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1

Let $\mathcal F$ be a field and suppose that $\mathcal F$ is also closed under countable disjoint unions. Show that $\mathcal F$ is a $\sigma$-field.

Take $A_1, A_2, \dots \in \mathcal F$. Then define the following disjoint unions

\[\begin{align} D_1 & = A_1 \\ D_2 & = A_1 \setminus A_2 \\ D_3 & = A_1 \setminus A_2 \setminus A_3 & \vdots \end{align}\]

Since $\mathcal F$ is closed under countable disjoint unions $\bigcup_{i=1}^\infty D_i \in \mathcal F$. But

\[\bigcup_{i=1}^{\infty}D_i = \bigcup_{i=1}^\infty A_i \in \mathcal F.\]

So, $\mathcal F$ is closed under countable union and is a $\sigma$-field.

2

Let $\mathcal F$ be a class of subsets of $\Omega$ containing $\Omega$ and having the property that if $A,B \in \mathcal F$, then $A \setminus B = A \cap B^C \in \mathcal F$. Show that $\mathcal F$ is a field.

Proof

Notice that $\Omega \setminus \Omega = \varnothing \in \mathcal F$, so $\Omega, \varnothing \in \mathcal F$. ✅

Now take $A, B \in \mathcal F$. Then $A \setminus B = A \cap B^C \in \mathcal F$. Then,

\[\begin{align} A \setminus (A \cap B^C) & = A \cap \left( A \cap B^C \right)^C \\ & = A \cap \left( A^C \cup B \right) \\ & = \left( A \cap A^C \right) \cup \left( A \cap B \right) \\ & = \varnothing \cup A \cap B \\ & = A \cap B. ✅ \end{align}\]

Finally, take $A \in \mathcal F$. Then $\Omega \setminus A = A^C \in \mathcal F$. ✅

Thus, $\mathcal F$ is a field.

3

Let $\mathcal A$ be a class of subsets of $\Omega$ with properties that (i) $\Omega \in \mathcal A$, (ii) $A \in \mathcal A$ implies that $A^C \in \mathcal A$, and (iii) $\mathcal A$ is closed under finite disjoint unions, i.e., $A_1, \dots A_k \in \mathcal A$, disjoint, then $\bigcup_{i=1}^kA_i \in \mathcal A$. Show by example that $\mathcal A$ need not be a field.

We want to come up with an example where $A, B \in \mathcal A$.

Take $\Omega = \left\{ 1,2,3,4 \right\}$ and $\mathcal A = \left\{ \varnothing, \left\{ 1,2 \right\}, \left\{ 3,4 \right\}, \left\{ 1,3 \right\}, \left\{ 2,4 \right\},\Omega \right\}$. Notice that for all $A \in \mathcal A$ , $A^C \in \mathcal A$ and that $\mathcal A$ is closed under disjoint unions. However, $\left\{ 1,2 \right\} \cap \left\{ 1,3 \right\} = \left\{ 1 \right\} \notin \mathcal A$.

4

Let $\mathcal F_n$, $n = 1, 2, \dots$ be a sequence of fields on $\Omega$ and that $\mathcal F_n \subset \mathcal F_{n+1}$ for all $n$. Show that $\bigcup_{n=1}^\infty \mathcal F_n$ is also a field.

Proof

Since $\mathcal F_1$ is a field, $\varnothing, \Omega \in \mathcal F_1$. Hence, $\varnothing, \Omega \in \bigcup_{n=1}^\infty \mathcal F_n$. ✅

Take $A, B \in \bigcup_{n=1}^\infty \mathcal F_n$. Then, without loss of generality, $\exists k$ such that $A \in \mathcal F_k$ and $\exists m \geq k$ such that $B \in \mathcal F_m$. Since $\mathcal F_k \subset \mathcal F_m$, $A \in \mathcal F_m$ and $A \cup B \in \mathcal F_m$. So, $A \cap B \in \bigcup_{n=1}^\infty \mathcal F_n$. ✅

Take $A \in \bigcup_{n=1}^\infty \mathcal F_n$. Then $\exists k$ such that $A \in \mathcal F_k$. Since $\mathcal F_k$ is a field, $A^C \in \mathcal F_k$. Thus, $A^C \in \bigcup_{n=1}^\infty \mathcal F_n$.

If $\mathcal F_n$, $n=1, 2, \dots$, is a sequence of $\sigma$-fields on $\Omega$ and that $\mathcal F_n \subset \mathcal F_{n+1}$ for all $n$, argue that $\bigcup_{n=1}^\infty \mathcal F_n$ is always a field, but construct an example to show that $\bigcup_{n=1}^\infty \mathcal F_n$ need not be a $\sigma$-field.

We need a counter example where $\bigcup_{n=1}^\infty \mathcal F_n$ is closed under intersection, but not under countable union. Take $\mathcal \Omega = \mathbb N$. Define $\mathcal F_i$ as

\[\begin{align} \mathcal F_1 & = \left\{ \varnothing, \left\{ 1 \right\}, \mathbb N - \left\{ 1 \right\}, \Omega \right\} \\ \mathcal F_2 & = \left\{ \varnothing, \left\{ 1 \right\}, \left\{ 2 \right\}, \left\{ 1,2 \right\}, \mathbb N - \left\{ 1 \right\}, \mathbb N - \left\{ 2 \right\}, \mathbb N - \left\{ 1,2 \right\}, \Omega \right\} \\ & \vdots \end{align}\]

Notice that these are all $\sigma$-fields. Take $x_i = 2i$, then $x_1, x_2, \dots \in \bigcup_{i=1}^\infty \mathcal F_i$, however, $\bigcup_{i=1}^\infty x_i \notin \bigcup_{i=1}^\infty \mathcal F_i$. So, $\bigcup_{i=1}^\infty \mathcal F_i$ is not a $\sigma$-field.

5

Let $\mathcal B_1$, $\mathcal B_2$ be $\sigma$-fields on $\Omega$. Then show that

\[\sigma \langle \mathcal B_1 \cup \mathcal B_2 \rangle = \sigma \langle \left\{ B_1 \cap B_2: B_1 \in \mathcal B_1, B_2 \in \mathcal B_2 \right\} \rangle\]

We will first show that

\[\sigma \langle \mathcal B_1 \cup \mathcal B_2 \rangle \subset \sigma \langle \left\{ B_1 \cap B_2: B_1 \in \mathcal B_1, B_2 \in \mathcal B_2 \right\} \rangle.\]

Since $\mathcal B_1$ is a $\sigma$-field, $B_1 \in \mathcal B_1 \Rightarrow B_1^C \in \mathcal B_1$. Similarly for $\mathcal B_2$. Thus,

\[\sigma \langle \left\{ B_1 \cap B_2: B_1 \in \mathcal B_1, B_2 \in \mathcal B_2 \right\} \rangle = \langle \dots, \left\{ B_1 \cap B_2 \right\} , \left\{ B_1 \cup B_2^C \right\} , \left\{ B_1^C \cap B_2 \right\} \rangle.\]

Notice that these first two sets contain all of $B_1$ and the first and third set contain all of $B_2$. Thus, for any $B_1 \in \mathcal B_1$ and $B_2 \in \mathcal B_2$,

\[B_1 \cup B_2 \in \sigma \langle \left\{ B_1 \cap B_2: B_1 \in \mathcal B_1, B_2 \in \mathcal B_2 \right\} \rangle\]

since $\sigma$-fields are closed under countable unions.

Now we must show

\[\sigma \langle \mathcal B_1 \cup \mathcal B_2 \rangle \supset \sigma \langle \left\{ B_1 \cap B_2: B_1 \in \mathcal B_1, B_2 \in \mathcal B_2 \right\} \rangle.\]

Again take $B_1 \in \mathcal B_1 \Rightarrow B_1^C \in \mathcal B_1$ and $B_2, B_2^C \in \mathcal B_2$. Then $B_1^C \cup B_2^C \in \mathcal B_1 \cup \mathcal B_2$ and $(B_1^C \cup B_2^C)^C = B_1 \cap B_2 \in \mathcal B_1 \cup \mathcal B_2$. Thus,

\[\sigma \langle \mathcal B_1 \cup \mathcal B_2 \rangle \supset \sigma \langle \left\{ B_1 \cap B_2: B_1 \in \mathcal B_1, B_2 \in \mathcal B_2 \right\} \rangle.\]

Hence,

\[\sigma \langle \mathcal B_1 \cup \mathcal B_2 \rangle = \sigma \langle \left\{ B_1 \cap B_2: B_1 \in \mathcal B_1, B_2 \in \mathcal B_2 \right\} \rangle.\]