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 $$ E_t :=\{x \in X : f(x)<t\} $$ 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.

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

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