Exercise 1

Constructing a cover over the null set using only closed intervals:

$$ \begin{gather*} \text{Let} \; I_n = [x_n - \frac{\varepsilon}{2^{n+2}}, \:x_n +\frac{\varepsilon}{2^{n+2}}] \implies l(I_n) = \frac{\varepsilon}{2^{n+1}} \\ \text{Then} \; \bigcup_{n = 1}^{\infty}I_n \implies \sum_{n=1}^{\infty}{l\left(I_n\right)} \leq \frac{\varepsilon}{2} < \varepsilon \: \text{(equality when disjoint)}\\ \end{gather*} $$

Since the length of an interval only depends on its endpoints, the above construction works for open and open/closed intervals. Thus, any type of interval is sufficient to cover the null set.

Exercise 2

Just as the uncountability of $\mathbb{R}$ is shown in decimal expansion through a diagonal argument, one can do the same in any base. In this case, as the hint suggests, we use the ternary expansion, as we are dividing the interval into thirds. Since we are using a ternary system, only $0, 1, 2$ are allowed, just as in binary only $0, 1$ are allowed. Define each $x \in C$ as $x = \sum_{k=1}^{\infty}\frac{a_k}{3^k} = 0.a_1a_2...$, where $a_k \in \{0,1,2\}$. However, any $a_k$ in the ternary expansion of $x \in C$ must be only $0,2$, as the middle third of the interval was always removed. Assume that all $x \in C$ are countable and listed, and enumerate them as $x_n$. Then take the $n$th digit of each $x_n$ and replace it with the digit it is not; e.g. $0$ is now $2$. Now, this newest $x_n$ constructed is not listed, thus a contradiction. So $C$ is uncountable.

Exercise 3

$$ \begin{gather*} F(x) = \begin{cases} \sum_{n=1}^{\infty}\frac{a_n}{2^n}, &\text{if} \: x = \sum_{n=1}^{\infty}\frac{2a_n}{3^n} \in C, \: a_n \in \{0,1\} \\ \sup_{y\leq x, y \in C}F(y), & \text{if} \: x \in [0,1] \setminus C \end{cases} \\ \left|F(x) - F(x_0) \right| = \left|\sum_{n=1}^{\infty}{\frac{a_n}{2^n}} -\sum_{n=1}^{\infty}\frac{b_n}{2^n} \right| = \left| \sum_{n=1}^{\infty}{\frac{a_n-b_n}{2^n}} \right| \\ = \left| \sum_{n=1}^{N}{\frac{a_n-b_n}{2^n}} +\sum_{n=N+1}^{\infty}{\frac{a_{n}-b_n}{2^n}} \right| = \frac{1}{2^N}\left| \sum_{n=1}^{\infty}{\frac{a_{n+N}-b_{n+N}}{2^n}} \right| \\ \leq \frac{4}{2^N}<\varepsilon, \: \text{when } 2^{N-2} > \frac{1}{\varepsilon} \end{gather*} $$

The key step is that for any $N$, an $x$ can be found sufficiently close to $x_0$ such that the first $N$ digits of its ternary expansion are equal to $x_0$ due to the construction of $C$. Thus, the first $N$ terms of the series go to $0$. If this were not true, then we can only say $\left| \sum_{n=1}^{\infty}{\frac{a_n-b_n}{2^n}} \right| \leq 2\cdot\max_{n}{\left| a_n-b_n \right|} $, and the expression cannot be simplified further to meet the $\varepsilon$ bound. Additionally, by limiting ourselves to a ternary expansion, $\left|a_n - b_n\right| \leq 2$. This proof was adapted from stack exchange. The notation of this function does a disservice to what is otherwise a simple idea.

Exercise 4

$$ \begin{gather*} B \subseteq A\cup B \implies m^*(B) \leq m^*(A\cup B) \\ m^*(A) = 0 \implies m^*(A \cup B) \leq m^*(A) + m^*(B) = m^*(B) \\ m^*(B) \leq m^*(A \cup B), \: m^*(A \cup B) \leq m^*(B) \implies m^*(B) = m^*(A \cup B) \end{gather*} $$

Exercise 5

$$ \begin{gather*} \text{if} \: A \subset B \cup (A \Delta B) \\ \text{then} \: m^*(A) \leq m^*(B \cup (A \Delta B)) \leq m^*(B) + m^*(A \Delta B) = m^*(B) \\ \text{if} \: B \subset A \cup (B \Delta A) \\ \text{then} \: m^*(B) \leq m^*(A \cup (B \Delta A)) \leq m^*(A) + m^*(B \Delta A) = m^*(A) \\ m^*(A) \leq m^*(B), \: m^*(B) \leq m^*(A) \implies m^*(A) = m^*(B) \end{gather*} $$

Exercise 6

Formula for $m(A \cup B)$ in terms of only $m(A), \: m(B), \: m(A \cap B) $:

$$ \begin{gather*} m(A \cup B) = m(A \cup (B \setminus A)), \: (B \setminus A) \cup A = \emptyset \\ \implies m(A \cup (B \setminus A)) = m(A) + m(B \setminus A) \\ m(B \setminus A) = m(B \cap A^c) =m(B) - m(B \cap A) \\ \implies m(A \cup B) = m(A) + m(B) - m(B\cap A) \end{gather*} $$

Where the equality for $m(B \cap A^c)$ arises from the definition of a measurable set. The same chain of reasoning applies to $m(A \cup B \cup C)$, except now the resulting formula subtracts the intersections of any two sets and the intersection of all sets.

Exercise 7

If we can show that a countable union of $(a,b]$ generates an open set, then we know the $\sigma$-algebra of $(a,b]$ is a superset of the $\sigma$-algebra of open sets, the Borel set.

$$ \begin{gather*} (a,b) = \bigcup_{n=1}^{\infty}{(a,b-\frac{1}{n}]} \end{gather*} $$

Likewise, we know that the $\sigma$-algebra of $(a,b]$ is a subset of the Borel algebra since all intervals of the form $(a,b]$ can be constructed by a countable intersection of open sets. Thus, $\sigma((a,b]) \subseteq \mathcal{B}, \: \mathcal{B} \subseteq \sigma((a,b]) \implies \sigma((a,b]) = \mathcal{B}$. The same process can be done with $[a,b)$.

Exercise 8

$E \in \mathcal{M} $ if for open set $O \supset E$ and $\forall \varepsilon > 0$, $m^*(O\setminus E) < \varepsilon$:

$$ \begin{gather*} \text{Let } \varepsilon = \frac{1}{n} \implies m^*(O_n \setminus E ) = \frac{1}{n} \implies m^*\left(\bigcap_{n}^{\infty}O_n \setminus E\right) = 0 \end{gather*} $$

$E \in \mathcal{M} $ if for closed set $C \subset E$ and $\forall \varepsilon > 0$, $m^*(E\setminus C) < \varepsilon$:

$$ \begin{gather*} \text{Let } \varepsilon = \frac{1}{n} \implies m^*(E \setminus C_n ) = \frac{1}{n} \implies m^*\left(E \setminus \bigcup_{n}^{\infty}C_n\right)) = 0 \end{gather*} $$

The key observation is that $O,C$ are Borel sets, so they are automatically in the Lebesgue space. Then, by the assumption of $m^* < \varepsilon$, we know that $E$ can be approximated by Borel sets (it cannot necessarily be constructed by Borel sets, and therefore is not part of the Borel algebra, because there may exist some $N \subseteq E, \: N \not\subset \mathcal{B}$). However, if we can show $E \setminus N \in \mathcal{B}$, then we know $E \in \mathcal{B} \cup N = \mathcal{M}$. The tricky part is showing that what remains from constructing an $\sigma$-algebra approximation to $E$ is the null set; i.e., a set of measure $0$. This is done by showing $\varepsilon \rightarrow 0$ within a countable number of unions/intersections.

Exercise 9

$$ \begin{gather*} P(A) = \sum_{i=1}^{\infty} P(A \mid H_i) P(H_i) = \sum_{i=1}^{\infty} \frac{P(A \cap H_i)}{P(H_i)} P(H_i) \\ = \sum_{i=1}^{\infty} P(A \cap H_i) = P\!\left( \bigcup_{i=1}^{\infty} (A \cap H_i) \right) \\ = P\!\left( A \cap \bigcup_{i=1}^{\infty} H_i \right) = P(A \cap \Omega) \\ = P(A). \end{gather*} $$

Union distributes over intersection; intersection distributes over union.

Exercise 10

$$ \begin{gather*} B = (B\setminus (A \cap B)) \cup (A \cap B) \implies P(B) = P(B\setminus (A \cap B)) + P(A \cap B)\\ \implies P(B) - P(A \cap B) = P(B \setminus (A \cap B)) \\ P(A^c| B) = \frac{P(A^c\cap B)}{P(B)} = \frac{P(B\setminus (A \cap B))}{P(B)} = \frac{P(B) - P(A \cap B)}{P(B)} \\= 1 - \frac{P(A)P(B)}{P(B)} = 1 - P(A) =P(A^c) \end{gather*} $$

The first part shows when the measure of set difference can be split into the difference of measures. This is only true when dealing with subsets, as the added/subtracted set is disjoint from the superset it came from, thus additive.
With the second part, the set operations for $(A^c \cap B)$ can be derived by drawing a picture.

Exercise 11

Number of paths = $2^5 = 32$.

$$ \begin{gather*} \frac{524.88}{500} = (1.2)^2(0.9)^3 \implies \binom{5}{3} = \frac{5!}{2!3!} = 10 \\ \frac{\binom{5}{5}+\binom{5}{4}}{2^5} = \frac{6}{32} \approx 19\% \end{gather*} $$

Exercise 12

If $\omega_m$ is a binary variable, there are $2^m$ possible combinations. There is only one valid $\sigma$-algebra for a discrete probability space, such as that of a binary random variable: the power set.

$$ \begin{gather*} \Omega = 2^m, \: \mathcal{F} = 2^{\Omega} \implies \mathcal{F} = 2^{2^{m}} \end{gather*} $$

Exercise 13

$$ \begin{gather*} m = 0, \: 1: \: \mathcal{F_0} = \{\emptyset,\Omega\}, \: \mathcal{F}_1 = \{\emptyset, A_1, A^c, \Omega \} \implies |\mathcal{F}_1| = \left( |\mathcal{F}_0| \right)^2 \\ m = k, \: k + 1: \: \mathcal{F}_{k+1} = 2^{2^{k+1}} = 2^{2^{k} \cdot 2} = \left( 2^{2^k} \right)^2 = \left( \mathcal{F}_k \right)^2 \\ \therefore \: \mathcal{F}_m \text{ is an increasing sequence} \end{gather*} $$

Exercise 14

Recall that the measure, $\mu$, on measurable space $(\Omega, \sigma)$ is $P = \frac{m_\Omega}{m(\Omega)} = \frac{|A \subseteq \Omega|}{2^\Omega}$. The specific $A$ is given by: $|A_k| = |A_m| = 2^{N-1}$, as one binary choice out of all $\omega_0,\omega_1,...,\omega_N$ is already determined, $\{ \omega_m, \omega_k\}$, respectively. By assumption, $m \neq k$. Their intersection is all possible combinations of these two determined flips out of $2^{2N}$ total flips: $2^{2N -2}$

$$ \begin{gather*} \mathcal{G}_m = \{ \emptyset, A_m, A_m^c, \Omega \}, \: \mathcal{G}_k = \{ \emptyset, A_k, A_m^c, \Omega \} \\ P(\emptyset \cap \{ a : a \in \mathcal{G}_k \}) = P(\emptyset) = P(a) \cdot 0 = P(a)P(\emptyset), \: \text{by definition} \\ P(A_m \cap A_k) = \frac{2^{2N-2}}{2^{2N}} = \frac{1}{4} =P(A_m)P(A_k), \\ P(A_m \cap A_k^c) = P(A_m)P(A_k^c), \: \text{by Exercise 10} \\ P(\Omega \cap \{ a : a \in \mathcal{G}_k \}) = P(a) = P(a) \cdot 1 = P(a)P(\Omega), \: \text{by definition}\\ \implies P(A_m^c \cap \{a:a \in \mathcal{G}_k \}) = P(A_m)P(a), \: \text{by symmetry} \end{gather*} $$

Thus, by Definition 2.11, the $\sigma$-algebras are independent. Technically, only independence of $A_m, A_k$ had to be shown as the rest are either follow by definition, corollary, or symmetry.

Exercise 1#

Exercise 2#

Exercise 3#

Exercise 4#

Exercise 5#

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Exercise 7#

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Exercise 9#

Exercise 10#

Exercise 11#

Exercise 12#

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