Probability III: Week-1

Probability Space

• Defined as a triplet \((\Omega, \mathcal{F}, P)\).
• \(\mathcal{F} \subseteq \mathcal{P}(\Omega)\) is a σ-field.

Properties of a σ-field

• \(\varnothing, \Omega \in \mathcal{F}\).
• If \(A \in \mathcal{F}\), then \(A^{c} \in \mathcal{F}\).
• If \(A_i \in \mathcal{F}\), then \(\bigcup A_i \in \mathcal{F}\).

Countable vs Uncountable Sample Space

• If \(\Omega\) is countable, then \(\mathcal{F} = \mathcal{P}(\Omega)\).
• \(\{\varnothing, \Omega\}\) and \(\mathcal{P}(\Omega)\) are trivial σ-fields.
• If \(\Omega\) is uncountable, \(\mathcal{P}(\Omega)\) is too large a σ-field.

Generated σ-field

• Start with a suitable collection of events \(\mathcal{C}\).
• Consider the smallest σ-field of \(\Omega\) that contains \(\mathcal{C}\).
• This is called \(\sigma(\mathcal{C})\), the σ-field generated by \(\mathcal{C}\).

Question 1

Show that \(\sigma(\mathcal C)\) exists and is unique, where \(\mathcal C\) is a collection of subsets of \(\Omega\) and \(\sigma(\mathcal C)\) denotes the smallest \(\sigma\)-field on \(\Omega\) that contains \(\mathcal C\).

1) Rigorous solution (existence and uniqueness)

Step 0 — set-up and notation

• Let \(\Omega\) be a nonempty set and \(\mathcal C\subseteq\mathcal P(\Omega)\) be any collection of subsets of \(\Omega\).
• Call

  \(\displaystyle \mathfrak S :=\{\,\mathcal F\subseteq\mathcal P(\Omega)\;:\;\mathcal F\text{ is a }\sigma\text{-field and }\mathcal C\subseteq\mathcal F\,\}\)

  the family of all \(\sigma\)-fields on \(\Omega\) that contain \(\mathcal C\).
• We will prove:
  • (A) \(\mathfrak S\neq\varnothing\).
  • (B) The intersection \(\bigcap_{\mathcal F\in\mathfrak S}\mathcal F\) is a \(\sigma\)-field containing \(\mathcal C\).
  • (C) That intersection is the unique smallest \(\sigma\)-field containing \(\mathcal C\).

(A) Non-emptiness of \(\mathfrak S\)

• \(\mathcal P(\Omega)\) (the power set of \(\Omega\)) is a \(\sigma\)-field:
  • it contains \(\varnothing\) and \(\Omega\),
  • is closed under complements, and
  • is closed under arbitrary unions.
• Since \(\mathcal C\subseteq\mathcal P(\Omega)\) trivially, we have \(\mathcal P(\Omega)\in\mathfrak S\). Therefore \(\mathfrak S\neq\varnothing\).

• Remark: alternatively, \(\{\varnothing,\Omega\}\) is a \(\sigma\)-field but need not contain \(\mathcal C\); \(\mathcal P(\Omega)\) is the convenient witness showing \(\mathfrak S\) is nonempty.

(B) The intersection of all members of \(\mathfrak S\) is a \(\sigma\)-field containing \(\mathcal C\)

• Define

  \(\displaystyle \mathcal G := \bigcap_{\mathcal F\in\mathfrak S}\mathcal F.\)

• We check the three axioms of a \(\sigma\)-field for \(\mathcal G\), and also that \(\mathcal C\subseteq\mathcal G\).
• • \(\varnothing,\Omega\in\mathcal G\).

    For every \(\mathcal F\in\mathfrak S\), \(\mathcal F\) is a \(\sigma\)-field so \(\varnothing,\Omega\in\mathcal F\). Hence \(\varnothing,\Omega\) belong to the intersection \(\mathcal G\).

  ```
  • Closed under complements.

    Let \(A\in\mathcal G\). Then \(A\in\mathcal F\) for every \(\mathcal F\in\mathfrak S\). Because each \(\mathcal F\) is a \(\sigma\)-field, \(A^c\in\mathcal F\) for every \(\mathcal F\in\mathfrak S\). Thus \(A^c\in\mathcal G\). So \(\mathcal G\) is closed under complements.

  • Closed under countable unions.

    Let \((A_n)_{n\ge1}\) be a sequence with each \(A_n\in\mathcal G\). Then for every \(\mathcal F\in\mathfrak S\) we have \(A_n\in\mathcal F\) for all \(n\), and since \(\mathcal F\) is a \(\sigma\)-field, \(\bigcup_{n=1}^\infty A_n\in\mathcal F\). This holds for every \(\mathcal F\in\mathfrak S\), hence \(\bigcup_{n=1}^\infty A_n\in\mathcal G\). Thus \(\mathcal G\) is closed under countable unions.

  • \(\mathcal C\subseteq\mathcal G\).

    By definition of \(\mathfrak S\), every \(\mathcal F\in\mathfrak S\) contains \(\mathcal C\). Therefore every element of \(\mathcal C\) lies in every \(\mathcal F\in\mathfrak S\), hence \(\mathcal C\subseteq\bigcap_{\mathcal F\in\mathfrak S}\mathcal F=\mathcal G\).

• So \(\mathcal G\) is a \(\sigma\)-field containing \(\mathcal C\). In particular \(\mathcal G\in\mathfrak S\).

```

(C) Minimality and uniqueness

• Claim: \(\mathcal G\) is the smallest \(\sigma\)-field containing \(\mathcal C\). "Smallest" means: if \(\mathcal H\) is any \(\sigma\)-field with \(\mathcal C\subseteq\mathcal H\), then \(\mathcal G\subseteq\mathcal H\).
• Proof:

  By definition \(\mathfrak S\) is the family of all \(\sigma\)-fields that contain \(\mathcal C\). Hence every such \(\mathcal H\) is an element of \(\mathfrak S\). Since \(\mathcal G\) is the intersection of all elements of \(\mathfrak S\), it follows that \(\mathcal G\subseteq\mathcal H\). Thus \(\mathcal G\) is contained in every \(\sigma\)-field that contains \(\mathcal C\); equivalently it is the smallest such \(\sigma\)-field.

• Uniqueness:

  Suppose \(\mathcal G_1\) and \(\mathcal G_2\) are two smallest \(\sigma\)-fields containing \(\mathcal C\). By minimality of \(\mathcal G_1\), \(\mathcal G_1\subseteq\mathcal G_2\). By minimality of \(\mathcal G_2\), \(\mathcal G_2\subseteq\mathcal G_1\). Hence \(\mathcal G_1=\mathcal G_2\). So the smallest \(\sigma\)-field containing \(\mathcal C\) is unique.

Conclusion: \(\displaystyle \sigma(\mathcal C):=\bigcap_{\mathcal F\in\mathfrak S}\mathcal F\) exists and is uniquely determined.

Question 2

If \(\mathcal C\) is a collection of singletons, show that \(\sigma(\mathcal C)\) is the countable / co-countable \(\sigma\)-field; i.e. every event in \(\sigma(\mathcal C)\) is either countable or the complement of a countable set.

1) Rigorous solution

Interpretation / precise assumption

• We take \(\displaystyle \mathcal C=\{\{\omega\}: \omega\in\Omega\}\), the collection of all singletons of \(\Omega\).
• Define

  \(\displaystyle \mathcal F := \{A\subseteq\Omega:\ A\text{ is countable or }A^c\text{ is countable}\}\),

  the family of countable or co-countable subsets of \(\Omega\). We will prove \(\sigma(\mathcal C)=\mathcal F\).

Step 1 — \(\mathcal F\) is a \(\sigma\)-field

• Null and whole space:

  \(\varnothing\in\mathcal F\) because \(\varnothing\) is countable. Also \(\Omega\in\mathcal F\) because \(\Omega^c=\varnothing\) is countable.

```
• Closed under complements:

  If \(A\in\mathcal F\) then either \(A\) is countable (so \(A^c\) is co-countable) or \(A^c\) is countable (so \(A\) is co-countable). In either case \(A^c\in\mathcal F\).

• Closed under countable unions:
  • Let \((A_n)_{n\ge1}\) be a sequence in \(\mathcal F\).
  • If at least one \(A_m\) has countable complement then

    \(\big(\bigcup_n A_n\big)^c=\bigcap_n A_n^c\subseteq A_m^c\),

    so the intersection is countable and the union is co-countable, hence in \(\mathcal F\).
  • If no \(A_n\) has countable complement then every \(A_n\) is countable; a countable union of countable sets is countable, so \(\bigcup_n A_n\) is countable and hence in \(\mathcal F\).
```

Therefore \(\mathcal F\) satisfies all three axioms and is a \(\sigma\)-field.

Step 2 — \(\mathcal F\) contains every singleton, so contains \(\mathcal C\)

• For any \(\omega\in\Omega\), the singleton \(\{\omega\}\) is finite and therefore countable; so \(\{\omega\}\in\mathcal F\).
• Hence \(\mathcal C\subseteq\mathcal F\), and so \(\mathcal F\) is a \(\sigma\)-field containing \(\mathcal C\).

Step 3 — Minimality: every \(\sigma\)-field containing \(\mathcal C\) must contain \(\mathcal F\)

• Let \(\mathcal G\) be any \(\sigma\)-field with \(\mathcal C\subseteq\mathcal G\).
• Because \(\mathcal G\) contains every singleton and is closed under countable unions, it contains every countable set (as a countable union of singletons).
• By closure under complements it also contains every co-countable set. Thus \(\mathcal F\subseteq\mathcal G\).

Step 4 — Conclude equality

• \(\mathcal F\) is a \(\sigma\)-field containing \(\mathcal C\) and is contained in every \(\sigma\)-field that contains \(\mathcal C\).
• Therefore \(\displaystyle \sigma(\mathcal C)=\mathcal F\): the \(\sigma\)-field generated by all singletons is exactly the countable / co-countable \(\sigma\)-field.

Special case — \(\Omega\) countable

• If \(\Omega\) is countable then every subset of \(\Omega\) is countable, so \(\mathcal F=\mathcal P(\Omega)\).
• Hence when \(\Omega\) is countable the \(\sigma\)-field generated by the singletons is the discrete \(\sigma\)-field \(\mathcal P(\Omega)\).

Question 3

Let \(\mathcal C_1\) and \(\mathcal C_2\) be two collections of events. If \(\mathcal C_1\subseteq\sigma(\mathcal C_2)\) and \(\mathcal C_2\subseteq\sigma(\mathcal C_1)\), show that \(\sigma(\mathcal C_1)=\sigma(\mathcal C_2)\).

1) Rigorous solution

Step 0 — recall a basic fact

• For any collection \(\mathcal C\subseteq\mathcal P(\Omega)\) we have the idempotence property

  \(\sigma(\sigma(\mathcal C))=\sigma(\mathcal C)\),

  because \(\sigma(\mathcal C)\) is already a \(\sigma\)-field containing \(\mathcal C\) and \(\sigma(\mathcal C)\) is the smallest such \(\sigma\)-field.

Step 1 — show \(\sigma(\mathcal C_1)\subseteq\sigma(\mathcal C_2)\)

• Assume \(\mathcal C_1\subseteq\sigma(\mathcal C_2)\).
• Apply the monotone operator \(\sigma(\cdot)\) to both sides to obtain

  \(\sigma(\mathcal C_1)\subseteq\sigma\big(\sigma(\mathcal C_2)\big).\)

• By idempotence \(\sigma(\sigma(\mathcal C_2))=\sigma(\mathcal C_2)\). Hence

  \(\sigma(\mathcal C_1)\subseteq\sigma(\mathcal C_2).\)

Step 2 — show \(\sigma(\mathcal C_2)\subseteq\sigma(\mathcal C_1)\)

• By symmetry, from \(\mathcal C_2\subseteq\sigma(\mathcal C_1)\) we get

  \(\sigma(\mathcal C_2)\subseteq\sigma\big(\sigma(\mathcal C_1)\big)=\sigma(\mathcal C_1).\)

Step 3 — conclude equality

• Combining inclusions,

  \(\sigma(\mathcal C_1)\subseteq\sigma(\mathcal C_2)\subseteq\sigma(\mathcal C_1)\),

  so \(\sigma(\mathcal C_1)=\sigma(\mathcal C_2)\).
• Justifications used: monotonicity of \(\sigma(\cdot)\) (intersection-of-all-\(\sigma\)-fields construction preserves inclusion) and idempotence (minimality of generated \(\sigma\)-fields).

Question 4

Let \(\Omega=\{0,1\}^{\mathbb N}\). Show that

\[ \mathcal C=\{\,A\times\Omega:\ A\subseteq\{0,1\}^n,\ n\in\mathbb N\,\}\cup\{\varnothing\} \]

(the class of finite-dimensional cylinder sets together with \(\varnothing\)) is closed under complements and finite unions.

Detailed solution (rigorous)

Notation & precise meaning

• Write an element \(\omega\in\Omega\) as a sequence \(\omega=(\omega_1,\omega_2,\dots)\) with \(\omega_i\in\{0,1\}\).
• For each \(n\in\mathbb N\) and each \(A\subseteq\{0,1\}^n\) define the corresponding cylinder set

  \(C(A,n):=\{\,\omega\in\Omega:\;(\omega_1,\dots,\omega_n)\in A\,\}.\)

• Equivalently \(C(A,n)=A\times\{0,1\}^{\{n+1,n+2,\dots\}}\), but we work with the coordinate description \(C(A,n)=\{\omega:(\omega_1,\dots,\omega_n)\in A\}\).
• The family \(\mathcal C\) is the set of all such \(C(A,n)\) together with \(\varnothing\).
• We must prove two things:
  • (I) Closed under complements: for every \(n\) and \(A\subseteq\{0,1\}^n\), \(C(A,n)^c\in\mathcal C\).
  • (II) Closed under finite unions: for any two cylinder sets \(C(A,m),C(B,n)\in\mathcal C\) their union belongs to \(\mathcal C\). From the pairwise case finite unions follow by induction.

(I) Proof — complements

• Fix \(n\) and \(A\subseteq\{0,1\}^n\). By definition

  \(C(A,n)=\{\omega\in\Omega : (\omega_1,\dots,\omega_n)\in A\}.\)

• Hence

  \(C(A,n)^c=\{\omega\in\Omega : (\omega_1,\dots,\omega_n)\notin A\} =\{\omega\in\Omega : (\omega_1,\dots,\omega_n)\in A^c\}=C(A^c,n).\)

• Since \(A^c\subseteq\{0,1\}^n\), \(C(A^c,n)\) is a cylinder set of the same form and therefore belongs to \(\mathcal C\). This holds for every \(A,n\).
• Also \(\varnothing^c=\Omega\) and \(\Omega\) is the cylinder corresponding to \(A=\{0,1\}^n\) for any \(n\). Thus \(\mathcal C\) is closed under complements.

(II) Proof — finite unions

• Let \(C(A,m)\) and \(C(B,n)\) be given with \(A\subseteq\{0,1\}^m\), \(B\subseteq\{0,1\}^n\). Put \(k=\max\{m,n\}\).
• We produce subsets \(A',B'\) of \(\{0,1\}^k\) such that

  \(C(A,m)=C(A',k),\qquad C(B,n)=C(B',k).\)

• Define for \(x=(x_1,\dots,x_k)\in\{0,1\}^k\) the projection \(\pi_m(x)=(x_1,\dots,x_m)\in\{0,1\}^m\). Set

  \(A' := \{x\in\{0,1\}^k:\ \pi_m(x)\in A\},\qquad B' := \{x\in\{0,1\}^k:\ \pi_n(x)\in B\}.\)

  (By convention if \(m=k\) then \(\pi_m\) is the identity and \(A'=A\), and similarly for \(n\).)
• Claim: \(C(A,m)=C(A',k)\).
• Proof of claim:

  \(\omega\in C(A,m)\) iff \((\omega_1,\dots,\omega_m)\in A\). Writing \(x=(\omega_1,\dots,\omega_k)\in\{0,1\}^k\) (note \(k\ge m\)), this is equivalent to \(\pi_m(x)\in A\), i.e. \(x\in A'\). Thus \(\omega\in C(A',k)\). The reverse implication is identical. So equality holds. The same argument gives \(C(B,n)=C(B',k)\).

• Therefore

  \(C(A,m)\cup C(B,n)=C(A',k)\cup C(B',k)=C(A'\cup B',k).\)

• Since \(A'\cup B'\subseteq\{0,1\}^k\), the right-hand side is a cylinder set belonging to \(\mathcal C\). Thus the union of the two arbitrary cylinder sets is again in \(\mathcal C\). By induction on the number of sets this argument extends to any finite union.

Conclusion

• Combining (I) and (II) we conclude that \(\mathcal C\) is closed under complements and finite unions, as required. (Trivially \(\varnothing\in\mathcal C\).)

Question 5

If \(\Omega=\mathbb R\) or \([0,\infty)\) or \([0,1]\) and \(\mathcal C\) is the collection of open subsets of \(\Omega\), then \(\mathcal B:=\sigma(\mathcal C)\) is the Borel \(\sigma\)-field. (Explain / justify.)

1) Rigorous solution

Goal

• Show that the \(\sigma\)-field generated by the open sets (the topology) is the standard Borel \(\sigma\)-field on the given space; and give convenient countable generators.

Definitions and setup

• Let \(\tau\) denote the topology of open sets on \(\Omega\) (the usual Euclidean topology when \(\Omega\subset\mathbb R\)).
• Define \(\mathcal B(\Omega):=\sigma(\tau)\), the smallest \(\sigma\)-field containing every open set. By definition this is the Borel \(\sigma\)-field on \(\Omega\).
• We will show standard alternative descriptions (useful for proofs and practice):
  • \(\mathcal B(\mathbb R)=\sigma\big(\{(a,b):a
  • \(\mathcal B(\mathbb R)=\sigma\big(\{(a,b):a
  • If \(\Omega=[0,\infty)\) or \(\Omega=[0,1]\) with the subspace topology, then \(\mathcal B(\Omega)=\{A\cap\Omega:A\in\mathcal B(\mathbb R)\}\) (subspace Borel \(\sigma\)-field).

(A) Open intervals generate \(\mathcal B(\mathbb R)\)

• Let \(\mathcal I:=\{(a,b):a
• Every open set \(U\subset\mathbb R\) is a union of open intervals: for each \(x\in U\) there exists an open interval \((a_x,b_x)\subset U\) with \(x\in(a_x,b_x)\), and \(U=\bigcup_{x\in U}(a_x,b_x)\).
• Hence \(\tau\subseteq\sigma(\mathcal I)\) because \(\sigma(\mathcal I)\) contains all unions of elements of \(\mathcal I\). Since \(\sigma(\tau)\) is the smallest \(\sigma\)-field containing \(\tau\), we have \(\mathcal B(\mathbb R)=\sigma(\tau)\subseteq\sigma(\mathcal I)\).
• Conversely, every open interval \((a,b)\) is an open set, so \(\mathcal I\subseteq\tau\), hence \(\sigma(\mathcal I)\subseteq\sigma(\tau)=\mathcal B(\mathbb R)\).
• Combining inclusions gives equality: \(\mathcal B(\mathbb R)=\sigma(\mathcal I)\).

(B) Rational-endpoint intervals give a countable generator

• Let \(\mathcal I_{\mathbb Q}:=\{(q_1,q_2):q_1,q_2\in\mathbb Q,\ q_1
• Claim. \(\sigma(\mathcal I_{\mathbb Q})=\mathcal B(\mathbb R)\).
• Proof. Every interval with real endpoints can be written as a union of rational-endpoint intervals:

  \((a,b)=\bigcup\{(q_1,q_2): q_1,q_2\in\mathbb Q,\ a Thus \(\mathcal I\subseteq\sigma(\mathcal I_{\mathbb Q})\), so \(\mathcal B(\mathbb R)=\sigma(\mathcal I)\subseteq\sigma(\mathcal I_{\mathbb Q})\). Conversely \(\mathcal I_{\mathbb Q}\subseteq\mathcal I\subseteq\mathcal B(\mathbb R)\), so \(\sigma(\mathcal I_{\mathbb Q})\subseteq\mathcal B(\mathbb R)\). Hence equality holds.

• This proves that the Borel \(\sigma\)-field has a countable generating family (useful for separability arguments and measure construction).

(C) Closed sets, half-open intervals, rays all generate the same Borel σ-field

• Because complements of open sets are closed, \(\sigma(\text{closed sets})=\sigma(\text{open sets})\).
• Rays \( (-\infty,a)\), \( (-\infty,a]\), \( (a,\infty)\), etc., generate the Borel σ-field. For instance

  \((a,b)=(-\infty,b)\cap (a,\infty)=(-\infty,b)\cap\big((-\infty,a)\big)^c,\)

  so \(\sigma(\{(-\infty,a):a\in\mathbb R\})\) contains all open intervals and hence equals \(\mathcal B(\mathbb R)\).
• In particular the countable family \(\{(-\infty,q):q\in\mathbb Q\}\) already generates \(\mathcal B(\mathbb R)\).

(D) Subspace cases \([0,\infty)\) and \([0,1]\)

• Equip \(\Omega\) with the subspace topology inherited from \(\mathbb R\). Let \(\tau_\Omega\) be the open sets in \(\Omega\); these are precisely \(U\cap\Omega\) with \(U\) open in \(\mathbb R\).
• Then

  \(\mathcal B(\Omega):=\sigma(\tau_\Omega)=\sigma\big(\{U\cap\Omega:U\text{ open in }\mathbb R\}\big).\)

• One standard identity is

  \(\mathcal B(\Omega)=\{A\cap\Omega: A\in\mathcal B(\mathbb R)\}.\)

  Proof sketch: every \(U\cap\Omega\) is of the form \(A\cap\Omega\) with \(A\in\mathcal B(\mathbb R)\) (since \(U\) is open hence Borel), so \(\sigma(\tau_\Omega)\subseteq\{A\cap\Omega:A\in\mathcal B(\mathbb R)\}\). Conversely, the right-hand side is a σ-field on \(\Omega\) that contains every \(U\cap\Omega\), hence it must contain \(\mathcal B(\Omega)\). Thus equality holds.

• Therefore the Borel σ-fields on \([0,\infty)\) and \([0,1]\) are exactly the restrictions of \(\mathcal B(\mathbb R)\) to those sets.

Conclusion

• The σ-field generated by the open sets on \(\Omega=\mathbb R\), \([0,\infty)\), or \([0,1]\) is the (standard) Borel σ-field \(\mathcal B(\Omega)\).
• Moreover, in \(\mathbb R\) it admits convenient countable generators such as \(\{(q_1,q_2):q_1,q_2\in\mathbb Q\}\) or \(\{(-\infty,q):q\in\mathbb Q\}\).

Question 6

On \(\mathbb R\), show that the \(\sigma\)-fields generated by the following families are all the same (namely the Borel \(\sigma\)-field):

• open intervals \((a,b),\ -\infty
• closed intervals \([a,b]\),
• half-open intervals \((a,b]\).

1) Rigorous solution (detailed, step-by-step, every inclusion justified)

Notation

• \(\displaystyle \mathcal O:=\{(a,b):a
• \(\displaystyle \mathcal C:=\{[a,b]:a\le b,\ a,b\in\mathbb R\}\).
• \(\displaystyle \mathcal H:=\{(a,b]:a
• We write \(\sigma(\mathcal O),\sigma(\mathcal C),\sigma(\mathcal H)\) for the \(\sigma\)-fields they generate. We prove \(\sigma(\mathcal O)=\sigma(\mathcal C)=\sigma(\mathcal H)\).

Step A. \(\sigma(\mathcal O)=\sigma(\mathcal C)\)

(A1) \(\sigma(\mathcal C)\subseteq\sigma(\mathcal O)\)

• Take a closed interval \([a,b]\in\mathcal C\). Its complement is

  \([a,b]^c=(-\infty,a)\cup(b,\infty)\),

  a union of open sets, hence open.
• Therefore \([a,b]\) is the complement of an open set, so \([a,b]\in\sigma(\mathcal O)\) (since \(\sigma(\mathcal O)\) contains all opens and is closed under complements).
• Monotonicity of \(\sigma(\cdot)\) gives \(\sigma(\mathcal C)\subseteq\sigma(\mathcal O)\).

(A2) \(\sigma(\mathcal O)\subseteq\sigma(\mathcal C)\)

• Fix an open interval \((a,b)\). For \(n\in\mathbb N\) set

  \(I_n := \big[a+\tfrac{1}{n},\,b-\tfrac{1}{n}\big]\) (when \(b-a>0\)).

• Each \(I_n\in\mathcal C\), and

  \((a,b)=\bigcup_{n=1}^\infty I_n\).

  Justification: any \(x\in(a,b)\) has distance to endpoints \(>0\), so choose \(n\) with \(1/n<\min\{x-a,b-x\}\) to get \(x\in I_n\).
• Thus \((a,b)\) is a countable union of closed intervals so \((a,b)\in\sigma(\mathcal C)\). Hence \(\sigma(\mathcal O)\subseteq\sigma(\mathcal C)\).

Conclusion of Step A

• Combining (A1) and (A2) yields \(\sigma(\mathcal O)=\sigma(\mathcal C)\).

Step B. \(\sigma(\mathcal O)=\sigma(\mathcal H)\)

(B1) \(\sigma(\mathcal H)\subseteq\sigma(\mathcal O)\)

• Fix \((a,b]\in\mathcal H\). Then

  \((a,b] = \bigcap_{n=1}^\infty (a,\,b+\tfrac{1}{n}).\)

• Each \((a,b+1/n)\) is open, so the countable intersection lies in \(\sigma(\mathcal O)\) (closure under complements + unions gives closure under countable intersections).
• Hence \(\sigma(\mathcal H)\subseteq\sigma(\mathcal O)\).

(B2) \(\sigma(\mathcal O)\subseteq\sigma(\mathcal H)\)

• Fix an open interval \((a,b)\). Using density of \(\mathbb Q\), write

  \((a,b)=\bigcup_{q\in\mathbb Q,\ a

• Each \((a,q]\in\mathcal H\) and the union is countable, so \((a,b)\in\sigma(\mathcal H)\). Thus \(\sigma(\mathcal O)\subseteq\sigma(\mathcal H)\).

Conclusion of Step B

• Combining (B1) and (B2) yields \(\sigma(\mathcal O)=\sigma(\mathcal H)\).

Final conclusion

• From Steps A and B we have \(\sigma(\mathcal O)=\sigma(\mathcal C)=\sigma(\mathcal H)\). This common \(\sigma\)-field is the Borel \(\sigma\)-field \(\mathcal B(\mathbb R)\).

2) Related concepts, lemmas, and common usages

• Monotonicity: If \(\mathcal A\subseteq\mathcal B\) then \(\sigma(\mathcal A)\subseteq\sigma(\mathcal B)\).
• Idempotence: \(\sigma(\sigma(\mathcal A))=\sigma(\mathcal A)\).
• Countable basis / rational endpoints: Intervals with rational endpoints give a countable generator for \(\mathcal B(\mathbb R)\).
• G\(_\delta\) and F\(_\sigma\) sets: Countable intersections of opens are \(G_\delta\); countable unions of closed sets are \(F_\sigma\). The proofs above used expressing sets as \(F_\sigma\) or \(G_\delta\) to show inclusion in generated σ-fields.

3) Viz (short intuition)

• Closed intervals are complements of unions of rays (open sets) → they are Borel.
• Open intervals are unions of slightly smaller closed intervals → they are Borel when closed intervals are generators.
• Half-open intervals are countable intersections of open intervals (a \(G_\delta\)), and conversely open intervals are countable unions of half-open intervals with rational endpoints; so all three families generate the same σ-field.

4) Other useful remarks (exam-ready takeaways)

• Succinct exam-style proof lines you can memorize:
  • \((a,b)=\bigcup_{n\ge1}[a+1/n,b-1/n]\) → \(\mathcal O\subset\sigma(\mathcal C)\).
  • \([a,b]=\big((-\infty,a)\cup(b,\infty)\big)^c\) → \(\mathcal C\subset\sigma(\mathcal O)\).
  • \((a,b]=\bigcap_{n\ge1}(a,b+1/n)\) and \((a,b)=\bigcup_{q\in\mathbb Q\cap(a,b)}(a,q]\) → \(\sigma(\mathcal O)=\sigma(\mathcal H)\).
• Remark: the (very small) countable π-system \(\{(-\infty,q):q\in\mathbb Q\}\) already generates \(\mathcal B(\mathbb R)\) and is useful in measure-uniqueness arguments (π–λ theorem).