Problem 3

Exercise 8.3 (Countable complement measure space)   Let $X$ be uncountable. Define the countable complement measure space

$$\mathcal{M} = \{E\subseteq X \;\vert\; X\setminus E E \}$$ (8.5)

$$\mu \begin{align*}: \mathcal{M} \to [0,\infty] \quad E\mapsto \begin{cases} \mu(E) =... ...mu(E) = 1 \quad \text{$X\setminus E$\ at most countable} \end{cases}\end{align*}$$ (8.6)

1. Prove that $\mathcal{M}$ is a $\sigma$-algebra and that $\mu$ is a measure on $\mathcal{M}$.
2. Prove that $\mathcal{M}$ is the $\sigma$-algebra generated by $\mathcal{E}=\{\{x\} : x\in X\}$.

Proof. To see that $\mathcal{M}$ forms a $\sigma$-algebra, let $\{E_n\}_{n=1}^\infty$ lie in $\mathcal{M}$. Then to verify the union

$$E = \bigcup_{n=1}^\infty E_n$$ (8.7)

lies in $\mathcal{M}$, we show that $E$ is either at most countable or $X\setminus E$ is at most countable. If each $E_n$ is at most countable, then the union is certainly at most countable, so suppose $X\setminus E_k$ is at most countable. Then

$$X\setminus E=\bigcap_{n=1}^\infty X\setminus E_n\subseteq X\setminus E_k$$ (8.8)

Therefore, $X\setminus E$ is at most countable. Unions are included, and complements are included by the definition, so $\mathcal{M}$ forms a $\sigma$-algebra.

Now we show $\mu$ is a measure. Let $\{E_n\}_{n=1}^\infty$ be disjoint sets in $\mathcal{M}$. If each $E_n$ is at most countable, then their union is at most countable and $\mu(E_n)=0$, so we have

$$\mu\left( \bigcup_{n=1}^\infty E_n \right)=0=\sum_{n=1}^\infty 0 = \sum_{n=1}^\infty \mu(E_n)$$ (8.9)

Otherwise, at least one $X\setminus E_k$ is at most countable, so $\mu(E_k)=1$ and disjointness implies $\mu(E_n)=0$ for $n\neq k$ and that the union is at most countable. Then

$$\mu\left( \bigcup_{n=1}^\infty E_n \right)=1=\mu(E_k)=\sum_{n=1}^\infty \mu(E_n)$$ (8.10)

Therefore, $\mu$ respects countable additivity, so is a measure.

To show that $\mathcal{M}$ is the $\sigma$-algebra generated by the singletons $\mathcal{E}=\{\{x\} \;\vert\; x\in X\}$, let $\mathcal{E}'$ be a $\sigma$-algebra containing $\mathcal{E}$. Note that $\mathcal{E}'$ contains all countable unions, countable intersections, and complements of singletons. If $E\in \mathcal{M}$, then $E=\{x_1,\dots\}$ or $X\setminus E=\{x_1,\dots\}$. Both of these lie in $\mathcal{E}'$, so that $E\in \mathcal{E}'$, therefore, $\mathcal{M}\subseteq \mathcal{E}'$, indicating $\mathcal{M}$ is the $\sigma$-algebra generated by singletons. $\qedsymbol$