Exercises - About

Chapter 2B¶

Exercise 2B.3

Suppose $\mathcal S$ is the smallest $\sigma-algebra$ on $\mathbb R$ containing $\{(r,s] : r,s \in \mathbb Q\}$ . Prove that $\mathcal S$ is the collection of Borel subsets of $\mathbb R$ .

We prove that $\mathcal S = \mathcal B(\mathbb R)$ , where $\mathcal B(\mathbb R)$ is the smallest $\sigma-algebra$ on $\mathbb R$ containing all open subsets of $\mathbb R$ and it is called the collection of Borel subsets of $\mathbb R$ .

Step 1. $\mathcal S \subset \mathcal B(\mathbb R)$ .

Fix $r,s \in \mathbb Q$ . Observe that

(r,s] = \bigcap_{k=1}^{\infty} (r, s + 1/k).

(1)

Each interval $(r, s + 1/k)$ is open, hence belongs to $\mathcal B(\mathbb R)$ . Because $\mathcal B(\mathbb R)$ is a $\sigma-algebra$ , it follows that $(r,s] \in \mathcal B(\mathbb R)$ . Since $\mathcal S$ is the smallest $\sigma-algebra$ containing all such intervals, we conclude that

\mathcal S \subset \mathcal B(\mathbb R).

(2)

Step 2. Rational approximation of real numbers.

Let $x \in \mathbb R^+$ . Define $r_n(x)$ as the truncation of $x$ at $n$ digits after the decimal point. Then:

$r_n(x) \in \mathbb Q$ ,
$r_n(x)$ is increasing,
$|x - r_n(x)| \le 10^{-n}$ .

Define $s_n(x)$ as the truncation of $x$ at $n$ digits after the decimal point, followed by adding $10^{-n}$ . Then:

$s_n(x) \in \mathbb Q$ ,
$s_n(x)$ is decreasing,
$|x - s_n(x)| \le 10^{-n}$ .

Hence $r_n(x) \uparrow x$ and $s_n(x) \downarrow x$ . If $x < 0$ , we use $-r_n(-x) \downarrow x$ and $-s_n(-x) \uparrow x$ .

Step 3. Representation of open intervals.

Let $a < b$ .

If $a,b > 0$ , then

(a,b) = \bigcup_{n=1}^{\infty} (\, s_n(a), \, r_n(b) \,].

(3)

If $a < 0 < b$ , then

(a,b) = \bigcup_{n=1}^{\infty} (\, -r_n(-a), \, r_n(b) \,].

(4)

The case $a,b<0$ follows by the same argument, using the sequences $-r_n(-x)$ and $-s_n(-x)$ .

We now treat an interval with an infinite endpoint. Let $a \in \mathbb R$ .

If $a > 0$ , then

(a,+\infty) = \bigcup_{m=1}^{\infty} \ \bigcup_{n=1}^{\infty} (\, s_n(a), \, m \,].

(5)

If $a < 0$ , a similar construction using $-r_n(-a)$ yields a representation of $(a,+\infty)$ as a countable union of intervals $(r,s]$ with rational endpoints.

Similarly, the case $(-\infty,a)$ follow by the same argument, using $r_n(x)$ for $a>0$ and $-s_n(-x)$ for $a<0$ .

Thus every open interval of $\mathbb R$ is a countable union of elements of $\{(r,s] : r,s \in \mathbb Q\}$ , and hence belongs to $\mathcal S$ .

Step 4. Conclusion.

Every open subset of $\mathbb R$ is a countable union of open intervals. Therefore every open set belongs to $\mathcal S$ , which implies

\mathcal B(\mathbb R) \subset \mathcal S.

(6)

Combining this with Step 1 yields

\mathcal S = \mathcal B(\mathbb R).

(7)

$\blacksquare$

Exercise 2B.4

Suppose $\mathcal S$ is the smallest $\sigma$ -algebra on $\mathbb R$ containing $\{(r,n] : r \in \mathbb Q,\ n \in \mathbb Z\}$ . Prove that $\mathcal S$ is the collection of Borel subsets of $\mathbb R$ .

Exercise 2B.7

Prove that the collection of Borel subsets of $\mathbb R$ is translation invariant. More precisely, prove that if $B \subset \mathbb R$ is a Borel set and $t \in \mathbb R$ , then $t + B$ is a Borel subset of $\mathbb R$ .

Exercise 2B.9

Give an example of a measurable space $(X,\mathcal S)$ and a function $f : X \to \mathbb R$ such that $|f|$ is $\mathcal S$ -measurable but $f$ is not $\mathcal S$ -measurable.

Define $\mathcal S = \{ A \cup B : A \subseteq [0,+\infty),\ B \in \{\varnothing,(-\infty,0)\} \}.$

Step 1. $\mathcal S$ is a $\sigma$ -algebra on $\mathbb R$

$\varnothing = \varnothing \cup \varnothing \in \mathcal S$ , and
$\mathbb R = (-\infty,0) \cup [0,+\infty) \in \mathcal S.$
(18)
Let $C \in \mathcal S$ with $C = A \cup B$ , where $A \subseteq [0,+\infty)$ and $B \in \{\varnothing,(-\infty,0)\}$ . Then
$\mathbb R \setminus C = (\mathbb R \setminus A) \cap (\mathbb R \setminus B).$
(19)
Since
$\mathbb R \setminus A = (-\infty,0) \cup ([0,+\infty)\setminus A),$
(20)
and
$\mathbb R \setminus B = \begin{cases} \mathbb R & \text{if } B=\varnothing,\\ [0,+\infty) & \text{if } B=(-\infty,0), \end{cases}$
(21)
it follows that $\mathbb R \setminus C$ is again of the form $A' \cup B'$ with $A' \subseteq [0,+\infty)$ and $B' \in \{\varnothing,(-\infty,0)\}$ . Hence $\mathbb R \setminus C \in \mathcal S$ .
Let $(C_k)_{k\ge1}$ be a sequence of elements of $\mathcal S$ with $C_k = A_k \cup B_k$ . Then
$\bigcup_{k=1}^\infty C_k = \left( \bigcup_{k=1}^\infty A_k \right) \cup \left( \bigcup_{k=1}^\infty B_k \right).$
(22)
Since $A_k \subseteq [0,+\infty)$ for all $k$ and $\bigcup_k B_k \in \{\varnothing,(-\infty,0)\}$ , we have $\bigcup_k C_k \in \mathcal S$ .

Therefore, $\mathcal S$ is a $\sigma$ -algebra on $\mathbb R$ .

Step 2. Definition of the function.

Define $f : \mathbb R \to \mathbb R$ by

f(x)= \begin{cases} -1 & \text{if } x \le -1,\\ 1 & \text{if } x > -1. \end{cases}

(23)

Step 3. $f$ is not $\mathcal S$ -measurable.

Consider the Borel set $\{1\} \subset \mathbb R$ . Then

f^{-1}(\{1\}) = (-1,+\infty).

(24)

This set contains some negative values but not all of $(-\infty,0)$ . By definition of $\mathcal S$ , such a set does not belong to $\mathcal S$ . Hence $f$ is not $\mathcal S$ -measurable.

Step 4. $|f|$ is $\mathcal S$ -measurable.

For all $x \in \mathbb R$ , $|f(x)| = 1$ . Thus $|f|$ is constant, and therefore $\mathcal S$ -measurable.

Hence $|f|$ is $\mathcal S$ -measurable but $f$ is not.

$\blacksquare$

Exercise 2B.16

Suppose $(X,\mathcal S)$ is a measurable space and $A \in \mathcal S$ .

Define

\mathcal S_A = \{ E \in \mathcal S : A \subset E \text{ or } A \cap E = \varnothing \}.

(25)

(a) Prove that $\mathcal S_A$ is a $\sigma$ -algebra on $X$ .

(b) Prove that a function $f : X \to \mathbb R$ is $\mathcal S_A$ -measurable if and only if $f$ is $\mathcal S$ -measurable and $f$ is constant on $A$ .

(a) $\mathcal S_A$ is a $\sigma$ -algebra.

Step 1. $\varnothing, X \in \mathcal S_A$ .

Since $A \cap \varnothing = \varnothing$ , we have $\varnothing \in \mathcal S_A$ .
Since $A \subset X$ , we have $X \in \mathcal S_A$ .

Step 2. Closed under complements.

Let $E \in \mathcal S_A$ .

Then either $A \subset E$ or $A \cap E = \varnothing$ .

If $A \subset E$ , then $A \cap (X \setminus E) = \varnothing$ , hence $X \setminus E \in \mathcal S_A$ .
If $A \cap E = \varnothing$ , then $A \subset X \setminus E$ , hence $X \setminus E \in \mathcal S_A$ .

Thus $\mathcal S_A$ is closed under complements.

Step 3. Closed under countable unions.

Let $(E_k)_{k\ge1}$ be a sequence of elements of $\mathcal S_A$ .

We consider two cases.

If there exists $k$ such that $A \subset E_k$ , then $A \subset \bigcup_{k=1}^\infty E_k$ , so the union belongs to $\mathcal S_A$ .
Otherwise, for every $k$ we have $A \cap E_k = \varnothing$ . Then
$A \cap \left( \bigcup_{k=1}^\infty E_k \right) = \bigcup_{k=1}^\infty (A \cap E_k) = \varnothing,$
(26)
so the union belongs to $\mathcal S_A$ .

Therefore $\mathcal S_A$ is a $\sigma$ -algebra on $X$ .

(b) Characterization of $\mathcal S_A$ -measurable functions

( $\Rightarrow$ ) Assume $f$ is $\mathcal S_A$ -measurable.

Since $\mathcal S_A \subset \mathcal S$ , we have $f^{-1}((a,+\infty)) \in \mathcal S$ for every $a \in \mathbb R$ . Hence $f$ is $\mathcal S$ -measurable.

Now suppose $f$ is not constant on $A$ . Then there exist $a_1, a_2 \in A$ such that

f(a_1) \ne f(a_2).

(27)

Consider the Borel set $\{ f(a_1) \}$ . Since $f$ is $\mathcal S_A$ -measurable,

f^{-1}(\{f(a_1)\}) \in \mathcal S_A.

(28)

By definition of $\mathcal S_A$ , either

$A \subset f^{-1}(\{f(a_1)\})$ , or
$A \cap f^{-1}(\{f(a_1)\}) = \varnothing$ .

The second possibility is impossible because $a_1 \in A$ and $f(a_1)$ belongs to that singleton.

Hence

A \subset f^{-1}(\{f(a_1)\}).

(29)

This implies that for every $x \in A$ ,

f(x) = f(a_1).

(30)

In particular,

f(a_2) = f(a_1),

(31)

which contradicts the assumption that $f(a_1) \ne f(a_2)$ .

Therefore $f$ must be constant on $A$ .

( $\Leftarrow$ ) Assume $f$ is $\mathcal S$ -measurable and constant on $A$ .

Let $f(x)=c$ for all $x \in A$ .

To prove $f$ is $\mathcal S_A$ -measurable, it suffices to check that

f^{-1}((a,+\infty)) \in \mathcal S_A \quad \text{for all } a \in \mathbb R.

(32)

We distinguish two cases.

If $c < a$ , then $A \cap f^{-1}((a,+\infty)) = \varnothing$ . Since $f^{-1}((a,+\infty)) \in \mathcal S$ , it follows that it belongs to $\mathcal S_A$ .
If $a \le c$ , then $A \subset f^{-1}((a,+\infty))$ . Since the set lies in $\mathcal S$ , it belongs to $\mathcal S_A$ .

Thus $f$ is $\mathcal S_A$ -measurable.

Therefore, $f$ is $\mathcal S_A$ -measurable if and only if $f$ is $\mathcal S$ -measurable and $f$ is constant on $A$ .

$\blacksquare$

Exercise 2B.17

Suppose $X$ is a Borel subset of $\mathbb R$ and $f : X \to \mathbb R$ is a function such that

\{x \in X : f \text{ is not continuous at } x\}

(33)

is a countable set. Prove $f$ is a Borel measurable function.

Exercise 2B.18

Suppose $f : \mathbb R \to \mathbb R$ is differentiable at every element of $\mathbb R$ . Prove the $f'$ is a Borel measurable function from $\mathbb R$ to $\mathbb R$ .

Exercise 2B.20

Suppose $(X,S)$ is measurable space and $f,g : X \to \mathbb R$ are $S$ -measurable functions.
Prove that if $f(x)>0$ for all $x\in X$ , then $f^g$ (which is the function whose value at $x\in X$ is $f(x)^{g(x)}$ ) is an $S$ -measurable function.

Exercise 2B.22

Suppose $B \subset \mathbb R$ and $f : B \to \mathbb R$ an increasing function.
Prove that $f$ is continuous at every element of $B$ except for a countable subset of $B$ .