Week 2 Worksheet Solutions

It suffices to prove that $f^{- 1} (U)$ is a Borel subset of $R$ for every open set $U \subset R$ because the open sets generate the Borel σ-algebra. But continuity of $f$ is equivalent to openness of $f^{- 1} (U)$ and therefore $f^{- 1} (U)$ is Borel.
Let us check that $g^{- 1} ((a, \infty))$ is Borel. This is sufficient by a result in the notes. We have $\begin{aligned} g^{- 1} ((a, \infty)) & = {t \in R : g (t) > a} \\ = {t \in R : lim_{n \to \infty} f_{n} (t) > a} \\ = ⋃_{j \in N} ⋃_{M \in N} ⋂_{n \geq M} {t \in R : f_{n} (t) > a + \frac{1}{j}} \\ = ⋃_{j \in N} ⋃_{M \in N} ⋂_{n \geq M} f_{n}^{- 1} ((a + \frac{1}{j}, \infty)) \end{aligned}$ which belongs to the Borel σ-algebra because each $f_{n}$ is Borel measurable. (Why does the parameter $j$ enter into things?)
If $f$ is non-decreasing then $f (x) \leq f (y)$ whenever $x \leq y$ . Fix $a \in R$ . The set $f^{- 1} ((a, \infty))$ has the property that if it contains $r$ then it contains $s$ for every $s > r$ . If $f^{- 1} ((a, \infty))$ is empty is is certainly Borel. Suppose it is non-empty. If it does not have a lower bound then it must be $R$ by the property identified earlier. If it does have a lower bound, let $α$ be its infimum. The set can then only be either $[α, \infty)$ or $(α, \infty)$ .
All we require of $A$ is that it contain $f^{- 1} (C)$ for every $C \in C$ . Thus define $A$ to be the σ-algebra generated by the collection ${f^{- 1} (C) : C \in C}$ .
Given any set $X$ define $Σ (a) = sup {a (x_{1}) + \dots + a (x_{n}) : {x_{1}, \dots, x_{n}} \subset X}$ for every $a : X \to [0, \infty]$ .
1. For every ${x_{1}, \dots, x_{n}} \subset X$ we have $\begin{aligned} Σ (a) + Σ (b) & \geq a (x_{1}) + \dots + x (x_{n}) + b (x_{1}) + \dots + b (x_{n}) \\ = (a + b) (x_{1}) + \dots (a + b) (x_{n}) \end{aligned}$ so $Σ (a) + Σ (b) \geq Σ (a + b)$ . Conversely, for any finite subsets ${y_{1}, \dots, y_{r}}$ and ${z_{1}, \dots, z_{s}}$ of $X$ we have $Σ (a + b) \geq a (y_{1}) + \dots + a (y_{r}) + b (z_{1}) + \dots + b (z_{s})$ and the right-hand side is greater than $Σ (a) + Σ (b) - ϵ$ if the sets are chosen appropriately.
2. Fix ${(x_{1}, y_{1}), \dots, (x_{r}, y_{r})} \subset N \times N$ . We have $Σ (c) \geq Σ (b_{x_{1}}) + \dots + Σ (b_{x_{r}}) \geq a (x_{1}, y_{1}) + \dots + a (x_{r}, y_{r})$ and therefore $Σ (c) \geq Σ (a)$ .
  
  Conversely, for every $ϵ > 0$ there is $r_{1}, \dots, r_{s}$ with $\begin{aligned} Σ (c) - ϵ & \leq Σ (b_{r_{1}}) + \dots + Σ (b_{r_{s}}) - \frac{ϵ}{2} \\ \leq a (r_{1}, x_{1, 1}) + \dots + a (r_{1}, x_{1, t (1)}) + \\ \dots + a (r_{s}, x (s, 1)) + \dots + a (r_{s}, x (s, t (s))) \\ \leq Σ (a) \end{aligned}$ and since $ϵ > 0$ was arbitrary we are done.
1. We define $(t μ + (1 - t) ν) (E) = t μ (E) + (1 - t) ν (E)$ for all $E \in B$ . Certainly $t μ + (1 - t) ν$ assigns the value zero to the empty set. Fix $E_{1}, E_{2}, \dots$ a countable, pairwise disjoint sequence of sets in $B$ . We calculate $\begin{aligned} (t μ + (1 - t) ν) (⋃_{n = 1}^{\infty} E_{n}) \\ = & t μ (⋃_{n = 1}^{\infty} E_{n}) + (1 - t) ν (⋃_{n = 1}^{\infty} E_{n}) \\ = & \sum_{n = 1}^{\infty} t μ (E_{n}) + \sum_{n = 1}^{\infty} (1 - t) ν (E_{n}) \\ = & \sum_{n = 1}^{\infty} (t μ + (1 - t) ν) (E_{n}) \end{aligned}$ to get countable additivity.
2. Define $(\sum_{n = 1}^{\infty} μ_{n}) (E) = \sum_{n = 1}^{\infty} μ_{n} (E)$ for all $E \in B$ . It maps the empty set to zero. Given $E_{1}, E_{2}, \dots$ pairwise disjoint in $B$ we can calculate $\begin{aligned} (\sum_{n = 1}^{\infty} μ_{n}) (⋃_{j = 1}^{\infty} E_{j}) \\ = & \sum_{n = 1}^{\infty} μ_{n} (⋃_{j = 1}^{\infty} E_{j}) \\ = & \sum_{n = 1}^{\infty} \sum_{j = 1}^{\infty} μ_{n} (E_{j}) \\ = & \sum_{j = 1}^{\infty} \sum_{n = 1}^{\infty} μ_{n} (E_{j}) \\ = & \sum_{j = 1}^{\infty} (\sum_{n = 1}^{\infty} μ_{n}) (E_{j}) \end{aligned}$ with the interchange of the summations justified by Problem 5.
Put $B_{N} = A_{N} \cup A_{N + 1} \cup A_{N + 2} \cup \dots$ for all $N \in N$ . We have $μ (B_{N}) \leq \sum_{n = N}^{\infty} μ (A_{n})$ by subadditivity. The sequence $N \mapsto B_{N}$ is decreasing and $μ (B_{1}) < \infty$ so $μ (\underset{n}{lim sup} A_{n}) = μ (⋂_{N \in N} B_{N}) = lim_{N \to \infty} μ (B_{N})$ which is zero because convergence of the series implies the tail of the series converges to zero.
Fix a measure space $(X, B, μ)$ . The symmetric difference of two sets $A, B \subset X$ is the set $A △ B = A ∖ B \cup B ∖ A = A xor B$ of points that belong to $A$ or to $B$ but not to $A \cap B$ . Define $ρ : B \times B \to [0, \infty]$ by $ρ (A, B) = μ (A △ B)$ .
1. It is immediate that $ρ$ is symmetric because $A △ B = B △ A$ . The triangle inequality follows from $A △ C \subset A △ B \cup B △ C$ and subadditivity.
2. No, because $μ$ could be the measure $μ (E) = 0$ on $P (R)$ for which $ρ$ does not satisfy the third defining property of a metric! Slightly more interestingly, once we know that $λ$ is a measure on $Bor (R)$ we could take $A = Q$ and $B = C$ the middle-thirds Cantor set. Although they are distinct Borel sets, the pseudo-metric $ρ$ does not distinguish them.
Fix an uncountable set $X$ and let $B$ be the σ-algebra of subsets of $X$ that are either countable or cocountable. Define $μ (B) = {\begin{cases} 1 & B cocountable \\ 0 & B countable \end{cases}$ for all $B \in B$ .
1. The empty set, being countable is assigned zero measure. Fix a sequence $n \mapsto B_{n}$ of pairwise disjoint subsets of $B$ . As a countable union of countable sets is countable there is nothing to prove unless one $B_{n}$ is cocountable. But if one $B_{n}$ is cocountable the rest must all be countable as we started with a sequence of pairwise disjoint sets.
2. This is immediate as $μ$ only takes the values $0$ and $1$ .
3. Any measure of the given sort will assign non-zero measure to the countable set ${x (i) : i \in N}$ .