]] \mathcal{O}_Y:=\{O\cap Y\mid O\in\mathcal{O}_X\}=\iota^{-1}(\mathcal{O}_X). [[/math]]

Moreover, we get

[[math]] \B(Y)=\sigma(\mathcal{O}_Y)=\sigma(\iota^{-1}(\mathcal{O}_X))=\iota^{-1}(\sigma(\mathcal{O}_X))=\iota^{-1}(\B(X))=\{A\cap Y\mid A\subset \B(X)\}, [[/math]]

which proves the first part of the proposition. The second part is easily obtained from the fact that [math]\sigma[/math]-Algebras are stable under finite intersections.

■

We have the following examples of Borel [math]\sigma[/math]-Algebras.

Example

Let [math]X=\R_+[/math]. Then

[[math]] \B(\R_+)=\{A\subseteq \R_+\mid A\in\B(\R)\}. [[/math]]

Example

Let [math]X=\R^\times=\R\setminus\{0\}[/math]. Then

[[math]] \B(\R^\times)=\{A\in\B(\R)\mid 0\not\in A\}. [[/math]]

Example

[Borel sets on [math]\bar \R[/math]] Let us define [math]\bar \R=\R\cup\{-\infty,\infty\}[/math] and let us consider the map

[[math]] \begin{align*} f:\R&\longrightarrow (-1,1)\\ x&\longmapsto \frac{x}{\sqrt{x^2+1}} \end{align*} [[/math]]

We can now consider an extension [math]\tilde f[/math] of [math]f[/math], which is defined on [math]\bar \R[/math] such that [math]\tilde f\mid_\R=f[/math] with [math]\tilde f(-\infty)=-1[/math] and [math]\tilde f(\infty)=1[/math]. Moreover, we can consider [math]\bar \R[/math] as a metric space by the considering the distance given for all [math]x,y\in\bar \R[/math] as [math]\|\tilde f(x)-\tilde f(y)\|[/math]. We write therefore [math](\bar\R,\delta)[/math] as a metric space with the metric [math]\delta[/math]. Thus we can define the Borel [math]\sigma[/math]-Algebra of [math]\bar \R[/math] by the Borel sets, which are described by the metric topology of [math]\bar \R[/math]. This concept is important as we will work many times with the space [math]\bar \R[/math].

It is useful to note that [math]\bar \R[/math] describes a totally ordered set, since [math]\leq [/math] arises with the usual naturalness as in [math]\R[/math]. Moreover, the identity map

[[math]] \begin{align*} id:(\R,\delta\mid_\R)&\longrightarrow (\R,\|\cdot\|)\\ x&\longmapsto x \end{align*} [[/math]]

is a homeomorphism. Another useful observation is that [math](\bar \R,\delta)[/math] is a compact space and homeomorphic to the interval [math][-1,1][/math] and eventually [math]\R[/math] is an open subset of [math]\bar\R[/math].

General references

Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].