# Measure of $X$ an uncountable set.

Let $X$ be an uncountable set and let $\mathcal{A}$ be the collection of subsets $A$ of $X$ s.t. either $A$ or $A^c$ is countable. Define $\mu(A) = 0$ if $A$ is countable and $\mu(A) = 1$ if $A$ is uncountable. Prove $\mu$ is a measure.

To prove $\mu$ is a measure I want to show

• $\mu(\emptyset) = 0$
• $\mu(\cup_{i=1}^{\infty}{A_i}) = \sum_{i=1}^{\infty}{\mu(A_i)}$

Since $\emptyset$ is countable then $\mu(\emptyset) = 0$.

Considering $\mu(\cup_{i=1}^{\infty}{A_i}) \in \{0, 1\}$, then there cannot contain two sets $A_i$ and $A_j$ $i \neq j$ such that they are both uncountable. Is there some fact that two disjoint subsets of an uncountable set both cannot be uncountable?

• What are the cardinalities of the reals, of the positive reals,and of the negative reals? – kimchi lover Sep 11 '17 at 22:58
• If I denote $A$ as the positive reals and $B$ as the negative reals, then I have $\mu(A \cup B) = 1$ and $\mu(A) + \mu(B) = 1 + 1 = 2$ Am I looking at this correctly? – Zed1 Sep 11 '17 at 23:01
• That's what I am trying to prove is true but I believe @Mark Fischler clarified things. – Zed1 Sep 11 '17 at 23:03
• For the poisitive/negative reals it does not hold that $A$ nor $A^c$ is countable, so both are not elements of the given $\sigma$-algebra and, consequently, this is not a valid counter example. – kneidell Sep 11 '17 at 23:25

Hint: If $A^c$ and $B^c$ are countable then $A \cap B$ cannot be empty (because the uncountable set $A$ cannot be contained in the countable set $B^c$). So if $A_i$ is a pairwise disjoint family of members of $\cal A$, at most one of the $A_i$ is uncountable.

Hint: If $\{A_{i}\}_{i=1}^{\infty}$ is a collection of countable sets, then $\bigcup_{i=1}^{\infty}A_{i}$ is countable.

• That I understand, but what if there are two subsets $A_i, A_j$ that are both uncountable. Is this possible? – Zed1 Sep 11 '17 at 22:55
• What you have overlooked is the statement that either $A$ or $A^c$ is countable. Therefore that cannot be two disjoint subsets $\alpha,\beta$ such that both are uncountable, if if there were, then $\beta$ would be an uncountable subset of $\alpha^c$ which itself is countable. – Mark Fischler Sep 11 '17 at 22:59

Let $$A, B \subseteq X$$ be disjoint sets such that $$A^c$$ and $$B^c$$ are countable.

$$\emptyset = A\cap B \implies X = (A\cap B)^c = A^c\cup B^c$$

This implies that $$X$$ is countable as a union of two countable sets. Contradiction.

Let $$(A_n)_{n=1}^\infty$$ be a pairwise disjoint sequence of subsets of $$X$$.

If all $$A_n$$ are countable, then:

$$\mu\left(\underbrace{\bigcup_{n=1}^\infty A_n}_{\text{countable}}\right) = 0 = \sum_{n=1}^\infty \underbrace{\mu(A_n)}_{=0}$$

If $$A_{i}^c$$ is countable for some $$i \in \mathbb{N}$$, then the argument above implies that $$A_n$$ are countable for $$n \ne i$$:

$$\mu\left(\underbrace{\bigcup_{n=1}^\infty A_n}_{\text{uncountable}}\right) = 1 = \underbrace{\mu(A_i)}_{=1} + \sum_{n\in\mathbb{N}\setminus\{i\}}\underbrace{\mu(A_n)}_{=0}$$

Hence, $$\mu$$ is $$\sigma$$-additive.

If $\{A_n\}_{n\in\mathbb N}\subseteq{\cal A}$ are disjoint subsets of $X$, then $A_m\subseteq A_n^c$ for $m\neq n$.

So, if $A_k\in{\cal A}$ is uncountable for some $k\in\mathbb N$, it follows that $A_k^c$ is countable and therefore $A_j$, being a subset of $A_k^c$ ($j\neq k$), is countable.

Therefore, the family $\{A_n\}$ must have at most one uncountable subset of $X$.