# Confused by Monotone class theorem for functions

Monotone Class Theorem has two types. One is Monotone class theorem for sets and the other for functions. I have no doubt for sets.

Here is a reference of definition of Monotone Class Theorem for functions from Probability: Theory and Examples, Durrett Rick, 4.1-edition.

Theorem 6.1.3. Monotone class theorem. Let $\mathcal {A}$ be a $\pi$-system that contains $\Omega$ and let $\mathcal {H}$ be a collection of real-valued functions that satisfies:

(i) If $A ∈ \mathcal {A}$, then $1_A ∈ \mathcal {H}$.

(ii) If $f, g ∈ \mathcal {H}$, then $f + g$, and $cf ∈ \mathcal {H}$ for any real number $c$.

(iii) If $f_n ∈ \mathcal {H}$ are nonnegative and increase to a bounded function $f$, then $f ∈ \mathcal {H}$.

Then $\mathcal {H}$ contains all bounded functions measurable with respect to $\sigma (\mathcal {A})$.

Proof. The assumption $\Omega \in \mathcal{A}$, (ii), and (iii) imply that $\mathcal {G} = \{A : 1_A ∈ \mathcal {H}\}$ is a $\lambda$-system so by (i) and the $\pi − \lambda$ theorem, Theorem 2.1.2, $\mathcal {G} ⊃ \sigma(A)$. (ii) implies $\mathcal{H}$ contains all simple functions, and (iii) implies that $\mathcal{H}$ contains all bounded measurable functions.

I can't understand one sentence from the proof, "(ii) implies $\mathcal{H}$ contains all simple functions". Why?

P.S.

I think "measurable with respect to $\sigma(\mathcal{A})$" means $\sigma(\mathcal {A})$-measurable. I know $\mathcal{F}$-measurable mapping that is suppose $(\Omega, \mathcal{F})$ and $(E, \mathcal{E})$ are two measurable spaces, $f$ is a mapping from $\Omega \to E$. $f$ is a $\mathcal{F}$-measurable mapping if $f^{-1}(A) \in \mathcal{F}$ for $\forall A \in \mathcal{E}$. So $\sigma(\mathcal {A})$-measurable probably means those $f: (\Omega, \sigma (\mathcal{A})) \to ([0, +\infty), \mathcal{B}{[0, +\infty)})$ where $\mathcal{B}{[0, +\infty)}$ denotes Borel $\sigma$-algebra on $[0, +\infty)$.

Update:

The author didn't make the sentence clear. So I try to make it a little bit more complete. Let $\forall f$ is $\sigma(\mathcal{A})$-measurable. I suppose $g_n =\sum_{k=0}^{n2^n -1} \frac{k}{2^n} I_{\{\frac{k}{2^n} \le f^+ ＜ \frac{k+1}{2^n}\}} + n I_{f^+ \ge n}$ where $f^+ = f ∨ 0$. Then $g_n \in \mathcal{H}$ coz $f$ is $\sigma(\mathcal{A})$-measurable that is $\{x: \frac{k}{2^n} \le f^+ ＜ \frac{k+1}{2^n} \} \in \sigma(\mathcal{A})$ and $\{x: f^+ \ge n\} \in \sigma(\mathcal{A})$. Then $g_n$ increasingly converges to $f^+$ and so does $f^-$. $f = f^+ - f^- \in \mathcal{H}$. Please note that I can only get simple functions defined on $\{x: \frac{k}{2^n} \le f^+ ＜ \frac{k+1}{2^n} \}$ and $\{x: f^+ \ge n\}$ are in $\mathcal{H}$. I can't prove $\mathcal{H}$ will contains all simple functions. Am I right?

• what do you mean by Let $\forall f$ is $\sigma(\mathcal{A})$-measurable? – Conrado Costa Sep 6 '15 at 1:42
• What is the sentence you find the author didn't make clear? – Conrado Costa Sep 6 '15 at 1:43
• @ConradoCosta: "(ii) implies $\mathcal{H}$ contains all simple functions" in the proof. I'm not sure how it imply? – Bear and bunny Sep 6 '15 at 2:00
• It is written in my answer. Can I clarify any passage? – Conrado Costa Sep 6 '15 at 2:04
• @ConradoCosta: "$\forall f$ is $\sigma (A)$-measurable". Hmmm, I want to show $\mathcal{H}$ contains all $\sigma(A)$-measurable. So suppose $f$ is arbitrary satisfying $\sigma(A)$-measurable and show it has been contained in $\mathcal{H}$ . – Bear and bunny Sep 6 '15 at 2:04

A simple function is of the form $$f(x) = \sum_{i = 1}^N a_i 1_{A_i}$$ a finite linear combination of functions $1_{A_i}$ for $A_i \in \sigma(A)$.

We would like to prove that those functions are in $\mathcal{H}$.

So we consider $\mathcal{C} = \{B \in \sigma(A)\mid 1_B \in \mathcal{H}\}$

So $\mathcal{A} \subset \mathcal{C}$ by $(i)$.

Since $\Omega \in \mathcal{A}$, $\Omega \in \mathcal{C}$

Moreover, if $A \subset B \in \mathcal{C}$ then $B \setminus A \in \mathcal{C}$ by $(ii)$

Now take $A_1, A_2 , \ldots A_n, \ldots$ an increasing sequence $A_i \subset A_{i+1}$. Then $A = \cup_i A_i$ is in $\mathcal{C}$ by $(iii)$

This proves that $\mathcal{C}$ is a $\lambda$-system containing the $\pi$-system $\mathcal{A}$. By Dynkin $\pi-\lambda$ theorem (https://en.wikipedia.org/wiki/Dynkin_system) $\mathcal{C} = \sigma(\mathcal{A})$

So now we know that $1_{A_1}, \ldots, 1_{A_N}$ are in $\mathcal{H}$ for any finite choice of $A_i \in \sigma(\mathcal{A})$, by $(ii)$ $a_1 1_{A_1} + \ldots + a_N 1_{A_N}$ is in $\mathcal{H}$.

This is the claim of the author.

If you would like to see that any $\sigma(\mathcal{A})$-measurable function is in $\mathcal{H}$ It suffices to consider $f = f^+ - f^-$ and to prove that $f^+,f^-$ are in $\mathcal{H}$.

The argument is essentially the one you presented.

Let $f$ be $\sigma(\mathcal{A})$-measurable. define $$g_n =\sum_{k=0}^{n2^n -1} \frac{k}{2^n} 1_{\{\frac{k}{2^n} < f^+ \le \frac{k+1}{2^n}\}} + n 1_{f^+ > n}$$ where $f^+ = f \vee 0$. Then $g_n \in \mathcal{H}$ once $\{x: \frac{k}{2^n} < f^+ \le \frac{k+1}{2^n} \} \in \sigma(\mathcal{A})$ and $\{x: f^+ > n\} \in \sigma(\mathcal{A})$ (because $f$ is $\sigma(\mathcal{A})$-measurable). Then $g_n$ increasingly converges to $f^+$. Since $f$ is bounded so is $f^+$ and therefore by $(iii)$ $f^+ \in \mathcal{H}$. The argument for $f^{-}$ is analogous.

this concludes the proof.

• I got trapped by myself. All functions satisfying (i), (ii) and (iii) are included in $\mathcal{H}$ or all elements in $\mathcal{H}$ satisfy (i), (ii) and (iii) coz the first one is much bigger I think. – Bear and bunny Sep 5 '15 at 16:39
• I think that none of the above is true. For instance $f = 1_A + 1_B \in \mathcal{H}$ and $f$ does not satisfies $(i)$ so neither "All functions satisfying (i), (ii) and (iii) are included in $\mathcal{H}$ nor all elements in $\mathcal{H}$ satisfy (i), (ii) and (iii). $\mathcal{H}$ has basic functions $(i)$, linear combinations of basic functions $(ii)$ and bounded limits of linear combinations of basic functions $(iii)$. – Conrado Costa Sep 6 '15 at 1:10
• I see. But from (i) you just now $\mathcal {H}$ has basic functions. But you can't say it contains all simple functions coz you don't know what kind of domain you needed and some domain may not be included in $\mathcal{A}$. So I try to show those domain I needed, that is $\{x: \frac{k}{2^n} \le f^+ \le \frac{k+1}{2^n} \} \in \sigma(\mathcal{A})$ and $\{x: f^+ \ge n\} \in \sigma(\mathcal{A})$, will be included in $\sigma(A)$. – Bear and bunny Sep 6 '15 at 2:20
• Then I can use simple functions defined on those domains I proved. But these domains are just some of simple functions on power set of $\Omega$. Not " contains all simple functions". – Bear and bunny Sep 6 '15 at 2:22
• I think there is something wrong with this proof. With my proof, the Monotone Class Theorem for functions still works. It is not a problem of the theorem. Just a problem of one sentence in the proof. – Bear and bunny Sep 6 '15 at 2:25