I am a newby in stochastic processes, which I study in the context of the modelling of security prices (financial mathematics). I'd appreciate your help.

Question. Consider the sample space $\displaystyle \Omega=\{-3,-2,-1,1,2,3\}$ and the algebraF=$\displaystyle \{\phi,\{-3,-2\}, \{-1,-1\}, \{2,3\}, \{-3,-2,-1,1\},\{-3,-2,2,3\}, \{-1,1,2,3\}, \Omega\}$.

For each of the following random variables, determine whether it isF-measurable:

(i) $\displaystyle X(\omega)=\omega^2$

(ii) $\displaystyle X(\omega)=max(\omega,2)$.

Find a random variable that isF-measurable.

My attempt at the answer.

I look back at the definition of F-measurable: "the random variable X is said to beF-measurable with respect to the algebraFif the function $\displaystyle \omega\rightarrow{X(\omega)}$ is constant on any subset in the partition corresponding toF(Pliska, Introduction to Mathematical Finance).

Therefore I need to check whether

(i) $\displaystyle X(\omega)=\omega^2$ is true for every element of every subset above. Obviously, it is only good for the subset {-1,1}; in any other subset, each individual $\displaystyle \omega$ is not equal to itself squared, so $\displaystyle X(\omega)$ is not the same on each subset excpet for {-1,1}.

(ii)similarly, I apply this function (max, 2) to each component of the subsets listed above, and most of them fail: even if $\displaystyle X(\omega)=max(\omega,2)$ for {-3,-2} and {-1,1}, in {2,3} I have $\displaystyle X(2)=2 $$\displaystyle X(3)=3$ and X(2) does not equal to X(3).

So, neither (i) nor (ii) areF-measurable.

To find an F-measurable variable, I borrow idea from (ii):

$\displaystyle X(\omega)=max(\omega,3)$. I think it isF-measurable...

PS Is there 'curly F' in Latex code?...