# Interpretation of measure theory question

I am struggling with the following question:

Let $\Omega = \{1,2,3,4\}$, $\mathscr E = \{\{1\}\{1,2\}\}$

(a) Find $\sigma (\mathscr E)$ (sigma algebra)

(b) By inventing an analog of a pmf or cdf, device an efficient method to characterise the probability measures on $\sigma (\mathscr E)$

(c) How would you describe what it means for a function $f:\Omega \to \bar{\mathbb R}$ to be $\sigma (\mathscr E) / \mathscr B(\bar{\mathbb R})$ measurable? Hint: Start investigation by writing down the canonical representation of a function in $S_+(\Omega,\sigma(\mathscr E))$

$\mathbf{Solution~Attempt}$:

(a) $\sigma (\mathscr E)=\{\Omega,\emptyset, \{1\},\{1,2\},\{2\},\{2,3,4\} ,\{3,4\}, \{1,3,4\} \}$

(b) I'm not quite sure what this means, what I know is that we want to define $\mathbb P:\sigma(\mathscr E)\to[0,1]$ where: $$\mathbb P (A)=\sum_{\omega \in A} \rho(w) ~~~~~\forall A \in \sigma(\mathscr E)$$ Where: $$\rho(\omega) = \mathbb P (\{\omega\}) ~~~~~\forall \omega \in \Omega$$

So now the difficulty is in finding $\rho$, but since we are not told anything about the likelihood of any of the outcomes, what is the point this question is trying to make?

$\mathbf{UPDATE}$ : I am wondering what the relationship between defining the pmf and a partition is. For example, in this case, the partition that generates the same sigma field as $\mathscr E$ is $\Theta=\{\{1\}, \{2\}, \{3,4\}\}$. So would defining a probabiltiy measure on $\sigma(\mathscr E)$ just mean defining a function that assigns each element of the partition that generates the same sigma algebra some non zero probability, in such a way that the probability assigned to each member of the partition adds up to 1?

(c) I'm quite confused about the $S_+$ notation here? What I do know is that for a function $f$ to be $\sigma (\mathscr E) / \mathscr B(\bar{\mathbb R})$ measurable. Then we need $f^{-1}(\bar{F}) \in \sigma(\mathscr E)~~~\forall \bar{F}\in \mathscr B(\bar{\mathbb R})$. So here since the singletons $\{3\}$,$\{4\}$ are not in $\sigma(\mathscr E)$ we need to define a function around this restriction.

## 1 Answer

(a) The sigma algebra generated by $\mathscr E$ must be closed under countable unions and complements, but is as small as possible. For your $\mathscr E$, think about what can be obtained as unions and complements of only the sets $\{1\}$ and $\{1,2\}$; you'll see that the set $\{3\}$ cannot be obtained, nor can $\{4\}$.

Hint for (b) and (c): as far as $\sigma(\mathscr E)$ is concerned, the elements $\omega=1$ and $\omega=2$ can be distinguished, but the elements $3$ and $4$ cannot. So $3$ and $4$ can be combined into a single "atom".

• But what does it mean to device an efficient strategy by using the analog of a pmf? So i know that this probability measure must assign a constant probability to the elements 3 and 4, but im not really sure how to go from here.] Mar 23, 2015 at 7:45
• I'm assuming that 'efficient' means 'without extraneous detail'. An inefficient way to characterize a probability measure would be to specify its value on every possible set in $\mathscr E$. The analogy with a pmf is that a pmf completely defines the distribution of a discrete random variable $X$ without specifying $P(X\in A)$ for every $A$. For your problem, find the minimal specification you can get away with, and use it to define all possible probability measures. Mar 23, 2015 at 21:11