Which sigma algebra is this function a measure of?

  • #1
303
36

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?
 

Answers and Replies

  • #2
Math_QED
Science Advisor
Homework Helper
2019 Award
1,692
718
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.
 
  • Informative
Likes fishturtle1
  • #3
303
36
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.
 
  • #4
303
36
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}##. []
 
Last edited:

Related Threads on Which sigma algebra is this function a measure of?

Replies
2
Views
4K
  • Last Post
Replies
1
Views
2K
Replies
0
Views
970
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
15
Views
5K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
11
Views
3K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
12
Views
2K
Top