Is the Monotone Class Theorem Applicable to the Borel Sets?

[fishturtle1]

Homework Statement

In older books the Borel sets are often introduced as the smallest family $\mathcal{M}$ of sets which is stable under countable intersections of decreasing and countable unions of increasing sequences of sets, and which contain all open sets $\mathcal{O}$. Use problem 3.14 to show that $\mathcal{M} = \mathcal{B}(\mathbb{R})$.

Relevant Equations

$\textbf{Definition.}$ A family $\mathcal{M} \subset \mathcal{P}(X)$ which contains $X$ and is stable under countable unions of increasing sets and countable intersections of decreasing sets
$$(A_n)_{n\in\mathbb{N}} \subset \mathcal{M}, A_1 \subset \dots \subset A_n \subset A_{n+1} \uparrow A = \bigcup_{n \in \mathbb{N}} A_n \Rightarrow A \in \mathcal{M}$$

$$(B_n)_{n\in\mathbb{N}} \subset \mathcal{M}, B_1 \supset \dots \supset B_n \supset B_{n+1} \downarrow B = \bigcap_{n\in\mathbb{N}} B_n\Rightarrow B \in \mathcal{M}$$

is called a $\textit{monotone class}$.

$\textbf{Definition.}$ Let $\mathcal{F} \subset \mathcal{P}(X)$. We define $m(\mathcal{F})$ to be the smallest monotone class containing $\mathcal{F}$. That is, if $\mathcal{M}$ is a monotone class containing $\mathcal{F}$, then $m(\mathcal{F}) \subseteq \mathcal{M}$.

Problem 3.14: There are four parts:
i) Mimic the proof of theorem 3.4. to show that there is a minimal monotone class $m(\mathcal{F})$ such that $\mathcal{F} \subset m(\mathcal{F})$.

ii) If $\mathcal{F}$ is stable w.r.t. complements, then so is $m(\mathcal{F})$.

iii) If $\mathcal{F}$ is stable w.r.t. intersection, then so is $m(\mathcal{F})$.

iv) Use i), ii), iii) to prove the following: Let $\mathcal{F} \subset \mathcal{P}(X)$ which is stable under the formation of intersections and complements. If $\mathcal{M} \supset \mathcal{F}$ is a monotone class, then $M \supset \sigma(\mathcal{F})$.

Proof: Let $A, B \in \mathcal{O}$ and $x \in A \cap B$. Then there exists $\varepsilon_A, \varepsilon_B > 0$ such that $B_{\varepsilon_A}(x) \subset A$ and $B_{\varepsilon_B}(x) \subset B$. Let $\varepsilon = \min\lbrace\varepsilon_A, \varepsilon_B\rbrace$. Then $B_\varepsilon(x) \subset A \cap B$. This shows $A \cap B \in \mathcal{O}$ i.e. $\mathcal{O}$ is stable under formation of intersections.

But $(0, 1) \in \mathcal{O}$ and $(0, 1)^c = (-\infty, 0] \cup [1, \infty) \not\in \mathcal{O}$. So $\mathcal{O}$ is not stable under complements.

If I could show that $m(\mathcal{O})$ is a $\sigma$ algebra, then $m(\mathcal{O}) \supseteq \sigma(\mathcal{O})$. Moreover, any $\sigma$ algebra is also a monotone class, so $m(\mathcal{O}) \subseteq \sigma(\mathcal{O})$, which would complete the proof.Have I made a mistake on the complements part? Or do I need to choose a different generating set for $\mathcal{B}(\mathbb{R^n})$?

 

[fishturtle1]

I might be respond too soon, but I think the other inclusion would be true if we had $\mathcal{O}$ is closed under complements and intersection. By problem 3.14 this implies $m(\mathcal{O})$ is a $\sigma$-algebra. Since $m(\mathcal{O})$ contains $\mathcal{O}$, we'd be able to conclude $m(\mathcal{O}) \supseteq \sigma(\mathcal{O})$.

But yeah, $\mathcal{O}$ is not stable under complements. If there were a set $A$ such that $\sigma(A) = \mathcal{B}(\mathbb{R}^n)$ that is stable under complements and intersection, I think that would work. But does one exist?

(Also I added the definition for monotone class to the OP, sorry about that).

Edit: I also added all four parts of problem 3.14. I interpreted the problem statement as "Use the conclusion of problem 3.14.", but maybe there is something in parts i), ii). or iii) in 3.14. that I should be using. As in, I think if I can just show $m(\mathcal{O})$ is closed under complements, then this would be enough.

 

[member 587159]

Hi. I tried this problem and did some googling around and found this pdf with solutions from the author:
http://www.motapa.de/measures_integrals_and_martingales/solutions-mims-2ed.pdf

However, the argument he presents there for this exercise is flawed. He applies the monotone class theorem but we can't apply it because $\mathcal{O}$ is not closed under complementation.

I'm thinking about a way to fix this argument but I have to study exams myself now so it might take a while before I can give an answer. I bookmarked this thread for now. When I get some spare time, I'll think about the problem.

 

[fishturtle1]

To be completely transparent, I checked the above link too, before posting here. (and yeah, I could not figure out the complementation part). I'm going to try a little more on this one, i'll post if I come up with something.

Thank you for your time, and best of luck on exams!

 

[member 587159]

I took the liberty to mail the author. He was so kind to give a very swift reply to fix the argument. Here is the sketched argument he gave (I already filled in some minor details for you). If you want any more details, feel free to ask me:

Let $\mathcal{O}$ be the set of open subsets and $\mathcal{C}$ the set of closed subsets. Then $\mathcal{O} \cup \mathcal{C}$ is stable under complements. Further, any closed set is the increasing union of open sets and thus $m(\mathcal{O}) = m(\mathcal{O}\cup \mathcal{C})$. Now, argue that $m(\mathcal{O}\cup \mathcal{C})$ is a $\sigma$-algebra because it is a complement-stable monotone class (use 3.14 (ii) for this). Thus $m(\mathcal{O})$ is a $\sigma$-algebra containing $\mathcal{O}$ and thus $\mathcal{B}(\mathbb{R}^n) \subseteq m(\mathcal{O})$.

 

[fishturtle1]

Thank you (and the author). I've tried to fill in the details (I am not sure how to show any closed set is the increasing union of open sets, instead I showed for any closed set $B$, we can find a sequence of open sets that decrease to $B$. Hopefully that's ok?) :

Proof: Let $\mathcal{O}$ be the set of open subsets and $\mathcal{C}$ be the set of closed subsets. Let $A \in \mathcal{O} \cup \mathcal{C}$. Then $A^c$ is closed or open and so $A^c \in \mathcal{O} \cup \mathcal{C}$. Hence, $\mathcal{O} \cup \mathcal{C}$ is stable under complements. By 3.14ii), we have $m(\mathcal{O} \cup \mathcal{C})$ is stable under complements. Let $F, G \in \mathcal{O} \cup \mathcal{C}$. Then $F \cap G$ is open or closed. So $\mathcal{O} \cup \mathcal{C}$ is stable under intersections. By 3.14iii), $m(\mathcal{O} \cup \mathcal{C})$ is stable under intersections.

Next, we'll show $m(\mathcal{O}) = m(\mathcal{O} \cup \mathcal{C})$. Since $\mathcal{O} \subset \mathcal{O} \cup \mathcal{C}$, we have $m(\mathcal{O}) \subseteq m(\mathcal{O} \cup \mathcal{C})$. Let $B \in \mathcal{C}$. Let $x_1, x_2, \dots$ be the list of all elements in $B$ such that for all $\varepsilon > 0$, $B_\varepsilon(x_i) \not\subset B$. Define a sequence of open sets $(B_n)_{n \in \mathbb{N}}$ as
$$B_n = B \cup \bigcup_{i \in I} (x_i - \frac1n, x_i + \frac1n)$$
Then $B_n \downarrow B$ and so $B \in m(\mathcal{O})$. It follows that $\mathcal{O} \cup \mathcal{C} \subset m(\mathcal{O})$. So $m(\mathcal{O} \cup \mathcal{C}) \subseteq m(\mathcal{O})$. We may conclude $m(\mathcal{O}) = m(\mathcal{O} \cup \mathcal{C})$.

Next, we'll show $m(\mathcal{O} \cup \mathcal{C})$ is a $\sigma$-algebra. We've shown $m(\mathcal{O} \cup \mathcal{C})$ is stable under complements. By definition of monotone class, $X \in m(\mathcal{O} \cup \mathcal{C})$. Hence $\emptyset = X^c \in m(\mathcal{O} \cup \mathcal{C})$. Let $(A_n)_{n\in\mathbb{N}} \subset m(\mathcal{O} \cup \mathcal{C})$. Define a sequence $(D_n)_{n\in\mathbb{N}} \subset m(\mathcal{O} \cup \mathcal{C})$ as
$$D_n = \bigcap_{k=1}^{n} A_k^c$$
Then $D_n \downarrow \bigcap_{k=1}^{\infty}A_k^c \in m(\mathcal{O} \cup \mathcal{C})$. Taking the complement of $\bigcap_{k=1}^{\infty}A_k^c$ gives us $m(\mathcal{O} \cup \mathcal{C})$ is closed under countable union. Thus, $m(\mathcal{O} \cup \mathcal{C})$ is a $\sigma$-algebra.

We have $m(\mathcal{O})$ is a $\sigma$-algebra containing $\mathcal{O}$ hence $m(\mathcal{O}) \supseteq \mathcal{B}(\mathbb{R}^n)$. Since $\mathcal{B}(\mathbb{R}^n)$ is a monotone class containing $\mathcal{O}$, we have $m(\mathcal{O}) \subseteq \mathcal{B}(\mathbb{R}^n)$. This completes the proof. []

 

[member 587159]

Thank you (and the author). I've tried to fill in the details (I am not sure how to show any closed set is the increasing union of open sets, instead I showed for any closed set $B$, we can find a sequence of open sets that decrease to $B$. Hopefully that's ok?) :

Proof: Let $\mathcal{O}$ be the set of open subsets and $\mathcal{C}$ be the set of closed subsets. Let $A \in \mathcal{O} \cup \mathcal{C}$. Then $A^c$ is closed or open and so $A^c \in \mathcal{O} \cup \mathcal{C}$. Hence, $\mathcal{O} \cup \mathcal{C}$ is stable under complements. By 3.14ii), we have $m(\mathcal{O} \cup \mathcal{C})$ is stable under complements. Let $F, G \in \mathcal{O} \cup \mathcal{C}$. Then $F \cap G$ is open or closed. So $\mathcal{O} \cup \mathcal{C}$ is stable under intersections. By 3.14iii), $m(\mathcal{O} \cup \mathcal{C})$ is stable under intersections.

Next, we'll show $m(\mathcal{O}) = m(\mathcal{O} \cup \mathcal{C})$. Since $\mathcal{O} \subset \mathcal{O} \cup \mathcal{C}$, we have $m(\mathcal{O}) \subseteq m(\mathcal{O} \cup \mathcal{C})$. Let $B \in \mathcal{C}$. Let $x_1, x_2, \dots$ be the list of all elements in $B$ such that for all $\varepsilon > 0$, $B_\varepsilon(x_i) \not\subset B$. Define a sequence of open sets $(B_n)_{n \in \mathbb{N}}$ as
$$B_n = B \cup \bigcup_{i \in I} (x_i - \frac1n, x_i + \frac1n)$$
Then $B_n \downarrow B$ and so $B \in m(\mathcal{O})$. It follows that $\mathcal{O} \cup \mathcal{C} \subset m(\mathcal{O})$. So $m(\mathcal{O} \cup \mathcal{C}) \subseteq m(\mathcal{O})$. We may conclude $m(\mathcal{O}) = m(\mathcal{O} \cup \mathcal{C})$.

Next, we'll show $m(\mathcal{O} \cup \mathcal{C})$ is a $\sigma$-algebra. We've shown $m(\mathcal{O} \cup \mathcal{C})$ is stable under complements. By definition of monotone class, $X \in m(\mathcal{O} \cup \mathcal{C})$. Hence $\emptyset = X^c \in m(\mathcal{O} \cup \mathcal{C})$. Let $(A_n)_{n\in\mathbb{N}} \subset m(\mathcal{O} \cup \mathcal{C})$. Define a sequence $(D_n)_{n\in\mathbb{N}} \subset m(\mathcal{O} \cup \mathcal{C})$ as
$$D_n = \bigcap_{k=1}^{n} A_k^c$$
Then $D_n \downarrow \bigcap_{k=1}^{\infty}A_k^c \in m(\mathcal{O} \cup \mathcal{C})$. Taking the complement of $\bigcap_{k=1}^{\infty}A_k^c$ gives us $m(\mathcal{O} \cup \mathcal{C})$ is closed under countable union. Thus, $m(\mathcal{O} \cup \mathcal{C})$ is a $\sigma$-algebra.

We have $m(\mathcal{O})$ is a $\sigma$-algebra containing $\mathcal{O}$ hence $m(\mathcal{O}) \supseteq \mathcal{B}(\mathbb{R}^n)$. Since $\mathcal{B}(\mathbb{R}^n)$ is a monotone class containing $\mathcal{O}$, we have $m(\mathcal{O}) \subseteq \mathcal{B}(\mathbb{R}^n)$. This completes the proof. []

I didn't check your proof in detail but the idea looks correct. However, at one point you work with intervals while the context is $\mathbb{R}^n$ and not $\mathbb{R}$.

So you should fix that if you want a proof that works for higher dimensional Euclidean spaces.

To see a closed set $F$ is the decreasing intersection of opens (or equivalently, taking complements any open is the increasing union of closeds), use that

$$F=\bigcap_{k=1}^\infty\{x \in \mathbb{R}^n: \Vert x-y \Vert < 1/k \mathrm{\ for \ some \ y \in F} \}$$

 

[fishturtle1]

I didn't check your proof in detail but the idea looks correct. However, at one point you work with intervals while the context is $\mathbb{R}^n$ and not $\mathbb{R}$.

So you should fix that if you want a proof that works for higher dimensional Euclidean spaces.

To see a closed set $F$ is the decreasing intersection of opens (or equivalently, taking complements any open is the increasing union of closeds), use that

$$F=\bigcap_{k=1}^\infty\{x \in \mathbb{R}^n: \Vert x-y \Vert < 1/k \mathrm{\ for \ some \ y \in F} \}$$

That makes sense, I've made the correction. Thank you.

 

[member 587159]

FYI, the author has edited the PDF and has now included the fixed solution: http://www.motapa.de/measures_integrals_and_martingales/solutions-mims-2ed.pdf