Which sigma algebra is this function a measure of?

[fishturtle1]

Homework Statement

Let $X = \mathbb{R}$. For which $\sigma$-algebras is the following set function a measure:

$\nu(A) = \begin{cases} 0 & \vert A \vert < \infty \\ 1 & \vert A^c \vert < \infty \\ \end{cases}$

Relevant Equations

$nu$ is a measure on $\sigma$-algebra $\mathcal{A}$ if

1) $\nu(\emptyset) = 0$ and
2) For any sequence of pairwise disjoint sets $(A_n)_{n\in\mathbb{N}} \subset \mathcal{A}$ we have $\nu\left(\bigcup_{n\in\mathbb{N}} A_n \right) = \sum_{n\in\mathbb{N}} \nu(A_n)$.

Suppose $\nu$ is a measure on some $\sigma$-algebra $\mathcal{A}$. Then we must have for all $A \in \mathcal{A}$ either $A$ or $A^c$ is finite, but not both. Because otherwise $\nu(A)$ is undefined or not well defined.

I've verified that $\lbrace \emptyset, X \rbrace$ and $\lbrace \emptyset, X, A , A^c \rbrace$ are suitable $\sigma$-algebras but I'm not sure how to rule out anything bigger? Can I please have a hint how to proceed?

For larger $\sigma$-algebras $\mathcal{B}$, I consider when $B, C \in \mathcal{B}$ such that $B, C$ infinite , $B^c, C^c$ finite, and $B, C$ disjoint. Then

$\nu(B \cup C) = 1 \neq 1 + 1 = \nu(B) + \nu(C)$. But I can't think of a specific example of this happening?

 

[member 587159]

For larger $\sigma$-algebras $\mathcal{B}$, I consider when $B, C \in \mathcal{B}$ such that $B, C$ infinite , $B^c, C^c$ finite, and $B, C$ disjoint. Then

$\nu(B \cup C) = 1 \neq 1 + 1 = \nu(B) + \nu(C)$. But I can't think of a specific example of this happening?

Can't happen. If $B^c$ and $C^c$ is finite and $B,C$ disjoint, then $B^c \cup C^c = (B\cap C)^c=\emptyset^c=\mathbb{R}$ is finite, which is absurd.

 

[fishturtle1]

Edit to OP: $\nu$ is a measure on $\lbrace \emptyset, A , A^c, \mathbb{R} \rbrace$ where $A$ or $A^c$ is finite but not both.

I'm sorry I have to come back in an hour or two.

 

[fishturtle1]

Proof: Consider any finite $\sigma$-algebra $\mathcal{A}$ such that for all $A \in \mathcal{A}$, either $A$ or $A^c$ is finite, but not both. By definition of $\nu$, $\nu(\emptyset) = 0$. Suppose $A, B \in \mathcal{A}$ are infinite sets. Then $\vert (A\cap B)^c \vert = \vert A^c \cup B^c \vert \le \vert A^c \vert + \vert B^c \vert < \infty$. Since $A \cap B \in \mathcal{A}$ we have $A \cap B$ is an infinite set. In particular, $A \cap B \neq \emptyset$ for all infinite sets $A, B \in \mathcal{A}$.

Let $(A_ n)_{n\in\mathbb{N}}\subset \mathcal{A}$ be a sequence of pairwise disjoint sets. We have two cases:

1) For all $n$, $A_n$ is finite. Then, there's only a finite number of $A_n's$ such that $A_n \neq \emptyset$. So, $\bigcup_{n\in\mathbb{N}} A_n$ is finite and
$$\nu\left(\bigcup_{n\in\mathbb{N}} A_n\right) = 0 = 0 + 0 + \dots = \sum_{n\in\mathbb{N}} \nu(A_n)$$

2) There is some $m$ such that $A_m$ is infinite. Well, by previous statements, this can be the only set that in our sequence that is infinite. So,

$$\nu\left(\bigcup_{n\in\mathbb{N}} A_n\right) = 1 = 1 + 0 + 0 + \dots = \nu(A_m) + \sum_{n\neq m}\nu(A_n) = \sum_{n\in\mathbb{N}} \nu(A_n)$$

We can conclude $\nu$ is a measure on $\mathcal{A}$. []

Still thinking about if $\mathcal{A}$ can have infinite elements..

Proof: We show that $\mathcal{A}$ was necessarily finite in the previous proof. Suppose $\mathcal{A}$ is an infinite set such that for all $A \in \mathcal{A}$ we have $A$ or $A^c$ is finite but not both. Moreover, assume by contradiction that $\nu$ is a measure on $\mathcal{A}$. Let $(A_n)_{n\in\mathbb{N}} \subset \mathcal{A}$ be a sequence of distinct finite sets such that $A_n \neq \emptyset$ (which we can surely choose). Define a sequence $(B_n)_{n\in\mathbb{N}} \subset \mathcal{A}$ by $B_n := A_n \setminus (A_{n-1} \cup \dots A_1)$. Then $(B_n)_{n\in\mathbb{N}}$ is a sequence of pairwise disjoint finite sets. Then

$$\nu\left(\bigcup_{n\in\mathbb{N}} B_n \right) = 1 \neq 0 + 0 + \dots = \sum_{n\in\mathbb{N}}\nu(B_n)$$

which contradicts our assumption that $\nu$ is a measure on $\mathcal{A}$. []