## Measurable Spaces – Problem (49/365)

Let $\mu$ be the Lebesque-Stieltjes measure generated by a continuous function. Show that if the set $A$ is at most countable, then $\mu(A) = 0$. A Lebesque-Stieltjes measure is a countably additive measure and is given by a generalized distribution function $G(x)$ such that $\mu([a,b)) = G(b) - G(a)$ that is continuous on the right. So,

$\displaystyle A = \cup_{i=1}^\infty \{ a_i \} \\ \mu(A) = \sum_{i=1}^\infty \mu(\{ a_i \}) \text{ by countable additivity}\\ \mu(\{ a_i \}) = \lim_{n \rightarrow \infty} \mu [a_i, a_i + \frac{1}{n}) \\ = \lim_{n \rightarrow \infty} G(a_i + \frac{1}{n}) - G(a_i) \\ = 0 \\ \text{Thus } \mu(A) = 0$

## Measurable Spaces – Problem (48/365)

Another question on distribution functions. Show that each of the functions

$\displaystyle G(x,y) = 1 \text{ if } x+y \ge 0 \text{ and } 0 \text{ otherwise} \\ G(x,y) = [x+y] \text{, the integral part of } x+y$

is continuous on the right but is not a distribution function in $\mathbb{R}^2$.

Take the first function. To show that it is continuous on the right, let $\epsilon > 0$ and let $(x,y) \in \mathbb{R}^2$. We need to show that there exists $\delta > 0$ such that for all $(a,b) > (x,y)$ and within a distance of $\delta$ the following holds: $| G(x,y) - G(a,b) | < \epsilon$. If $G(x,y) = 0$, then let $\delta$ be the distance to the nearest point $(p,q)$ where $p+q = 0$. We see that in this case, picking a point $(a,b) > (x,y)$ within $\delta$ of $(x,y)$ will take on a value of $0$ and the difference will be less that $\epsilon$. If $G(x,y) = 1$, then $x \ge y$ and we can easily pick $(a,b) > (x,y)$ such that $a > b$, meaning it will also take on a value of $1$ and satisfy $\epsilon$. Thus, the first function is continuous on the right.

However, it is not a distribution function because it does not satisfy the requirement described in the last post that $\Delta_{a_1b_1}\Delta_{a_2b_2} \ge 0$ because if $(a_1,a_2) = (1,2)$ and $(b_1,b_2) = (-4,3)$ the difference function evaluates to $-1$.

The second function $G(x,y) = [x+y]$ is not a distribution function because $G(\infty, \infty) \ne 1$. But it is continuous on the right because if we pick a point $(x,y)$ the function is constant on the interval $[[x+y],[x+y+1])$ and it is open on the right meaning we can always find a delta on the right to satify any $\epsilon > 0$.

## Measurable Spaces – Problem (47/365)

The previous post involved distribution functions over the real numbers but it’s also possible to have distribution functions over $R^n$. A problem asks to show that if we have the distribution function

$\displaystyle F(x_1, \dots, x_n) = P((-\infty, x_1] \times \dots \times (\infty, x_n])$

And a difference function

$\displaystyle \Delta_{a_i,b_i} : \mathbb{R}^n \rightarrow \mathbb{R} \\ \Delta_{a_i,b_i} F(x_1, \dots, x_n) = F(x_1, \dots, x_{i-1}, b_i, x_{i+1}, \dots) - F(x_1, \dots, x_{i-1}, a_i, x_{i+1}, \dots) \\ \text{for } a_i \le b_i$

Then show that

$\displaystyle \Delta_{a_1,b_1} \dots \Delta_{a_n,b_n} F(x_1,\dots,x_n) = P(a,b] \\ \text{where } a = (a_1, \dots, a_n), b = (b_1, \dots, b_n)$

Just to make clear the notation above (which confused me for a while), take the example where $n=2$, then

$\displaystyle \Delta_{a_2,b_2}F(x_1,x_2) = F(x_1, b_2) - F(x_2, a_2) \\ \Delta_{a_1,b_1}\Delta_{a_2,b_2}F(x_1,x_2) = [F(b_1, b_2) - F(b_1, a_2)] - [F(a_1, b_2) - F(a_1, a_2)] \\ \text{this is actually } \\ = P(x_1 \le b_1, x_2 \le b_2) + P(x_1 \le a_1, x_2 \le a_2) \\ \text{ } - P(x_1 \le b_1, x_2 \le a_2) - P(x_1 \le a_1, x_2 \le b_2) \\ = P(a_1 < x_1 \le b_1, a_2 < x_2 \le b_2) \\ = P(a,b]$

So this is true for $n=2$. I won’t do the general case.

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## Measurable Spaces – Problem (46/365)

The rest of the chapter on $\sigma$-algebras goes through the construction of various other measurable spaces such as those on the space of 1) continuous functions, 2) functions continuous on the right, and 3) direct products of measurable spaces. The next chapter introduces methods of introducing probability measures on measurable spaces.

The way we do this is to start with a distribution function $F(x)$ from which we derive a unique probability measure where $P(a,b]) = F(b) - F(a)$. Here is a problem.

Let $F(x) = P(-\infty, x)$, then verify that $P(a,b] = F(b) - F(a)$.

$\displaystyle P(a,b] = P\left( (-\infty,b] \setminus (-\infty, a] \right) \\ = P(-\infty, b] - P(-\infty, a] \text{ by additivity} \\ = F(b) - F(a)$

Verify that $P(a,b) = F(b-) - F(a)$ where $F(x-) = \lim_{y \uparrow x} F(y)$.

$\displaystyle P(a,b) = P \left( \cup_{n=1}^\infty P(a,b-\frac{1}{n}] \right) \\ = \lim_n P(a,b - \frac{1}{n}] \text{ because } P \text{ is countably additive over } \mathcal{B}(R) \\ = \lim_n F(b - \frac{1}{n}) - F(a) \\ = F(b-) - F(a)$

The proof for the following are similar: $P[a,b] = F(b) - F(a-)$, $P[a,b) = F(b-) - F(a-)$, and $P\{x\} = F(x) - F(x-)$.

## Measurable Spaces – Problem (45/365)

Show that the following is not a Borel set in $\mathcal{B}(R^{[0,1]})$ (this is the $\sigma$-algebra of functions over the domain $[0,1]$ unlike the previous onces we looked at where the domain was over the natural numbers).

$\displaystyle A = \{ x : x_t = 0 \text{ for at least one } t \in [0,1] \}$

If $A$ is a Borel set, then so is its complement $\bar{A}$. We know that all sets in $\mathcal{B}(R^{[0,1]})$ must have this form $\{ x : (x_{t_1}, x_{t_2}, \dots) \in B \}$ for some $B \in \mathcal{B}(R^\infty)$ and $t_1, t_2, \dots \in [0,1]$ (proved in book and is simple). The function $y_t = 1$ belongs to $\bar{A}$ and therefore $(y_{t_1}, \dots) \in B$. Consider

$\displaystyle z_t = y_t \text{ if } t \in (t_1, t_2, \dots) \\ z_t = 0 \text{ otherwise}$

Then, since $(z_{t_1}, \dots) = (y_{t_1}, \dots)$ the function $z_t$ belongs to $\{ x : (x_{t_1}, x_{t_2}, \dots) \in B \}$. But $z_t$ clearly does not belong to $\bar{A}$. Hence $\bar{A}$ is not a Borel set and neither is $A$.

## Measurable Spaces – Problem (44/365)

Following on from the previous post, is the following a Borel set?

$\displaystyle \{ x \in R^\infty : \lim_n x_n > a \}$

This is a Borel set because we can intersect the set of converging sequences and the set of sequences bounded from below.

$\displaystyle \{ x \in R^\infty : \lim_n x_n > a \} = \{ x \in R^\infty : x_n \rightarrow \} \cap \{ x \in R^\infty : \sup_n \inf_{k \ge n} x_k > a \}$

## Measurable Spaces – Problem (43/365)

Continuing the last post on showing that certain sets are Borel sets, today we ask if the following is a Borel set.

$\displaystyle \{ x \in R^\infty : x_n \rightarrow \}$

This is the set of all sequences converging to a finite limit. My initial thought was to use the result from last time where we showed that the set of sequences bounded from above or below by $a$

$\displaystyle \{ x \in R^\infty : x_n \rightarrow a \} = \{ x \in R^\infty : \sup \inf x_n \ge a \} \cap \{ x \in R^\infty : \inf \sup x_n \le a \}$

But we can’t then union all the sets of the above form for each possible limit because there are uncountably many choices. It would seem that we need a way to characterize limits without picking the value of the limit. Luckily, there is such a characterization for converging sequences of real numbers; namely, Cauchy sequences. A sequence $x_1, x_2, \dots$ is a Cauchy sequence, if for all $\epsilon > 0$, there is a positive integer $N$ such that for all $m,n > N$, $| x_m - x_n | < \epsilon$. All sequences of real numbers converging to a finite limit are also Cauchy sequences.

The Cauchy condition is true if and only if for all $m \ge 1$, $\lim_n | x_n - x_{n+m} | = 0$. We can write the set of all converging sequences as

$\displaystyle \cap_{m=1}^\infty \{ x \in R^\infty : | x_n - x_{n+m} | \rightarrow 0 \} \\ \cap_{m=1}^\infty \{ x \in R^\infty : \sup_n \inf_{k \ge n} | x_k - x_{k+m} | \ge 0 \} \cap \{ x \in R^\infty : \inf_n \sup_{k \ge n} | x_k - x_{k+m} | \le 0 \}$

As a result, the set of all sequences converging to a finite limit is measurable.

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## Measurable Spaces – Problem (42/365)

The chapter on measurable spaces introduces a $\sigma$-algebra over the real numbers $R = (-\infty, \infty)$. The Borel algebra, $\mathcal{B}(R)$, is the smallest $\sigma$-algebra $\sigma(\mathcal{A})$ where $\mathcal{A}$ is the algebra generated by finite disjoint sums of intervals of the form $(a,b]$. By the direct product of algebras we also get algebras over higher dimensions $\mathcal{B}(R^n) = \mathcal{B}(R) \otimes \dots \otimes \mathcal{B}(R)$. We also get a legal $\sigma$-algebra for the infinite direct product $\mathcal{B}(R^\infty) = \mathcal{B}(R) \otimes \mathcal{B}(R) \otimes \dots$.

The book asks to show that certain sets are members of $\mathcal{B}(R^\infty)$. Show that the following are Borel sets.

$\displaystyle \{ x \in R^\infty : \sup \inf x_n > a \} \\ \{ x \in R^\infty : \inf \sup x_n \le a \}$

Take the first case. Note that $\sup \inf x_n > a$ is not satisfied if for every $i$, $\inf \{ x_k : k \ge i \} \le a$. This can only happen if there are an infinite number of coordinates whose value is $\le a$. Let

$\displaystyle \text{let } I \subset \mathbb{N} \text{ finite subset of natural numbers} \\ S(I) = \cup_{i=1}^\infty \{ x \in \mathcal{B}(R^\infty) : x_i \le a \text{ if } i \in I \text{ and } x_i > a \text{ otherwise} \}$

The set $S(I)$ is a Borel set since we have constructed it as a countable union of Borel sets. Therefore, $\{ x \in R^\infty : \sup \inf x_n > a \} = \cup_{I} S(I)$ (this is also a countable union) is a Borel set. A similar argument is made for the other.

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## Measurable Spaces – Problem (41/365)

Let $\mathcal{D} = \{ D_1, D_2, \dots \}$ be a countable decomposition of $\mathcal{\Omega}$ and $\mathcal{A} = \sigma(\mathcal{D})$ be the $\sigma$-algebra generated by $\mathcal{D}$. Are there only countably many sets in $\mathcal{A}$?

The answer is no. We can show this by showing that the natural numbers is a strict subset of $\mathcal{A}$. Every natural number $n$ can be written as $\sum_{i=0}^\infty a_i 2^i$ where $a_i \in \{0,1\}$ and only a finite number of $a_i = 1$ (because if an infinite number of $a_i = 1$ the sum is $\infty$).

Note that since $\mathcal{D}$ is a decomposition we can write every set in $\mathcal{A}$ as a countable (because this is a $\sigma$-algebra) union of a subset of $\mathcal{D}$. This means we can encode every set in $\mathcal{A}$ as $\cup_{i=1}^\infty (A_i \cap D_i)$ where $A_i \in \{ \emptyset, \Omega \}$. First, since $\mathcal{D}$ is countable there is a bijection between the natural numbers and $\mathcal{D}$, however, $\mathcal{A}$ is not countable since we can have a countable number of $A_i = \Omega$.

## Measurable Spaces – Problem (40/365)

The past few problems looked at how probability works when we have an infinite sample space. It didn’t cover how one can actually assign probabilities to such spaces. That will be the next task. Before that, the book covers the topic of $\sigma$-algebras which form the algebra of events on top of which we can assign a measure.

Given a set $\Omega$, and a set of subsets $\mathcal{A}$, we say that $\mathcal{A}$ is an algebra if $\Omega \in \mathcal{A}$ and is closed under unions and complementation. A $\sigma$-algebra adds to that the requirement that it also be closed under countable unions. The pair $(\Omega, \mathcal{A})$ is called a measureable space.

Let $\mathcal{A}_1, \mathcal{A}_2$ be $\sigma$-algebras of $\Omega$. Are the following systems of sets $\sigma$-algebras?

$\displaystyle \mathcal{A}_1 \cap \mathcal{A}_2 = \{ A : A \in \mathcal{A}_1 \text{ and } A \in \mathcal{A}_2 \} \\ \mathcal{A}_1 \cup \mathcal{A}_2 = \{ A : A \in \mathcal{A}_1 \text{ or } A \in \mathcal{A}_2 \}$

The intersection of $\sigma$-algebras is also a $\sigma$-algebra because $\Omega \in \mathcal{A}_1 \cap \mathcal{A}_2$, and $A_1 \cup A_2 \cup \dots \in$ in the intersection is contained in both $\mathcal{A}_1$ and $\mathcal{A}_2$.

However, the union of $\sigma$-algebras is not always a $\sigma$-algebra. For instance, let $\mathcal{A} = \{ A, \bar{A}, \emptyset, \Omega_1 \}$, $\mathcal{B} = \{ B, \bar{B}, \emptyset, \Omega_2 \}$, then their union does not contain $A \cup B$.