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A very important notion is that of a monotone class. We will see that there are many things which can be deduced by using the monotone class lemma.

Definition (Monotone Class)

Let [math]E[/math] be some topological space and let [math]M\subset\mathcal{P}(E)[/math]. [math]M[/math] is called a monotone class if the following holds.

• [math]E\in M[/math].
• Let [math]A\in M[/math] and [math]B\in M[/math]. If [math]A\subset B[/math] [math]\Longrightarrow B\setminus A\in M[/math].
• Let [math](A_n)_{n\in\N}\in M[/math]. If [math]A_n\subset A_{n+1}[/math] [math]\Longrightarrow \bigcup_{n\in\mathbb{N}}A_n \in M[/math].

A [math]\sigma[/math]-Algebra is a monotone class

As for [math]\sigma[/math]-Algebras, we notice that an arbitrary intersection of monotone classes is again a monotone class. Thus, if [math]\mathcal{C}\in \mathcal{P}(E)[/math], we can define the monotone class generated by [math]\mathcal{C}[/math] as

[[math]] M(\mathcal{C})=\bigcap_{\mathcal{C}\subset M\atop M\text{mon.cl.}}M. [[/math]]

This is also by construction the smallest monotone class containing [math]\mathcal{C}[/math].

Theorem (Monotone Classes lemma)

Let [math]E[/math] be a topological space. If [math]\mathcal{C}\subset\mathcal{P}(E)[/math] is stable under finite intersection, i.e. for [math]A\in \mathcal{C}[/math] and [math]B\in\mathcal{C}\Longrightarrow A\cap B\in \mathcal{C}[/math], then

[[math]] \sigma(\mathcal{C})=M(\mathcal{C}). [[/math]]

Show Proof

It is obvious that, [math]M(\mathcal{C})\subset\sigma(\mathcal{C})[/math] since a [math]\sigma[/math]-Algebra is also a monotone class. Next we want to show that [math]M(\mathcal{C})[/math] is a [math]\sigma[/math]-Algebra to conclude that [math]\sigma(\mathcal{C})\subset M(\mathcal{C})[/math] and hence then [math]M(\mathcal{C})[/math] contains [math]\mathcal{C}[/math], i.e. [math]\sigma(\mathcal{C})[/math]. It is not difficult to see that a monotone class, which is stable under finite intersection, is a [math]\sigma[/math]-Algebra. Let us therefore show that [math]M(\mathcal{C})[/math] is stable under finite intersections. First, we fix [math]A\in\mathcal{C}[/math] and define

[[math]] M_A:=\{B\in\mathcal{C}\mid A\cap B\in M(\mathcal{C})\}. [[/math]]

Then we get that [math]\mathcal{C}\in M_A[/math] since [math]\mathcal{C}[/math] is stable under finite intersections and obviously [math]E\in M_A[/math]. We can also note that If [math]B,B'\in M_A[/math] and [math]B\subset B'[/math], with

[[math]] A\cap (B'\setminus B)=\underbrace{\underbrace{(A\cap B')}_{\in M(\mathcal{C})}\setminus \underbrace{(A\cap B)}_{\in M(\mathcal{C})}}_{\in M(\mathcal{C})} [[/math]]

then [math]B'\setminus B\in M_A[/math]. Moreover, if [math](B_n)_{n\in\N}\in M_A[/math] and [math]A\cap\bigcup_n\underbrace{(A\cap B_n)}_{\in M(\mathcal{C})}[/math] we get the implication

[[math]] \bigcup_{n\geq 1}(A\cap B_n)\in M(\mathcal{C})\Longrightarrow \bigcup_{n\geq 1}B_n\in M_A. [[/math]]

since [math](A\cap B_n)[/math] is increasing. Finally we can conclude the above facts. That means if [math]M_A[/math] is a monotone class containing [math]\mathcal{C}[/math], then [math]M_A=M(\mathcal{C})[/math], which shows that for all [math]A\in\mathcal{C}[/math] and [math]B\in M(\mathcal{C})[/math] we get [math]A\cap B\in M(\mathcal{C})[/math]. We can now apply the same idea another time. Fix [math]B\in M(\mathcal{C})[/math] and define

[[math]] M:=\{A\in M(\mathcal{C})\mid A\cap B\in M(\mathcal{C})\}. [[/math]]

Now, from above, we get that [math]M[/math] is a monotone class, i.e. [math]M\subset M(\mathcal{C})[/math] and thus for all [math]A,B\in M(\mathcal{C})[/math] we get [math]A\cap B\in M(\mathcal{C})[/math]. Hence it follows that [math]M(\mathcal{C})[/math] is stable under finite intersections and is therefore a [math]\sigma[/math]-Algebra.

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Corollary

Let [math](E,\A)[/math] be a measurable space and let [math]\mu,\nu[/math] be two measures on [math](E,\mathcal{A})[/math]. Moreover, assume that there exists a family of subsets [math]\mathcal{C}[/math], which is stable under finite intersections, such that [math]\sigma(\mathcal{C})=\mathcal{A}[/math] and [math]\mu(A)=\nu(A)[/math] for all [math]A\in\mathcal{C}[/math]. Then the following hold.

• If [math]\mu(E)=\nu(E) \lt \infty[/math], then we get [math]\mu=\nu[/math].
• If there exists an increasing family [math](E_n)_{n\in\N}[/math] with [math]E_n\in\mathcal{C}[/math] such that
  [[math]] E=\bigcup_{n\in\N}E_n [[/math]]
  and [math]\mu(E_n)=\nu(E_n) \lt \infty[/math], then it follows that [math]\mu=\nu[/math].

Show Proof

Let us first define the set [math]G:=\{A\in\A\mid \mu(A)=\nu(A)\}[/math]. By assumption we get that [math]\mathcal{C}\subset G[/math]. Moreover, we note that [math]G[/math] is a monotone class. Note at first that [math]E\in G[/math] by assumption since [math]\mu(E)=\nu(E)[/math]. Now let [math]A,B\in G[/math] such that [math]A\subset B[/math] and since

[[math]] \mu(B\setminus A)=\mu(B)-\mu(A)=\nu(B)-\nu(A)=\nu(B\setminus A), [[/math]]

we get that [math](B\setminus A)\in G[/math]. Now let [math](A_n)_{n\in\N}[/math] be an increasing sequence in [math]G[/math]. Then the fact

[[math]] \mu\left(\bigcup_{n\in\N}A_n\right)=\lim_{n\to\infty}\mu(A_n)=\lim_{n\to\infty}\nu(A_n)=\nu\left(\bigcup_{n\in\N}A_n\right) [[/math]]

implies that [math]\bigcup_{n\in\N}A_n\in G[/math]. Moreover, since [math]G[/math] is a monotone class containing [math]\mathcal{C}[/math], we get that [math]G[/math] contains [math]M(\mathcal{C})[/math]. On the other hand we know that [math]\mathcal{C}[/math] is stable under finite intersections and therefore it follows that [math]M(\mathcal{C})=\sigma(\mathcal{C})[/math] and that [math]G=\sigma(\mathcal{C})=\A[/math]. Now define for all [math]n\in\N[/math] and [math]A\in \A[/math] the two sequences

[[math]] \begin{align*} \mu_n(A)&=\mu(A\cap E_n)\\ \nu_n(A)&=\nu(A\cap E_n) \end{align*} [[/math]]

Now since [math]\mu(E)=\nu(E)[/math], we get the same for the sequence elements and obtain therefore that [math]\mu_n=\nu_n[/math]. Moreover, for [math]A\in \A[/math], we have

[[math]] \mu(A)=\lim_{n\to\infty}\mu(A\cap E_n)=\lim_{n\to\infty}\uparrow\nu(A\cap E_n)=\nu(A). [[/math]]

Hence we get that [math]\mu=\nu[/math].

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There are several applications of this corollary. Let us emphasize a first one, by giving already a small introduction to the Lebesgue measure. Assume that [math]\lambda[/math] is a measure on the measurable space [math](\R,\B(\R))[/math] such that [math]\lambda((a,b))=b-a[/math] for [math]a \lt b[/math] and let [math]\mathcal{C}[/math] be the class of intervals [math]E_n=(-n,n)[/math] for [math]n\geq 1[/math]. With the corollary above, it follows that [math]\lambda[/math] is unique. We will call [math]\lambda[/math] the Lebesgue measure. A second application is that a finite measure on [math](\R,\B(\R))[/math] is uniquely characterized, for [math]a\in\R[/math], by the values

[[math]] \mu((-\infty,a])=\mu((-\infty,a)). [[/math]]

General references

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