Week 1 Worksheet Solutions

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1. Define a sequence $B_n$ in $\mathcal{P}(X)$ by $B_1 = B$ and $B_2 = A$ and $B_n = \emptyset$ for all $n \ge 3$. The sequence is pairwise disjoint and therefore \[ \Xi(A \cup B) = \Xi \left( \, \bigcup_{n=1}^\infty B_n \right) = \sum_{n=1}^\infty \Xi(B_n) = \Xi(A) + \Xi(B) \] as desired.
2. Let's check that $\Lambda(A-t) \ge \Lambda(A)$ for all $A \subset \R$ and all $t \in \R$. Fix $\epsilon > 0$ and let $(a_n,b_n)$ be a sequence of intervals with \[ \Lambda(A) + \epsilon \ge \sum_{n=1}^\infty b_n - a_n \] and $A \subset \cup \{ (a_n,b_n) : n \in \N \}$. Certainly \[ A-t \subset \bigcup_{n=1}^\infty (a_n-t,b_n-t) \] so \[ \Lambda(A-t) \le \sum_{n=1}^\infty (b_n - t) - (a_n - t) \] giving $\Lambda(A-t) \le \Lambda(A) + \epsilon$. Since $\epsilon > 0$ was arbitrary we get $\Lambda(A-t) \le \Lambda(A)$. Replacing $A$ with $A-t$ and $t$ with $-t$ gives the converse inequality, establishing the desired equality.
3. We need the following fact, which can be proved by induction: if \[ [r,s] \subset (a_1,b_1) \cup \cdots \cup (a_n,b_n) \] then \[ s-r \le \sum_{k=1}^n b_k - a_k \] holds.
   
   Now fix a sequence $n \mapsto (a_n,b_n)$ of intervals that satisfies \[ \Lambda([0,1]) + \epsilon \ge \sum_{n=1}^\infty b_n - a_n \] and covers $[0,1]$. By compactness we can find numbers $r(1) \l \cdots \l r(k)$ with \[ [0,1] \subset \bigcup_{i=1}^k (a_{r(i)}, b_{r(i)}) \] so we can say that \[ \Lambda([0,1]) + \epsilon \ge \sum_{i=1}^k b_{r(i)} - a_{r(i)} \ge 1 \] using the fact above. As $\epsilon > 0$ was arbitrary we are done.
4. We verify directly the three defining properties. The first is that $X$ belongs to the collection, which is explicitly the case. The second is that complements of sets in $\{\emptyset, X\}$ also belong to $\{\emptyset, X\}$: we can check this directly for each set in the collection, because $X \setminus \emptyset = X$ and $X \setminus X = \emptyset$. Lastly, let $n \mapsto B_n$ be a sequence in $\{\emptyset, X\}$. Each $B_n$ is therefore either $\emptyset$ of $X$. If all are the empty set then their union is also empty and belongs to $\{\emptyset, X\}$. If any one of them is $X$ then their union is also $X$ and again the union belongs to $\{\emptyset, X \}$.
5. Fix a measurable space $(X,\B)$. Given a sequence $B_1,B_2,B_3,\dots$ in $\B$ define \[ \begin{aligned} \limsup_n B_n &= \bigcap_{N \in \N} \bigcup_{j \ge N} B_j \\ \liminf_n B_n &= \bigcup_{N \in \N} \bigcap_{j \ge N} B_j \end{aligned} \]
   1. A point belongs to the limit supremum if there are infinitely many $n \in \N$ such that $x \in B_n$. A points belongs to the limit inferior if there are only finitely many $n \in \N$ for which $B_n$ does not contain $x$.
   2. If $x$ belongs to the limit inferior then are certainly infinitely many $n$ for which $x$ belongs to $B_n$.
   3. They both belong to $\B$ because we cannot leave $\B$ by taking countable unions and countable intersections.
6. 1. $[0,1] = \R \setminus (-\infty,0) \cap \R \setminus (1,\infty)$ is Borel.
   2. $[0,1] \cup (3,5)$ is neither open nor closed.
   3. The set $\R \setminus \Q$ of irrational numbers is not open and therefore not a countable union of open sets. It is also not the countable union of closed sets $F_1,F_2,\dots$ because if it were then \[ \R = \bigcup_{n \in \N} F_n \cup \bigcup_{q \in \Q} \{q\} \]
   would be a countable union of closed sets each having empty interior, which is impossible. (See next question!)
7. 1. Since every singleton set $\{b\}$ is nowhere dense, every countable set is a countable union of nowhere dense sets and therefore meagre.
   2. Since $\R$ is a complete metric space, it is impossible by the Baire category theorem to write $\R$ as a countable union of nowhere dense sets.
   3. The whole set $\R$ is open, so it belongs to the Baire σ-algebra. Since \[ \left( \, \bigcup_{n=1}^\infty B_n \right) \triangle \left( \, \bigcup_{n=1}^\infty U_n \right) \subset \bigcup_{n=1}^\infty B_n \triangle U_n \] and a countable union of meagre sets is meagre, it is true that a countable union of Baire sets is a Baire set.
      
      It remains to show that the complement of a Baire set is also a Baire set. Suppose that $B$ is a Baire set. There is an open set $U$ such that $B \triangle U$ is meagre. Let $V$ be the complement of the closure of $U$. Then $V$ is open and \[ (\R \setminus B) \triangle (\R \setminus V) = B \triangle V \subset (B \triangle U) \cup \partial U \] so it suffices to prove that the frontier $\partial U$ of an open set is nowhere dense. As $\partial U$ is closed, so we just need to prove that $\partial U$ has empty interior. Suppose, to reach a contradiction, that $B(x,r)$ is contained in $\partial U$. By definition of $\partial U$ the ball $B(x,r)$ intersects $U$. But then the open intersection $U \cap B(x,r)$ must contain some $B(y,s)$ which is contradicts the definition of $\partial U$.
   4. Every open set is a Baire set. Since $\baire(\R)$ is a σ-algebra, it must therefore contain the smallest σ-algebra that contains the open sets i.e. the Borel σ-algebra.
8. 1. The cardinality of $\borel(\R)$ is the same as the cardinality of $\R$.
   2. The cardinality of the power set $\P(\mathcal{C})$ of the middle-thirds Cantor set $\mathcal{C}$ is strictly greater than the cardinality of $\R$ because $\mathcal{C}$ is uncountable. As the cardinality of $\mathcal{P}(\mathcal{C})$ is strictly greater than the cardinality of $\borel(\R)$ it is false that every subset of $\mathcal{C}$ is a Borel set.