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Suppose we have the probability space $(\Omega,\mathcal{A},P)$. Which of the following are right?

  1. $P$ is the probability measure defined on the events $\mathcal{A}$ as follows: $P:\mathcal{A}\rightarrow[0,1]$
  2. $P$ is the probability measure defined on the outcome space $\Omega$ as follows: $P:\Omega\rightarrow[0,1]$
  3. $X$ is a function $X:\mathcal{A}\rightarrow(E,\mathcal{E})$, where $(E,\mathcal{E})$ is a measurable space.
  4. $X$ is a function $X:\Omega\rightarrow(E,\mathcal{E})$, where $(E,\mathcal{E})$ is a measurable space.

Basically, I am unsure whether probability measures and random variables are defined on the state space $\Omega$, or the $\sigma-algebra$ $\mathcal{A}$, or both?

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up vote 2 down vote accepted

Probability measures assign values (probabilities) to sets in the $\sigma$-algebra $\mathcal{A}$. On the other hand, random variables are functions $f\colon \Omega\to E$ that are measurable in this sense: If $B \in \mathcal{E}$, then $f^{-1}(B) \in \mathcal{A}$.

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If random variables are functions $f\colon \Omega\to E$, why is it not sufficient for $f^{-1}(B) \in \Omega$ ($B \in \mathcal{E}$) for them to be measurable? Could we construct a random variable on $\mathcal{A}$ instead of $\Omega$, or does that make no sense? I guess I'm a bit confused on $\sigma$-algebras: why are they needed, why are probability measures defined on them instead of on $\Omega$? – lodhb Mar 26 '12 at 4:56
$f^{-1}(B) \subseteq \Omega$ makes sense, but not $f^{-1}(B)\in \Omega$. For any function, the preimage of a set is a subset of the domain, not an element of the domain. – Patrick Mar 26 '12 at 5:38
But $\Omega\subset\mathcal{A}$, so $f^{-1}(B) \in \mathcal{A}$ doesn't imply $f^{-1}(B) \subset \Omega$, right? – lodhb Mar 26 '12 at 9:02
Not quite. $\mathcal{A}$ is a bunch of subsets of $\Omega$, and $f^{-1}(B)\in \mathcal{A}$ means that $f^{-1}(B)$ is one of those subsets. – Patrick Mar 26 '12 at 22:03
We often want to know the probability that some particular "group of outcomes" will occur. For example, suppose we choose a number $x$ from $[0,1]$, and want to know the probability that $\frac{7}{16} < x < \frac23$. The probability is the measure of the set $(\frac{7}{16}, \frac23)$ which is a subset of the sample space $[0,1]$. – Patrick Mar 26 '12 at 22:14

1 and 4 are right. 2 and 3 are incorrect.

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Without an explanation this isn't contributing much. – ᴡᴏʀᴅs Jul 1 '15 at 14:20
@1999 I explicitly answered OP's question. – bcf Jul 1 '15 at 14:24
The question being from 2012, it's safe to assume the OP isn't struggling with that multiple choice question any more. – ᴡᴏʀᴅs Jul 1 '15 at 14:27

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