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 $B_n=\varnothing$ for all $n\ge 3$. The sequence is pairwise disjoint and therefore $$\mathrm{\Xi }(A\cup B)=\mathrm{\Xi }\left(\bigcup _{n=1}^{\mathrm{\infty }}{B}_n\right)=\sum _{n=1}^{\mathrm{\infty }}\mathrm{\Xi }(B_n)=\mathrm{\Xi }(A)+\mathrm{\Xi }(B)$$ as desired.
2. Let's check that $\mathrm{\Lambda }(A-t)\ge \mathrm{\Lambda }(A)$ for all $A\subset \mathbb{R}$ and all $t\in \mathbb{R}$. Fix $ϵ>0$ and let $(a_n,b_n)$ be a sequence of intervals with $$\mathrm{\Lambda }(A)+ϵ\ge \sum _{n=1}^{\mathrm{\infty }}{b}_n-a_n$$ and $A\subset \cup \{(a_n,b_n):n\in \mathbb{N}\}$. Certainly $$A-t\subset \bigcup _{n=1}^{\mathrm{\infty }}(a_n-t,b_n-t)$$ so $$\mathrm{\Lambda }(A-t)\le \sum _{n=1}^{\mathrm{\infty }}(b_n-t)-(a_n-t)$$ giving $\mathrm{\Lambda }(A-t)\le \mathrm{\Lambda }(A)+ϵ$. Since $ϵ>0$ was arbitrary we get $\mathrm{\Lambda }(A-t)\le \mathrm{\Lambda }(A)$. 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 $$\mathrm{\Lambda }([0,1])+ϵ\ge \sum _{n=1}^{\mathrm{\infty }}{b}_n-a_n$$ and covers $[0,1]$. By compactness we can find numbers $r(1)<\cdots <r(k)$ with $$[0,1]\subset \bigcup _{i=1}^k(a_{r(i)},b_{r(i)})$$ so we can say that $$\mathrm{\Lambda }([0,1])+ϵ\ge \sum _{i=1}^k{b}_{r(i)}-a_{r(i)}\ge 1$$ using the fact above. As $ϵ>0$ 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 $\{\varnothing ,X\}$ also belong to $\{\varnothing ,X\}$: we can check this directly for each set in the collection, because $X\setminus \varnothing =X$ and $X\setminus X=\varnothing$. Lastly, let $n\mapsto B_n$ be a sequence in $\{\varnothing ,X\}$. Each $B_n$ is therefore either $\varnothing$ of $X$. If all are the empty set then their union is also empty and belongs to $\{\varnothing ,X\}$. If any one of them is $X$ then their union is also $X$ and again the union belongs to $\{\varnothing ,X\}$.
5. Fix a measurable space $(X,\mathcal{B})$. Given a sequence $B_1,B_2,B_3,\dots$ in $\mathcal{B}$ define $$\begin{array}{rl}\underset{n}{lim sup}{B}_n& =\bigcap _{N\in \mathbb{N}}\bigcup _{j\ge N}{B}_j\\ \underset{n}{lim inf}{B}_n& =\bigcup _{N\in \mathbb{N}}\bigcap _{j\ge N}{B}_j\end{array}$$
   1. A point belongs to the limit supremum if there are infinitely many $n\in \mathbb{N}$ such that $x\in B_n$. A points belongs to the limit inferior if there are only finitely many $n\in \mathbb{N}$ 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 $\mathcal{B}$ because we cannot leave $\mathcal{B}$ by taking countable unions and countable intersections.
6. 1. $[0,1]=\mathbb{R}\setminus (-\mathrm{\infty },0)\cap \mathbb{R}\setminus (1,\mathrm{\infty })$ is Borel.
   2. $[0,1]\cup (3,5)$ is neither open nor closed.
   3. The set $\mathbb{R}\setminus \mathbb{Q}$ 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 $$\mathbb{R}=\bigcup _{n\in \mathbb{N}}{F}_n\cup \bigcup _{q\in \mathbb{Q}}\{q\}$$
   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 $\mathbb{R}$ is a complete metric space, it is impossible by the Baire category theorem to write $\mathbb{R}$ as a countable union of nowhere dense sets.
   3. The whole set $\mathbb{R}$ is open, so it belongs to the Baire σ-algebra. Since $$\left(\bigcup _{n=1}^{\mathrm{\infty }}{B}_n\right)\mathrm{△}\left(\bigcup _{n=1}^{\mathrm{\infty }}{U}_n\right)\subset \bigcup _{n=1}^{\mathrm{\infty }}{B}_n\mathrm{△}{U}_n$$ 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 $B\mathrm{△}U$ is meagre. Let $V$ be the complement of the closure of $U$. Then $V$ is open and $$(\mathbb{R}\setminus B)\mathrm{△}(\mathbb{R}\setminus V)=B\mathrm{△}V\subset (B\mathrm{△}U)\cup \partial U$$ 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 $\mathsf{Baire}(\mathbb{R})$ 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 $\mathsf{Bor}(\mathbb{R})$ is the same as the cardinality of $\mathbb{R}$.
   2. The cardinality of the power set $\mathcal{P}(\mathcal{C})$ of the middle-thirds Cantor set $\mathcal{C}$ is strictly greater than the cardinality of $\mathbb{R}$ because $\mathcal{C}$ is uncountable. As the cardinality of $\mathcal{P}(\mathcal{C})$ is strictly greater than the cardinality of $\mathsf{Bor}(\mathbb{R})$ it is false that every subset of $\mathcal{C}$ is a Borel set.