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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?

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  • $\begingroup$ What are the cardinalities of the reals, of the positive reals,and of the negative reals? $\endgroup$ Sep 11, 2017 at 22:58
  • $\begingroup$ 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? $\endgroup$
    – Zed1
    Sep 11, 2017 at 23:01
  • $\begingroup$ That's what I am trying to prove is true but I believe @Mark Fischler clarified things. $\endgroup$
    – Zed1
    Sep 11, 2017 at 23:03
  • $\begingroup$ 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. $\endgroup$
    – kneidell
    Sep 11, 2017 at 23:25

4 Answers 4

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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.

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

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  • $\begingroup$ That I understand, but what if there are two subsets $A_i, A_j$ that are both uncountable. Is this possible? $\endgroup$
    – Zed1
    Sep 11, 2017 at 22:55
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    $\begingroup$ 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. $\endgroup$ Sep 11, 2017 at 22:59
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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.

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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 $X$.

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