# Almost Sure

## 27 October 19

### The Functional Monotone Class Theorem

Filed under: Probability Theory — George Lowther @ 8:29 PM
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The monotone class theorem is a very helpful and frequently used tool in measure theory. As measurable functions are a rather general construct, and can be difficult to describe explicitly, it is common to prove results by initially considering just a very simple class of functions. For example, we would start by looking at continuous or piecewise constant functions. Then, the monotone class theorem is used to extend to arbitrary measurable functions. There are different, but related, monotone class theorems’ which apply, respectively, to sets and to functions. As the theorem for sets was covered in a previous post, this entry will be concerned with the functional version. In fact, even for the functional version, there are various similar, but slightly different, statements of the monotone class theorem. In practice, it is beneficial to use the version which most directly applies to the specific application. So, I will state and prove several different versions in this post.

Before going any further, we establish some notation. For a set ${E}$, use ${B(E)}$ to denote the collection of all bounded functions from ${E}$ to the real numbers ${{\mathbb R}}$. This is a vector space, since it is closed under taking linear combinations: ${af+bg\in B(E)}$ for ${f,g\in B(E)}$ and real numbers ${a,b}$. Furthermore, ${B(E)}$ is closed under multiplication, so that it contains the product ${fg}$ for all ${f}$ and ${g}$ in ${B(E)}$. In fact, ${B(E)}$ is an algebra, meaning that it is a vector space, is closed under multiplication, and contains the multiplicative identity ${1}$. Another property of ${B(E)}$ that is of importance here is that it is closed under bounded increasing limits and under bounded decreasing limits. A sequence ${f_1,f_2,\ldots\in B(E)}$ is bounded if ${\lvert f_n\rvert\le K}$ for all ${n}$ and some fixed ${K\in{\mathbb R}}$. By saying that ${B(E)}$ is closed under bounded increasing limits, we mean that for any such bounded sequence with ${f_{n+1}\ge f_n}$ for all ${n}$, then ${\lim_nf_n\in B(E)}$. Similarly, if the sequence is decreasing, so that ${f_{n+1}\le f_n}$ for all ${n}$, then ${\lim_nf_n\in B(E)}$, so ${B(E)}$ is closed under bounded decreasing limits.

For a ${\sigma}$-algebra ${\mathcal E}$ on a set ${E}$, we use ${\mathcal E_b}$ to denote the collection of all bounded and ${\mathcal E}$-measurable functions from ${E}$ to ${{\mathbb R}}$. Here, and throughout this post, for real-valued functions we use the standard Borel ${\sigma}$-algebra on the codomain ${{\mathbb R}}$. So, ${\mathcal E_b}$ consists of all ${f\in B(E)}$ such that ${f^{-1}(A)\in\mathcal E}$ for Borel sets ${A\subseteq{\mathbb R}}$. By standard properties of measurable functions, ${\mathcal E_b}$ is a vector subspace of ${B(E)}$, as well a sub-algebra and is closed under bounded increasing limits and bounded decreasing limits.

The ${\sigma}$-algebra generated by a collection of functions ${\mathcal K\subseteq B(E)}$ will be denoted by ${\sigma(\mathcal K)}$. This is the smallest ${\sigma}$-algebra on ${E}$ with respect to which every ${f\in\mathcal K}$ is measurable or, equivalently, is the smallest ${\sigma}$-algebra on ${E}$ satisfying ${\mathcal K\subseteq\sigma(\mathcal K)_b}$. Alternatively, ${\sigma(\mathcal K)}$ is the ${\sigma}$-algebra generated by ${f^{-1}(A)}$ over ${f\in\mathcal K}$ and Borel sets ${A\subseteq{\mathbb R}}$.

In all versions of the monotone class theorem, we consider a simple’ collection of real-vaued functions ${\mathcal K\subseteq B(E)}$ and a larger class of functions ${\mathcal H}$ containing ${\mathcal K}$, which satisfies some basic properties including closure under bounded increasing limits. The aim is to conclude that ${\mathcal H}$ contains all bounded ${\sigma(\mathcal K)}$-measurable functions. That is, ${\sigma(\mathcal K)_b\subseteq\mathcal H}$. In the first version of the theorem stated below, it is only required that the non-negative functions in ${\mathcal H}$,

$\displaystyle \mathcal H_+=\left\{f\in\mathcal H\colon f\ge0\right\}$

is closed under bounded increasing limits. In conjunction with the vector space property, it is not difficult to show that this is equivalent to ${\mathcal H}$ being closed under bounded increasing (and decreasing) limits. However, the weaker condition on ${\mathcal H_+}$ is sometimes useful, so the result is usually stated in this form. The proofs of all the versions of the monotone class theorem given here are included further down in the post.

Theorem 1 (Monotone Class Theorem) Let ${\mathcal K\subseteq B(E)}$ be closed under multiplication. Let ${\mathcal H}$ be a vector subspace of ${B(E)}$ satisfying

• ${\mathcal K\subseteq\mathcal H}$.
• ${1\in\mathcal H}$.
• ${\mathcal H_+}$ is closed under bounded increasing limits.

Then, ${\sigma(\mathcal K)_b\subseteq\mathcal H}$.

As an immediate application of the theorem, we can use it to show that a finite Borel measure on ${{\mathbb R}^+}$ is uniquely determined by its Laplace transform. Recall that the Laplace transform of a measure ${\mu}$ is a map,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle\mathcal L_\mu\colon{\mathbb R}^+\rightarrow{\mathbb R},\smallskip\\ &\displaystyle\mathcal L_\mu(a)=\int_0^\infty e^{-ax}\,d\mu(x). \end{array}$

The monotone class theorem quickly shows the uniqueness of measures with a given transform.

Lemma 2 Finite Borel measures on ${{\mathbb R}^+}$ are uniquely determined by their Laplace transforms.

Proof: We need to show that, for any two finite Borel measures ${\mu}$ and ${\nu}$ on ${{\mathbb R}^+}$ with the same Laplace transforms, then ${\mu=\nu}$. Let ${\mathcal H}$ be the space of bounded Borel measurable functions ${f\colon{\mathbb R}^+\rightarrow{\mathbb R}}$ such that ${\mu(f)=\nu(f)}$. By linearity of the integrals, ${\mathcal H}$ is a vector space and, by monotone convergence, ${\mathcal H}$ is closed under bounded increasing limits. Next, let ${\mathcal K}$ consist of the functions ${f_a\colon{\mathbb R}^+\rightarrow{\mathbb R}}$, for ${a\ge0}$, defined as ${f_a(x)=e^{-ax}}$. Using ${f_af_b=f_{a+b}}$, we see that ${\mathcal K}$ is closed under multiplication. Also, the constant function ${1=f_0}$ is in ${\mathcal K}$. Equality of Laplace transforms,

$\displaystyle \mu(f_a)=\mathcal L_\mu(a)=\mathcal L_\nu(a)=\nu(f_a),$

shows that ${\mathcal K\subseteq\mathcal H}$. By theorem 1, ${\mathcal H}$ contains all bounded ${\sigma(\mathcal K)}$-measurable functions from ${{\mathbb R}^+}$ to ${{\mathbb R}}$.

Finally, for any ${a > 0}$, the intervals ${[x,\infty)=f_a^{-1}([0,f_a(x)])}$ generate the Borel ${\sigma}$-algebra on ${{\mathbb R}^+}$. So, ${\sigma(\mathcal K)=\mathcal B({\mathbb R}^+)}$. Hence, every bounded Borel measurable ${f\colon{\mathbb R}^+\rightarrow{\mathbb R}}$ is in ${\mathcal H}$, giving ${\mu(f)=\nu(f)}$, and ${\mu=\nu}$ as required. ⬜

The version of the monotone class theorem stated above also has the immediate consequence that ${\sigma}$-algebras on a set ${E}$ are in one-to-one correspondence with subalgebras of ${B(E)}$ which are closed under bounded monotone convergence. This suggests that measure theory could alternatively be constructed on commutative algebras rather than ${\sigma}$-algebras of sets.

Lemma 3 For ${\mathcal H\subseteq B(E)}$, the following are equivalent,

1. ${\mathcal H=\mathcal E_b}$ for some ${\sigma}$-algebra ${\mathcal E}$ on ${E}$.
2. ${\mathcal H}$ is an algebra and ${\mathcal H_+}$ is closed under bounded increasing limits.

Proof: As mentioned further up, the facts that ${\mathcal E_b}$ is an algebra and is closed under bounded increasing limits are standard properties of measurable functions. We prove the converse, so suppose that the second statement holds. Taking ${\mathcal K=\mathcal H}$ and applying theorem 1 gives ${\sigma(\mathcal H)_b\subseteq\mathcal H}$. The reverse inequality holds from the definition of ${\sigma(\mathcal H)}$. So, ${\mathcal H=\mathcal E_b}$ with ${\mathcal E=\sigma(\mathcal H)}$. ⬜

For the second version of the monotone class theorem considered here, we take ${\mathcal K}$ to be a collection of indicator functions, ${1_A}$, for a collection ${\mathcal A}$ of sets ${A\subseteq E}$. Using the identity ${1_A1_B=1_{A\cap B}}$, the property that ${\mathcal K}$ is closed under multiplication is equivalent to ${\mathcal A}$ being closed under pairwise intersections. That is, ${\mathcal A}$ is a ${\pi}$-system.

Theorem 4 (Monotone Class Theorem) Let ${\mathcal A}$ be a ${\pi}$-system on a set ${E}$, and ${\mathcal H}$ be a vector subspace of ${B(E)}$ satisfying

• ${1_A\in\mathcal H}$ for all ${A\in\mathcal A}$.
• ${1\in\mathcal H}$.
• ${\mathcal H_+}$ is closed under bounded increasing limits.

Then, ${\sigma(\mathcal A)_b\subseteq\mathcal H}$.

Even though it is weaker than, and a direct consequence of, theorem 1 above, this is one of the more common versions of the monotone class theorem used in practise. It is also rather easier to prove than the more general theorem 1 above. I include a direct proof theorem 4 below, and will use it to prove all other versions of the monotone class theorem stated in this post.

Fubini’s theorem allowing the orders of multiple integrals to be commuted is a consequence of the monotone class theorem. For ${\sigma}$-algebras ${\mathcal E}$ and ${\mathcal F}$ on sets ${E}$ and ${F}$ respectively, the product ${\sigma}$-algebra ${\mathcal E\otimes\mathcal F}$ on the product space ${E\times F}$ is, by definition, the ${\sigma}$-algebra generated by products ${A\times B}$ over ${A\in\mathcal E}$ and ${B\in\mathcal F}$. I state Fubini’s theorem for finite measures, although it is straightforward to extend to ${\sigma}$-finite measures using monotone convergence.

Lemma 5 (Fubini) Let ${(E,\mathcal E,\mu)}$ and ${(F,\mathcal F,\nu)}$ be finite measure spaces. Then, for every bounded ${\mathcal E\otimes\mathcal F}$-measurable function ${f\colon E\times F\rightarrow{\mathbb R}}$,

• ${\int f(x,y)\,d\mu(x)}$ is ${\mathcal F}$-measurable w.r.t. ${y}$.
• ${\int f(x,y)\,d\nu(y)}$ is ${\mathcal E}$-measurable w.r.t. ${x}$.
• ${\iint f(x,y)\,d\mu(x)d\nu(y)=\iint f(x,y)\,d\nu(y)d\mu(x)}$.

Proof: Let ${\mathcal H}$ be the space of bounded ${\mathcal E\otimes\mathcal F}$-measurable functions ${f\colon E\times F\rightarrow{\mathbb R}}$ for which the conclusion of the lemma holds. By linearity of the integrals, ${\mathcal H}$ is a vector space, and, by monotone convergence, ${\mathcal H_+}$ is closed under taking limits of uniformly bounded increasing sequences of functions. Next, let ${\mathcal A}$ be the collection of ${A\times B}$ over ${A\in\mathcal E}$ and ${B\in\mathcal F}$, which is a ${\pi}$-system generating the ${\sigma}$-algebra ${\mathcal E\otimes\mathcal F}$. For any ${A\times B\in\mathcal A}$, writing ${1_{A\times B}(x,y)=1_A(x)1_B(y)}$ gives,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle\int 1_{A\times B}(x,y)d\mu(x)=1_B(y)\int 1_A(x)d\mu(x)=\mu(A)1_B(y)\,\smallskip\\ &\displaystyle\int 1_{A\times B}(x,y)d\nu(y)=1_A(x)\int 1_B(y)d\nu(y)=\nu(B)1_A(x). \end{array}$

These are measurable and, integrating w.r.t. ${\nu}$ and ${\mu}$ respectively gives,

$\displaystyle \iint 1_{A\times B}(x,y)d\mu(x)d\nu(y)=\mu(A)\nu(B)=\iint 1_{A\times B}(x,y)d\nu(y)d\mu(x).$

So, ${1_{A\times B}\in\mathcal H}$. In particular, ${1=1_{E\times F}\in\mathcal H}$. Then, theorem 4 says that ${\mathcal H}$ contains all bounded ${\mathcal E\otimes\mathcal F}$-measurable functions ${f\colon E\times F\rightarrow{\mathbb R}}$. ⬜

In the next version of the monotone class theorem, we relax the conditions on ${\mathcal H}$ so that it is no longer required to be a vector space. This does, however, require us to enforce closure under both bounded increasing and bounded decreasing sequences. It also requires us to enforce that ${\mathcal K}$ is a vector space and contains ${1}$, in addition to being closed under multiplication. That is, ${\mathcal K}$ is an algebra.

Theorem 6 (Monotone Class Theorem) Let ${\mathcal K}$ be a subalgebra of ${B(E)}$ and ${\mathcal H\subseteq B(E)}$ satisfy,

• ${\mathcal K\subseteq\mathcal H}$.
• ${\mathcal H}$ is closed under bounded increasing and bounded decreasing limits.

Then, ${\sigma(\mathcal K)_b\subseteq\mathcal H}$.

As an example, once the integrals of continuous real-valued functions on a bounded interval ${[a,b]}$ have been constructed (by, e.g., limits of Riemann sums), the integrals of all bounded measurable functions are uniquely determined by the dominated convergence property.

The monotone class theorem as given in theorem 6 also has a version starting from a ${\pi}$-system of subsets of ${E}$, rather than a general algebra ${\mathcal K}$. Given a ${\pi}$-system ${\mathcal A}$, the identity

$\displaystyle \left(\sum_ia_i1_{A_i}\right)\left(\sum_jb_j1_{B_j}\right)=\sum_{i,j}a_ib_j1_{A_i\cap B_j}$

shows that the linear span of indicator functions ${1_A}$ over ${A\in\mathcal A}$ is closed under multiplication. If ${E\in\mathcal A}$, then this means that the linear span is an algebra, so the following is a consequence of theorem 6.

Theorem 7 (Monotone Class Theorem) Let ${\mathcal A}$ be a ${\pi}$-system on a set ${E}$, such that ${E\in\mathcal A}$, and ${\mathcal H\subseteq B(E)}$ satisfy,

• the linear span of ${\{1_A\colon A\in\mathcal A\}}$ is contained in ${\mathcal H}$.
• ${\mathcal H}$ is closed under bounded increasing and bounded decreasing limits.

Then, ${\sigma(\mathcal A)_b\subseteq\mathcal H}$.

As an example, for an interval ${I=[a,b]\subseteq{\mathbb R}}$, we can let ${\mathcal A}$ consist of the closed subintervals of ${I}$. For any function ${f\colon I\rightarrow{\mathbb R}}$ of the form

$\displaystyle f=\sum_{i=1}^nc_i 1_{[a_i,b_i]}$

for real numbers ${c_i}$ and ${a_i\le b_i}$ in ${I}$, the integral is

$\displaystyle \int_a^b f(x)dx = \sum_{i=1}^n c_i(b_i-a_i).$

The monotone class theorem shows that dominated convergence uniquely determines the extension of the integral to all bounded measurable functions ${f\colon I\rightarrow{\mathbb R}}$.

#### Proof of the Monotone Class Theorem

I will give proofs of the various alternative versions of the monotone class theorem stated in theorems 1, 4, 6 and 7 above. In all cases, we have a ${\sigma}$-algebra ${\mathcal E}$ and want to show that ${\mathcal H}$ contains ${\mathcal E_b}$. The method is to first show that ${\mathcal H}$ contains ${1_A}$ for ${A\in\mathcal E}$. Then, ${\mathcal H}$ will contain all `simple’ functions, defined as linear combinations of such indicator functions. We then finish off by approximating arbitrary bounded measurable functions by increasing limits of simple functions.

The following method of approximating arbitrary nonnegative measurable functions as increasing limits of simple functions is standard.

Lemma 8 Let ${\mathcal E}$ be a ${\sigma}$-algebra on a set ${E}$, and ${\mathcal S\subseteq B(E)}$ denote the linear span of ${\{1_A\colon A\in\mathcal E\}}$. Then, for any ${\mathcal E}$-measurable function ${f\colon E\rightarrow{\mathbb R}^+}$, there exists an increasing sequence ${f_n\in\mathcal S_+}$ with ${\lim_nf_n=f}$.

Proof: For any finite subset ${U\subseteq{\mathbb R}^+}$ write ${U=\{u_1 < u_2 < \ldots < u_n\}}$. Letting ${A_i=f^{-1}([u_i,\infty))\in\mathcal E}$, write

$\displaystyle f_U=\sum_{i=1}^n(u_i-u_{i-1})1_{A_i}\in\mathcal S_+.$

For convenience, I am using ${u_0=0}$. For any ${x\in E}$, we see that ${x\in A_i}$ if and only if ${u_i\le f(x)}$, so

$\displaystyle f_U(x)=\max\{u_i\colon u_i\le f(x)\}.$

It follows that ${f_U\le f_V}$ whenever ${U\subseteq V}$. Hence, if we let ${U_n}$ be an increasing sequence of finite subsets of ${{\mathbb R}^+}$ such that ${\bigcup_nU_n}$ is dense in ${{\mathbb R}^+}$, and set ${f_n=f_{U_n}\in\mathcal S_+}$, then ${f_n}$ increases to ${f}$. For example, we can choose

$\displaystyle U_n=\left\{2^{-n}m\colon m=1,2,\ldots,4^n\right\}.$

Lemma 8 has the following immediate consequence.

Corollary 9 Let ${\mathcal H\subseteq B(E)}$ be a vector space such that ${\mathcal H_+}$ is closed under bounded increasing limits. Let ${\mathcal E}$ be a ${\sigma}$-algebra such that ${1_A\in\mathcal H}$ for all ${A\in\mathcal E}$. Then, ${\mathcal E_b\subseteq\mathcal H}$.

Proof: As ${\mathcal H}$ is a vector space, it contains the linear span, ${\mathcal S}$, of the indicator functions ${1_A}$ over ${A\in\mathcal E}$. For any nonnegative ${f\in\mathcal E_b}$, lemma 8 gives a sequence ${f_n\in\mathcal S_+\subseteq\mathcal H_+}$ increasing to ${f}$. Hence, ${f\in\mathcal H}$. Finally, for any ${f\in\mathcal E_b}$, then ${f_+=\max(f,0)}$ and ${f_-=\max(-f,0)}$ are in ${\mathcal H}$ and, by the vector space property, ${f=f_+-f_-}$ is also in ${\mathcal H}$. ⬜

The proof of the monotone class theorem as stated in theorem 4 is now a straightforward application of the ${\pi}$-system lemma

Proof of Theorem 4: Let ${\mathcal E}$ be the collection of sets ${A\subseteq E}$ such that ${1_A\in\mathcal H}$. By assumption of the theorem, ${\mathcal A\subseteq\mathcal E}$. We show that ${\mathcal E}$ is a d-system. First, as ${1\in\mathcal H}$ by assumption, then ${E\in\mathcal E}$. Next, for any ${A\subseteq B}$ in ${\mathcal E}$ then,

$\displaystyle 1_{B\setminus A}=1_B-1_A\in\mathcal H$

so, ${B\setminus A\in\mathcal E}$. Finally, if ${A_n}$ is an increasing sequence in ${\mathcal E}$ with limit ${A=\bigcup_nA_n}$, then ${1_{A_n}}$ is an increasing sequence in ${\mathcal H_+}$ with limit ${1_A}$. As ${\mathcal H_+}$ is closed under bounded increasing limits, ${A\in\mathcal E}$, so ${\mathcal E}$ is a d-system as claimed.

The ${\pi}$-system lemma now states that ${\sigma(\mathcal A)\subseteq\mathcal E}$. In particular, ${1_A\in\mathcal H}$ for all ${A\in\sigma(\mathcal A)}$. Corollary 9 gives ${\sigma(\mathcal A)_b\subseteq\mathcal H}$. ⬜

We move on to the proofs of the alternative versions of the monotone class theorem, all of which will follow from theorem 4. First, it is helpful to use a bit of notation: say that a set ${\mathcal H\subseteq B(E)}$ is a monotone class if it is closed under bounded increasing and bounded decreasing limits. The monotone class generated by an arbitrary ${\mathcal K\subseteq B(E)}$ is the smallest monotone class containing ${\mathcal K}$, and will be denoted by ${\mathcal M(\mathcal K)}$.

Lemma 10 Let ${\mathcal K}$ be a vector subspace of ${B(E)}$. Then, ${\mathcal M(\mathcal K)}$ is a vector space.

Proof: Fixing ${a\in{\mathbb R}}$, let ${\mathcal H}$ be the collection of ${f\in\mathcal M(\mathcal K)}$ such that ${af\in\mathcal M(\mathcal K)}$. Note that, if ${f_n\in\mathcal H}$ is a bounded and increasing or decreasing sequence, then so is ${af_n}$. Setting ${f=\lim_n f_n}$, we see that ${af=\lim_n af_n\in\mathcal M(\mathcal K)}$, so ${f\in\mathcal H}$. This shows that ${\mathcal H}$ is a monotone class. By the vector space property, ${\mathcal K\subseteq\mathcal H}$ and, hence, ${\mathcal H=\mathcal M(\mathcal K)}$. This shows that ${af\in\mathcal M(\mathcal K)}$ for all ${f\in\mathcal M(\mathcal K)}$ and ${a\in{\mathbb R}}$.

Next, for any ${g\in\mathcal M(\mathcal K)}$, let ${\mathcal H_g}$ denote the collection of ${f\in\mathcal M(\mathcal K)}$ such that ${f+g\in\mathcal M(\mathcal K)}$. If ${f_n\in\mathcal H_g}$ is a bounded increasing or decreasing sequence, then so is ${f_n+g}$. Setting ${f=\lim_nf_n}$, we have ${f+g=\lim_n(f_n+g)\in\mathcal M(\mathcal K)}$. and, hence, ${\mathcal H_g}$ is a monotone class. By the vector space property, ${\mathcal K\subseteq\mathcal H_g}$ for any ${g\in\mathcal K}$ and, therefore, ${\mathcal H_g=\mathcal M(\mathcal K)}$. We have shown that, if ${f}$ is in ${\mathcal M(\mathcal K)}$, then ${f+g\in\mathcal M(\mathcal K)}$ for all ${g\in\mathcal K}$. Hence, ${\mathcal K\subseteq\mathcal H_f}$. This gives ${\mathcal H_f=\mathcal M(\mathcal K)}$, so that ${f+g\in\mathcal M(\mathcal K)}$ for all ${f}$ and ${g}$ in ${\mathcal M(\mathcal K)}$. ⬜

Lemma 10 allows us to quickly establish theorem 7 as a consequence of theorem 4.

Proof of Theorem 7: Let ${\mathcal K}$ be the linear span of ${1_A}$ over ${A\in\mathcal A}$. As ${\mathcal H}$ is a monotone class containing ${\mathcal K}$, we have ${\mathcal M(\mathcal K)\subseteq\mathcal H}$. Then, by lemma 10, ${\mathcal M(\mathcal K)}$ is a vector space. As ${1=1_E\in\mathcal K}$, theorem 4 gives

$\displaystyle \sigma(\mathcal A)_b\subseteq\mathcal M(\mathcal K)\subseteq\mathcal H$

as required. ⬜

We turn to the proofs of theorems 1 and 6. Here, we are given ${\mathcal K\subseteq\mathcal H\subseteq B(E)}$ satisfying some conditions, and need to show that ${\sigma(\mathcal K)_b\subseteq\mathcal H}$. The idea is to show that the indicator functions ${1_A}$ are in ${\mathcal H}$ for all sets ${A}$ in a ${\pi}$-system generating ${\sigma(\mathcal K)}$. Unlike in the case of theorems 4 and 7, we are not given this at the outset. Instead, there is a bit more work to be done. When ${\mathcal K}$ is an algebra, the idea is roughly as follows, for any ${f\in\mathcal K}$.

1. By the algebra property, ${p\circ f}$ is in ${\mathcal K}$, for any real polynomial ${p}$.
2. The Stone–Weierstrass theorem states that every continuous function ${c\colon{\mathbb R}\rightarrow{\mathbb R}}$ can be uniformly approximated by polynomials on bounded intervals, from which it follows that ${c\circ f\in\mathcal H}$.
3. For any closed ${S\subseteq{\mathbb R}}$, the indicator function ${1_C}$ is a decreasing limit of continuous functions, so ${1_{f^{-1}(C)}=1_C\circ f}$ is in ${\mathcal H}$.

Using this approach, we prove the following.

Lemma 11 Let ${\mathcal K}$ be a subalgebra of ${B(E)}$, and let ${\mathcal A}$ be the collection of ${A\subseteq E}$ such that there exists a decreasing sequence ${f_n\in\mathcal K_+}$ with ${1_A=\lim_nf_n}$. Then,

• ${\mathcal A}$ is a ${\pi}$-system.
• ${f^{-1}(S)\in\mathcal A}$ for all ${f\in\mathcal K}$ and closed ${S\subseteq{\mathbb R}}$.
• ${\sigma(\mathcal K)\subseteq\sigma(\mathcal A)}$.

Proof: Suppose that ${A}$ and ${B}$ are in ${\mathcal A}$. Then, there exists (nonnegative) decreasing sequences ${f_n,g_n\in\mathcal K_+}$ such that ${1_A=\lim_nf_n}$ and ${1_B=\lim_ng_n}$. Therefore, ${f_ng_n}$ is a decreasing sequence in ${\mathcal K_+}$ with

$\displaystyle 1_{A\cap B}=\lim_{n\rightarrow\infty}(f_ng_n).$

So, ${A\cap B\in\mathcal A}$, showing that ${\mathcal A}$ is a ${\pi}$-system.

Next, fixing ${f\in\mathcal K}$ and closed ${S\subseteq{\mathbb R}}$, we show that ${f^{-1}(S)\in\mathcal A}$. Choose a sequence of continuous functions ${c_n\colon{\mathbb R}\rightarrow{\mathbb R}^+}$ decreasing to ${1_S}$. For example, if ${S}$ is nonempty, we can take

$\displaystyle c_n(x) = \max(1-n d(x,S),0)$

where ${d(x,S)=\min_{y\in S}\lvert x-y\rvert}$ is the distance from ${x}$ to ${S}$. If ${S}$ is empty then, trivially, we can take ${c_n=0}$.

The next step is to replace the continuous functions ${c_n}$ by polynomials. As ${f\in B(E)}$, it satisfies ${\lvert f\rvert\le K}$ for some nonnegative real ${K}$. We apply the Stone–Weierstrass theorem to find a sequence of polynomials ${p_n}$ with ${\lvert p_n-c_n\rvert\le 2^{-n}}$ on ${[-K,K]}$. Then, on this interval,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle p_n-p_{n+1}&\displaystyle\ge c_n-c_{n+1}-2^{-n}-2^{-n-1}\smallskip\\ &\displaystyle\ge -2^{-n}-2^{-n-1}=3.(2^{-n-1}-2^{-n}). \end{array}$

So, ${q_n\equiv p_n+3.2^{-n}}$ is a sequence of polynomials, decreasing on ${[-K,K]}$, to the limit ${1_S}$. As ${f}$ is contained in the algebra ${\mathcal K}$, the composition ${q_n\circ f}$ is also in ${\mathcal K}$. As the image of ${f}$ is contained in ${[-K,K]}$, ${q_n\circ f}$ is decreasing with limit

$\displaystyle \lim_n q_n\circ f=1_S\circ f=1_{f^{-1}(S)}.$

Hence, ${f^{-1}(S)\in\mathcal A}$ as required.

Finally, by the above, for any ${f\in\mathcal K}$ and closed ${S\subseteq{\mathbb R}}$, ${f^{-1}(S)\in\sigma(\mathcal A)}$. As the Borel ${\sigma}$-algebra is generated by closed sets, ${f}$ is ${\sigma(\mathcal A)}$-measurable, showing that ${\sigma(\mathcal K)\subseteq\sigma(\mathcal A)}$. ⬜

Lemma 11 enables us to reduce theorems 1 and 6 to the simpler version of the monotone class theorem given by theorem 4, which has already been proven above. It is convenient to start by establishing the following result, which will imply theorems 1 and 6.

Lemma 12 Let ${\mathcal K\subseteq\mathcal H\subseteq B(E)}$, where ${\mathcal K}$ is an algebra and ${\mathcal H}$ is a vector space such that ${\mathcal H_+}$ is closed under bounded increasing limits. Then, ${\sigma(\mathcal K)_b\subseteq\mathcal H}$.

Proof: Let ${\mathcal A}$ be the collection of ${A\subseteq E}$ such that there exists a sequence ${f_n\in\mathcal K}$ decreasing to ${1_A}$. As ${f_1-f_n}$ is a bounded increasing sequence in ${\mathcal H_+}$, its limit, ${f_1-1_A}$ is in ${\mathcal H}$, so ${1_A\in\mathcal H}$. Lemma 11 states that ${\mathcal A}$ is a ${\pi}$-system so, by the last statement of lemma 11, and by theorem 4,

$\displaystyle \sigma(\mathcal K)_b\subseteq\sigma(\mathcal A)_b\subseteq\mathcal H$

as required. ⬜

Theorem 1 is an immediate consequence of the lemma.

Proof of Theorem 1: As ${\mathcal H}$ is a vector space containing ${\mathcal K}$ and ${1}$, it contains the linear span of ${\mathcal K\cup\{1\}}$, which I denote as ${\mathcal K^\prime}$. As ${\mathcal K^\prime}$ is an algebra, lemma 12 gives

$\displaystyle \sigma(\mathcal K)_b\subseteq\sigma(\mathcal K^\prime)_b\subseteq\mathcal H$

as required. ⬜

Theorem 6 is also an immediate consequence.

Proof of Theorem 6: As ${\mathcal H}$ is a monotone class, it contains ${\mathcal M(\mathcal K)}$. Lemma 10 says that ${\mathcal M(\mathcal K)}$ is a vector space. Applying lemma 12 with ${\mathcal M(\mathcal K)}$ in place of ${\mathcal H}$,

$\displaystyle \sigma(\mathcal K)_b\subseteq\mathcal M(\mathcal K)\subseteq\mathcal H$

as required. ⬜