# Constructing a Measure Space where $\mu(A) = 1$ if $A^C$ is Uncountable

I'm working out of Salamon's Measure and Integration in preparation for studying probability, and I came across the following exercise.

Let $$X$$ be uncountable and let $$\mathcal{A} \subset 2^X$$ be the set of all subsets $$A \subset X$$ such that either $$A$$ or $$A^C$$ is countable. Define: $$\mu(A) := \begin{cases} 0 &\mbox{if } A \ \text{is countable} \\ 1 &\mbox{if } A^C \ \text{is countable} \end{cases}$$ where $$A \in \mathcal{A}$$. Show that $$(X, \mathcal{A}. \mu)$$ is a measure space. Describe the measurable functions and their integrals.

I was able to attempt to show that $$\mathcal{A}$$ was a $$\sigma$$-algebra with the following argument:

($$\mathcal{A}$$ is a $$\sigma$$-algebra.) From $$X$$, we may construct a countable set $$S = \bigcup_\limits{j \geq 1} S_j$$ where each $$S_j$$ contains $$j$$ elements from $$X$$. Thus if $$X^C = S$$, then we may say $$X \in \mathcal{A},$$ and $$\mathcal{A}$$ is nontrivially nonempty. Let $$T \in \mathcal{A},$$ and suppose $$T$$ is countable. Then $$T^C \in \mathcal{A},$$ as $$(T^C)^C = T$$ is countable. Finally, to show closure under countable union, we note that if $$Y_j$$ is a countable collection of sets in $$\mathcal{A}$$, then their union must be at most countable, and hence in $$\mathcal{A}$$. Thus, $$\mathcal{A}$$ is a $$\sigma$$-algebra.

(The triple is a measure space.) It is seen that all sets have finite measure in this space. It remains to show that this measure is countably additive with respect to disjoint sets. Let $$A_j$$ be a countable collection of disjoint sets in $$\mathcal{A}.$$ The measure counts the number of sets whose complements are countable.

I was then stuck there. How can I show that $$\mu$$ is countable additive? How can I also describe what the measurable functions are?

(1) $$\mu$$ is a measure: Since $$\varnothing$$ is countable, then $$\mu(\varnothing)=0$$. Let's take $$E_1,E_2, \cdots \in \mathcal{A}$$ pairwise disjoint. If every $$E_n$$ is countable then $$\mu\left( \bigcup_{n=1}^{\infty} E_n\right) = 0 = \sum_{n=1}^{\infty} 0 =\sum_{n=1}^{\infty} \mu(E_n)$$ If there is at least one $$j \in \mathbb{N}$$ such that $$E_j^C$$ is countable, we claim that this must be the only one with this property. Indeed, if $$k \neq j$$ is such that $$E_k^C$$ is countable, then $$(E_k^C) \cup (E_j^C)= (E_k \cap E_j)^C=\emptyset^C= X$$, but $$X$$ is uncountable by hypothesis and hence it cannot be written as a union of two countable sets. Then $$E_j$$ is the only uncountable set in the union $$\bigcup_{n=1}^{\infty} E_n$$, making such union uncountable. Hence $$\mu\left( \bigcup_{n=1}^{\infty} E_n\right) = 1 = \mu(E_j)=\sum_{n=1}^{\infty} \mu(E_n)$$ We have that in fact $$\mu$$ is a measure in $$(X, \mathcal{A})$$.

(2) what the measurable functions are: Let $$\ f : X \to \mathbb{C}$$ be a function. We claim that $$f$$ is measurable if and only if there is $$\lambda \in \mathbb{C}$$ such that $$f(x) = \lambda$$ for all but countably many $$x \in X$$.

Considering real and imaginary parts, we see that it suffices to prove the claim for a function $$\ f : X \to \mathbb{R}$$. For $$t \in [-\infty, \infty]$$ set $$E_t :=\{x \in X : f(x) Next, let $$\lambda := \sup_{t\in \mathbb{R}}\{E_t \text{ is countable }\}.$$ If $$\lambda = −\infty$$, then $$E_t^C$$ is countable for all $$t \in \mathbb{R}$$. So $$X = \bigcup_{n=0}^\infty E_{-n}^C$$ is countable, a contradiction.

So $$\lambda \neq -\infty$$. Choose a sequence $$(t_n)_{n=0}^{\infty}$$ in $$\mathbb{R}$$ such that $$t_n < \lambda$$ for all $$n$$ and $$\lim_{n\to \infty} t_n = \lambda$$. Then $$E_λ = \bigcup_{n=0}^\infty E_{t_n}$$ is countable. This implies $$λ \neq \infty$$. Therefore there is a sequence $$(s_n)_{n=0}^{\infty}$$ in $$\mathbb{R}$$ such that $$s_n > \lambda$$ for all $$n$$ and $$\lim_{n\to \infty} s_n = \lambda$$. Then $$E_{s_n}^C$$ is countable for all $$n$$. So $$\{x \in X : f(x) \neq \lambda \} = E_\lambda \cup \{x \in X: f(x)>\lambda\} = E_\lambda \cup \left( \bigcup_{n=0}^\infty E_{s_n}^C \right)$$ is countable.

Hint: If $$(A_n)_n$$ is a family of pairwise disjoint, countable or cocountable sets, then at most one of them is cocountable.

• How is this so though? – Sean Roberson Nov 3 '18 at 19:22
• @SeanRoberson: What is? – tomasz Nov 4 '18 at 0:17
• The claim that at most one of this disjoint union is cocountable. But it was already shown by @AlonsoDelfin – Sean Roberson Nov 4 '18 at 1:05