Let $\Omega = \{a,b,c,d\}$ have just four points with $\sigma-$algebra $\mathcal{F}= 2^{\Omega}$ and probability assignment $P[A] = \sum_{i=1}^{4} \frac{1}{10} \textbf{1}_{A}(\omega_{i})$ to events $A\in\mathcal{F}$ where, $\omega_{1}=a,\omega_{2}=b,\omega_{3}=c,\omega_{4}=d$. Define a collection of sets by $$\mathcal{G} = \{\phi,\{a\},\{b,c\},\{d\},\{a,b,c\},\{a,d\},\{b,c,d\},\Omega\}$$
a) Give explicitly a real random variable X that generates $\mathcal{G} = \sigma(X)$.
My approach - Since, $\sigma-$algebra represents information, knowing the value of $X(\omega)$ should tell us something about $\omega$ or which $\omega$ was chosen.
Since, $\{a,b,c\}$ are together, $X(a)=X(b)=X(c)$, and $\{b,c,d\}$ are together, $X(b)=X(c)=X(d)$. So, $X(\omega)$ has to be the same value i.e. $X(a)=X(b)=X(c)=X(d) = 10$ is one example of $X$.
Is this the correct way to think about this?
b) Give explicitly a real random variable Y that takes only two distinct values, for which $\mathcal{F} = \sigma(X,Y)$.
$$Y(a)= $$ $$Y(b) = $$ $$Y(c) = $$ $$Y(d) = $$
How do I go about this?
Your answer for a) is not correct. A constant random variable generates the trivial sigma algebra consisting of only the empty set and the whole space.
What you need is a random variable with $X(b)=X(c)$ and $X(a), X(b),X(d)$ are all distinct. So $X(a)=0, X(b)=X(c)=1$ and $X(d)=2$ will serve the purpose.
For b) just take any $Y$ such that $Y(b) \neq Y(c)$.