Week 2 Worksheet Solutions

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1. 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.
2. Let us check that $g^{-1}((a,\infty))$ is Borel. This is sufficient by a result in the notes. We have \[ \begin{aligned} g^{-1}((a,\infty)) &= \{ t \in \R : g(t) > a \} \\ &= \{ t \in \R : \lim_{n \to \infty} f_n(t) > a \} \\ &= \bigcup_{j \in \N} \bigcup_{M \in \N} \bigcap_{n \ge M} \{ t \in \R : f_n(t) > a + \tfrac{1}{j} \} \\ &= \bigcup_{j \in \N} \bigcup_{M \in \N} \bigcap_{n \ge M} f_n^{-1}((a + \tfrac{1}{j},\infty)) \end{aligned} \] which belongs to the Borel σ-algebra because each $f_n$ is Borel measurable. (Why does the parameter $j$ enter into things?)
3. If $f$ is non-decreasing then $f(x) \le f(y)$ whenever $x \le y$. 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 $\alpha$ be its infimum. The set can then only be either $[\alpha,\infty)$ or $(\alpha,\infty)$.
4. All we require of $\mathscr{A}$ is that it contain $f^{-1}(C)$ for every $C \in \mathscr{C}$. Thus define $\mathscr{A}$ to be the σ-algebra generated by the collection $\{ f^{-1}(C) : C \in \mathscr{C} \}$.
5. Given any set $X$ define \[ \Sigma(a) = \sup \{ a(x_1) + \cdots + a(x_n) : \{ x_1,\dots,x_n\} \subset X \} \] for every $a : X \to [0,\infty]$.
   1. For every $\{ x_1,\dots,x_n \} \subset X$ we have \[ \begin{aligned} \Sigma(a) + \Sigma(b) &\ge a(x_1) + \cdots + x(x_n) + b(x_1) + \cdots + b(x_n) \\ &= (a+b)(x_1) + \cdots (a+b)(x_n) \end{aligned} \] so $\Sigma(a) + \Sigma(b) \ge \Sigma(a+b)$. Conversely, for any finite subsets $\{y_1,\dots,y_r\}$ and $\{ z_1,\dots,z_s \}$ of $X$ we have \[ \Sigma(a+b) \ge a(y_1) + \cdots + a(y_r) + b(z_1) + \cdots + b(z_s) \] and the right-hand side is greater than $\Sigma(a) + \Sigma(b) - \epsilon$ if the sets are chosen appropriately.
   2. Fix $\{ (x_1,y_1),\dots,(x_r,y_r) \} \subset \N \times \N$. We have \[ \Sigma(c) \ge \Sigma(b_{x_1}) + \cdots + \Sigma(b_{x_r}) \ge a(x_1,y_1) + \cdots + a(x_r,y_r) \] and therefore $\Sigma(c) \ge \Sigma(a)$.
      
      Conversely, for every $\epsilon > 0$ there is $r_1,\dots,r_s$ with \[ \begin{aligned} \Sigma(c) - \epsilon &\le \Sigma(b_{r_1}) + \cdots + \Sigma(b_{r_s}) - \tfrac{\epsilon}{2} \\ & \le a(r_1,x_{1,1}) + \cdots + a(r_1,x_{1,t(1)}) + \\ & \qquad \cdots + a(r_s,x(s,1)) + \cdots + a(r_s,x(s,t(s))) \\ & \le \Sigma(a) \end{aligned} \] and since $\epsilon > 0$ was arbitrary we are done.
6. 1. We define \[ (t \mu + (1-t) \nu)(E) = t \mu(E) + (1 - t)\nu(E) \] for all $E \in \mathscr{B}$. Certainly $t \mu + (1-t) \nu$ assigns the value zero to the empty set. Fix $E_1,E_2,\dots$ a countable, pairwise disjoint sequence of sets in $\mathscr{B}$. We calculate \[ \begin{aligned} & (t \mu + (1-t) \nu) \left( \,\bigcup_{n=1}^\infty E_n \right) \\ ={} & t \mu \left( \,\bigcup_{n=1}^\infty E_n \right) + (1-t) \nu \left( \,\bigcup_{n=1}^\infty E_n \right) \\ ={} & \sum_{n=1}^\infty t \mu ( E_n ) + \sum_{n=1}^\infty (1-t) \nu ( E_n ) \\ ={} & \sum_{n=1}^\infty (t \mu + (1-t) \nu)(E_n) \end{aligned} \] to get countable additivity.
   2. Define \[ \left( \, \sum_{n=1}^\infty \mu_n \right)(E) = \sum_{n=1}^\infty \mu_n(E) \] 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 \[ \begin{aligned} & \left(\, \sum_{n=1}^\infty \mu_n \right) \left( \, \bigcup_{j=1}^\infty E_j \right) \\ ={} & \sum_{n=1}^\infty \mu_n \left( \, \bigcup_{j=1}^\infty E_j \right) \\ ={} & \sum_{n=1}^\infty \sum_{j=1}^\infty \mu_n (E_j) \\ ={} & \sum_{j=1}^\infty \sum_{n=1}^\infty \mu_n (E_j) \\ ={} & \sum_{j=1}^\infty \left( \, \sum_{n=1}^\infty \mu_n \right) (E_j) \end{aligned} \] with the interchange of the summations justified by Problem 5.
7. Put $B_N = A_N \cup A_{N+1} \cup A_{N+2} \cup \cdots$ for all $N \in \N$. We have \[ \mu(B_N) \le \sum_{n=N}^\infty \mu(A_n) \] by subadditivity. The sequence $N \mapsto B_N$ is decreasing and $\mu(B_1) \l \infty$ so \[ \mu( \limsup_n A_n) = \mu \left( \, \bigcap_{N \in \N} B_N \right) = \lim_{N \to \infty} \mu(B_N) \] which is zero because convergence of the series implies the tail of the series converges to zero.
8. Fix a measure space $(X,\B,\mu)$. The symmetric difference of two sets $A,B \subset X$ is the set \[ A \symdiff B = A \setminus B \cup B \setminus A = A \textsf{ xor } B \] of points that belong to $A$ or to $B$ but not to $A \cap B$. Define \[ \rho : \mathscr{B} \times \mathscr{B} \to [0,\infty] \] by $\rho(A,B) = \mu(A \symdiff B)$.
   1. It is immediate that $\rho$ is symmetric because $A \triangle B = B \triangle A$. The triangle inequality follows from \[ A \triangle C \subset A \triangle B \cup B \triangle C \] and subadditivity.
   2. No, because $\mu$ could be the measure $\mu(E) = 0$ on $\mathcal{P}(\R)$ for which $\rho$ does not satisfy the third defining property of a metric! Slightly more interestingly, once we know that $\lambda$ is a measure on $\borel(\R)$ we could take $A = \Q$ and $B = \mathcal{C}$ the middle-thirds Cantor set. Although they are distinct Borel sets, the pseudo-metric $\rho$ does not distinguish them.
9. Fix an uncountable set $X$ and let $\B$ be the σ-algebra of subsets of $X$ that are either countable or cocountable. Define \[ \mu(B) = \begin{cases} 1 & B \textsf{ cocountable} \\ 0 & B \textsf{ countable} \end{cases} \] 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 $\mu$ only takes the values $0$ and $1$.
   3. Any measure of the given sort will assign non-zero measure to the countable set $\{ x(i) : i \in \N \}$.