Probability and Stochastics (Graduate Texts in Mathematics 261)

Erhan Cinlar

Chapter 1: Measure and Integration

Exercises

1.9   Partition generated $\sigma$-algebras.
$\quad \text{a)} \>$ Let $\mathcal{C} = \{A, B, C\}$ be a partition of $E$. List the elements of $\sigma \mathcal{C}$.
Solution
We start off with the elements that are members of all $\sigma$-algebras
$$\emptyset, E$$
Then we go on to list the elements of $\mathcal{C}$
$$A, B, C$$
We also need the complements of each of those
$$B \cup C, A \cup C, A \cup B$$
where we are using the fact that $\mathcal{C}$ is a partition to deduce that $E \setminus A = B \cup C$, etc. I would go on to list their unions but I would just be repeating elements (because $A \cup B \cup C = E$). So far what I have is
$$\sigma \mathcal{C} = \{\emptyset, A, B, C, B \cup C, A \cup C, A \cup B, E\}$$
This should be correct as it is closed under finite unions and complements, and is the smallest such algebra containing $\mathcal{C}$. $\square$

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$\quad \text{b)} \>$ Let $\mathcal{C}$ be a (countable) partition of $E$. Show that every element of $\sigma \mathcal{C}$ is a countable union of elements taken from $\mathcal{C}$. Hint: Let $\mathcal{E}$ be the collection of all sets that are countable unions of elements taken from $\mathcal{C}$. Show that $\mathcal{E}$ is a $\sigma$-algebra, and argue that $\mathcal{E} = \sigma \mathcal{C}$.
Solution
First let's use the fact that $\mathcal{C}$ is a countable partition of $E$, so we have $\mathcal{C} = \{C_i: i \in I\}$, such that $\bigcup_{i \in I} C_i = E$ and $i \neq j \implies C_i \cap C_j = \emptyset$. Let $\mathcal{E}$ be the collection of all sets that are countable unions of elements taken from $\mathcal{C}$. We proceed by showing that $\mathcal{E}$ is a $\sigma$-algebra.
Let $A \in \mathcal{E}$, so for some $J \subseteq I$ we have $A = \bigcup_{j \in J}C_j$. Let $B = \bigcup_{k \in I, k \notin J} C_k$. Clearly $B \in \mathcal{E}$, and propose that $E \setminus A = B$. Let $x \in E \setminus A$, since $x \in E$ and $\bigcup_{i \in I} C_i = E$ we have some $i \in I$ such that $x \in C_i$, and since $x \notin A$ we have $i \notin J$, so by our definition of $B$ we have $x \in B$. Alternatively, if $x \in B$ and $x \notin E \setminus A$ then $x \in A \cap B = \emptyset$, so we can not have this. We justify $A \cap B = \emptyset$ as follows: let $x \in A \cap B$. Then for some $j \in J$ we have $x \in C_j$ and for some $k \notin J$ we have $x \in C_k$. Since $\mathcal{C}$ is a partition and $k \neq j$ we have $C_j \cap C_k = \emptyset$. Anyways, since $E \setminus A = B \in \mathcal{E}$ for all $A \in \mathcal{E}$, we have shown $\text{1.3 a)}$.
Showing $\text{1.3 b)}$ is trivial, as the countable union of a countable union is a countable union.
Combining $\text{1.3 a)}$ and $\text{1.3 b)}$ shows that $\mathcal{E}$ is a $\sigma$-algebra.
To finish the proof we need to show that $\mathcal{E} = \sigma \mathcal{C}$. It is obvious that $\mathcal{E}$ contains $\mathcal{C}$ and we know that $\mathcal{E}$ is a $\sigma$-algebra so it is obvious that $\sigma \mathcal{C} \subseteq \mathcal{E}$. Because a $\sigma$-algebra is by definition closed under countable unions we must also have $\mathcal{E} \subseteq \sigma \mathcal{C}$, so $\mathcal{E} = \sigma \mathcal{C}$. $\square$

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$\quad \text{c)} \>$ Let $E = \R$, the set of all real numbers. Let $\mathcal{C}$ be the collection of all singleton subsets of $\R$, that is, each element of $\mathcal{C}$ is a set that consists of exactly one point in $\R$. Show that every element of $\sigma \mathcal{C}$ is either a countable set or the complement of a countable set. Incidentally, $\sigma \mathcal{C}$ is much smaller than $\mathcal{B}(\R)$; for instance, the interval $(0, 1)$ belongs to the latter but not to the former.
Solution
Let $\mathcal{E}$ be the set of all finite unions of singletons and their complements. We propose that $\sigma \mathcal{C} = \mathcal{E}$.
First we show that $\mathcal{E}$ is indeed a $\sigma$-algebra. By definition this means showing that

• $A \in \mathcal{E} \implies \R \setminus A \in \mathcal{E}$
  This is obvious from construction.
• $A_1, A_2, \dots \in \mathcal{E} \implies \bigcup_{n}A_n \in \mathcal{E}$
  Some examples
  $\{1,\frac{1}{2}\} \cup \{\pi\} = \{1, \frac{1}{2}, \pi\}$, which is countable
  $\{1,2,3, \dots,\} \cup \{2\}^c = \R$, which is the complement of $\emptyset$
  Let $I$ be a countable set and consider $\bigcup_{i \in I}A_i$. Let $I = U \cup V$, where this union is disjoint and if $u \in U$ then $A_u$ is countable and if $v \in V$ then $A_v$ is the complement of a countable set. Then
  $$\bigcup_{i \in I} A_i = \bigcup_{u \in U} A_u \cup \bigcup_{v \in V} A_v$$
  Since each $A_v$ is the complement of a countable set we can write $A_v = (B_v)^{c}$ where $B_v$ is a countable subset of $\R$. Plugging in
  $$= \bigcup_{u \in U}A_u \cup \bigcup_{v \in V} (B_v)^{c}$$
  applying DeMorgan's Law
  $$= \bigcup_{u \in U}A_u \cup \left(\bigcap_{v \in V} B_v\right)^{c}$$
  If $V$ is empty then $\bigcap_{v\in V}B_v = \R$ and so the whole union is just the countable $\bigcup_{u \in U} A_u$. Otherwise $\bigcap_{v\in V}B_v$ is countable and so we are left with the union of a countable set and the complement of a countable set, which is the complement of a countable set.

Anyways after having showed that $\mathcal{E}$ is a $\sigma$-algebra the rest of this problem is trivial. $\mathcal{C} \subset \mathcal{E}$ is obvious by construction, so by the definition of a $\sigma$-algebra being generated we have $\sigma \mathcal{C} \subset \mathcal{E}$. $\mathcal{E} \subset \sigma \mathcal{C}$ is trivial because $\sigma$-algebras are closed under finite unions and complements. $\quad \square$

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1.10   Comparisons. Let $\mathcal{C}$ and $\mathcal{D}$ be two collections of subsets of $E$. Show the following:
$\quad \text{a)} \quad$ If $\mathcal{C} \subset \mathcal{D}$ then $\sigma \mathcal{C} \subset \sigma \mathcal{D}$
$\quad \text{b)} \quad$ If $\mathcal{C} \subset \sigma \mathcal{D}$ then $\sigma C \subset \sigma D$
$\quad \text{c)} \quad$ If $C \subset \sigma \mathcal{D}$ and $\mathcal{D} \subset \sigma \mathcal{C}$, then $\sigma \mathcal{C} = \sigma \mathcal{D}$
$\quad \text{d)} \quad$ If $\mathcal{C} \subset \mathcal{D} \subset \sigma \mathcal{C}$, then $\sigma \mathcal{C} = \sigma \mathcal{D}$

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1.11   Borel $\sigma$-algebra on $\R$. Every open subset of $\R = (-\infty, +\infty)$, the real line, is a countable union of open intervals. Use this fact to show that $\mathcal{B}_\R$ is generated by the collection of all open intervals.

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1.12   Continuation. Show that every interval of $\R$ is a Borel set. In particular, $(-\infty, x)$, $(-\infty,x]$, $(x,y]$, $[x,y]$ are all Borel sets. For each $x$, the singleton $\{x\}$ is a Borel set.

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1.13   Continuation. Show that $\mathcal{B}_\R$ is also generated by any one of the following (and many others):
$\quad \text{a)} \quad$ The collection of all intervals of the form $(-\infty, x]$.
$\quad \text{b)} \quad$ The collection of all intervals of the form $(x,y]$.
$\quad \text{c)} \quad$ The collection of all intervals of the form $[x,y]$.
$\quad \text{d)} \quad$ The collection of all intervals of the form $(x, \infty)$.

Moreover, in each case, $x$ and $y$ can be limited to rationals.

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1.14   Lemma 1.7. Prove.

1.7   Lemma. Let $\mathcal{D}$ be a d-system on $E$. Fix $D$ in $\mathcal{D}$ and let

$$\hat{\mathcal{D}} = \{A \in \mathcal{D}: A \cap D \in \mathcal{D}\}$$

Then, $\hat{\mathcal{D}}$ is again a d-system.

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1.15   Trace spaces. Let $(E, \mathcal{E})$ be a measurable space. Fix $D \subset E$ and let

$$\mathcal{D} = \mathcal{E} \cap D = \{A \cap D: A \in \mathcal{E}\}.$$

Show that $\mathcal{D}$ is a $\sigma$-algebra on $D$. It is called the trace on $\mathcal{E}$ on $D$, and $(D,\mathcal{D})$ is called the trace of $(E,\mathcal{E})$ on $D$.

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