4. Measurable Spaces

Example (Trivial $\sigma$-field): Let $\Omega \neq \varnothing$ be an arbitrary set. Let

\sigmaF = \qty{\varnothing, \Omega}

. Then, $\sigmaF$ is a $\sigma$-field on $\Omega$.

1. Clearly, $\Omega \in \sigmaF$.
2. Let $A \in \sigmaF$. Then, there are only two cases, eiter $A = \varnothing$ or $A = \Omega$. In each case, $A^c \in \sigmaF$.
3. Consider a countable collection

   \qty{A_i : i \in \N} \subset \sigmaF

   . This collection is composed only of the sets $A_i = \varnothing$ or $A_i = \Omega$, $i \in \N$. Thus $$\bigcup_{i \in \N} A_i = \begin{cases} \Omega & \text{if } \exists i \text{ s.t. } A_i = \Omega \\ \varnothing & \text{otherwise} \end{cases}$$ and $\bigcup_{i \in \N} A_i \subset \sigmaF$.

Example (Power set): Let $\Omega \neq \varnothing$ be an arbitrary set. Let $\sigmaF$ be the family which consists of all possible subsets of $\Omega$, i.e.

\sigmaF = \qty{A : A \subset \Omega}

. Then, $\sigmaF$ is a $\sigma$-field on $\Omega$.

1. Since $\Omega \subset \Omega$, $\Omega \in \sigmaF$.
2. Let $A \in \sigmaF$, then by definition $A^c = \Omega \setminus A \subset \Omega$ and hence $A^c \in \sigmaF$.
3. Leet

   \qty{A_i : i \in \N} \subset \sigmaF

   . Then, by definition

   \bigcup_{i \in \N} A_i = \qty{\omega \in \Omega : \exists i \text{ s.t. } \omega \in A_i} \subset \Omega

   and $\bigcup_{i \in \N} A_i \subset \sigmaF$.

Example (Uncountable $\Omega$, countable $A, A^c$): Let $\Omega$ be an uncountable set. We consider the family

\sigmaF = \qty{A: A \subset \Omega \text{ s.t. } A \text{ is countable or } A^c \text{ is countable}}

Then, $\sigmaF$ is a $\sigma$-field on $\Omega$.

1. We have that $\Omega^c = \varnothing$. We know that $\#\varnothing = 0$, in particular $\varnothing$ is countable. Thus, $\Omega \in \sigmaF$.
2. Let $A \in \sigmaF$. Thus, $A$ is countable or $A^c$ is countable. Since $(A^c)^c = A$, $A^c \in \sigmaF$.
3. Let

   \qty{A_i : i \in \N} \subset \sigmaF

   . If there exists $j \in \N$ s.t. $A_j$ is uncountable, we have that

   \qty(\bigcup_{i \in \N} A_i)^c = \bigcap_{i \in \N} A_i^c = \bigcap_{i \in \N, \ i \neq j} A_i^c \cap A_j^c \subset A_j^c

   hence, using the fact that $A_j^c$ must be countable, we have

   \#\qty(\bigcup_{i \in \N} A_i)^c \leq \# A_j^c \leq \# \N

   and $\bigcup_{i \in \N} A_i \in \sigmaF$. If for all $i \in \N$ we have that $A_i$ is countable we rely on TODO: reference Proposition (Union of countable sets) and conclude that $\bigcup_{i \in \N} A_i$ must be countable as well and hence $\bigcup_{i \in \N} A_i \in \sigmaF$.

Example ($\sigma$-field restricted on subset): Let $\Omega$ be a non empty set and $\sigmaF$ be a $\sigma$-field on $\Omega$. Let $\Omega_0 \subset \Omega$ s.t. $\Omega_0 \neq \varnothing$. Then, the collection

\sigmaF \cap \Omega_0 = \qty{A \cap \Omega_0 : A \in \sigmaF}

is a $\sigma$-field on $\Omega_0$.

1. Clearly, $\Omega_0 = \Omega \cap \Omega_0 \in \sigmaF \cap \Omega_0$.
2. If $B \in \sigmaF \cap \Omega_0$, then $B = A \cap \Omega_0$ for some $A \in \sigmaF$ and $\Omega_0 \setminus B = (\Omega_0 \setminus A) \cup (\Omega_0 \setminus \Omega_0) = \Omega_0 \setminus A = A^c \cap \Omega_0 \in \sigmaF \cap \Omega_0$.
3. Suppose that

   \qty{B_i : i \in \N} \subset \sigmaF \cap \Omega_0

   . Therefore, for any $i \in \N$, $B_i = A_i \cap \Omega_0$ for some $A_i \in \sigmaF$. Then, since $\bigcup_{i \in \N} B_i = (\bigcup_{i\in\N} A_i) \cap \Omega_0$ and $\bigcup_{i\in\N} A_i \in \sigmaF$, it follows that $\bigcup_{i \in \N} B_i \in \sigmaF \cap \Omega_0$.

Example (Infinte $\Omega$, finite $A, A^c$): Let $\Omega$ be an infinite set. Define the family

\set{G} = \qty{A : A \subset \Omega \text{ s.t. } A \text{ is finite or } A^c \text{ is finite}}

Then, $\set{G}$ is not a $\sigma$-field on $\Omega$. To see this, let

\qty{\omega_i : i \in \N}

be a countably infinite sequence of distinct points of $\Omega$ and define the set

A = \qty{\omega_{2i} : i \in \N}

. We note that both $A$ and, since

\qty{\omega_{2i + 1} : i \in \N} \subset A^c

, $A^c$ are not finite. Let

A_i = \qty{\omega_{2i}}

s.t. $A = \bigcup_{i \in \N} A_i$. It is clear that $A_i \in \set{G}$ but $A \not\in \set{G}$, hence $\set{G}$ is not a $\sigma$-field on $\Omega$.

Example (Left-open real intervals): Let $\Omega = \R$ and define the family

\set{R} = \qty{A = (a,b]: a,b \in \R} \cup \qty{\varnothing}

Then, $\set{R}$ is not a $\sigma$-field on $\Omega$. To see this, let $A = (a,b] \in \set{R}$ with $A \neq \varnothing$, then $A^c = (-\infty,a] \cup (b,\infty) \not \in \set{R}$.

Example ($\sigma$-field restricted on subset continued): We show $\Omega_0 \in \sigmaF$ iff $\sigmaF \cap \Omega_0$ is a sub-$\sigma$-field of $\sigmaF$. We prove the direction “$\Omega_0 \in \sigmaF$” $\implies$ “$\sigmaF \cap \Omega_0$ is a sub-$\sigma$-field of $\sigmaF$” first. We know that

\sigmaF \cap \Omega_0 = \qty{A \cap \Omega_0 : A \in \sigmaF}

and from an exercise (TODO, make proposition) we know that if $A \in \sigmaF$ and $\Omega_0 \in \sigmaF$, then $A \cap \Omega_0 \in \sigmaF$, thus $\sigmaF \cap \Omega_0 \subset \sigmaF$. We already proved before that $\sigmaF \cap \Omega_0$ is a $\sigma$-field. For the direction “$\sigmaF \cap \Omega_0$ is a sub-$\sigma$-field of $\sigmaF$” $\implies$ “$\Omega_0 \in \sigmaF$” we note that $\Omega_0 \in \sigmaF \cap \Omega_0$ as $\Omega \cap \Omega_0 = \Omega_0$. Hence $\Omega_0 \in \sigmaF$ follows from $\sigmaF \cap \Omega_0 \subset \sigmaF$.

Example (

\set{G} = \qty{\varnothing}

): Let $\Omega \neq \varnothing$ and let

\set{G} = \qty{\varnothing}

. Then

\sigma(\set{G}) = \qty{\varnothing, \Omega}

, i.e. the trivial $\sigma$-field on $\Omega$. By 1. of Proposition 4.6 it is enough to show that

\qty{\varnothing, \Omega} \subset \sigma(\set{G})

since

\qty{\varnothing, \Omega}

is a $\sigma$-field that contains $\set{G}$. It is clear that

\qty{\varnothing, \Omega} \subset \sigma(\set{G})

is a $\sigma$-field, hence it must contain both, $\varnothing$ and $\varnothing^c = \Omega$.

Example (

\set{G} = \qty{\qty{1}}

): Let

\Omega = \qty{1,2,3}

and define

\set{G} = \qty{\qty{1}}

. Clearly, $\sigma(\set{G})$ must contain

\qty{1}

,

\qty{1}^c = \qty{2,3}

, $\Omega$ and $\varnothing$. Thus we claim

\sigma(\set{G}) = \qty{\varnothing, \qty{1}, \qty{2,3}, \Omega}

. More generally, if $A \subset \Omega$, then

\sigma(\qty{A}) = \qty{\varnothing, A, A^c, \Omega}

.

Example (

\set{G} = \qty{\qty{\omega}: \omega \in \Omega}

): Let $\Omega$ be an uncountable set and

\sigmaF = \qty{A: A \subset \Omega \text{ s.t. } A \text{ is countable or } A^c \text{ is countable}}

be the $\sigma$-field introduced in a previous example. We show that $\sigmaF = \sigma(\set{G})$ where

\set{G} = \qty{\qty{\omega}: \omega \in \Omega}

. Clearly $\set{G} \subset \sigmaF$ since each set

\qty{\omega}

has cardinality one. Thus, it remains to show that $\sigmaF \subset \sigma(\set{G})$. Let $A \in \sigmaF$. Then either $A$ or $A^c$ is countable. Suppose that $A$ is countable, then

A = \qty{\omega_i : i \in \N}

for some collection of singletons $\omega_i \in \Omega$. Therefore,

A = \bigcup_{i \in \N}\qty{\omega_i}

and thus $A \in \sigma(\set{G})$ since

\qty{\omega_i} \in \sigma(\set{G})

. If $A^c$ is countable, by the latter argument, $A^c \in \sigma(\set{G})$ and hence $A = (A^c)^c \in \sigma(\set{G})$.