Listing all sets in sigma-algebra of a random variable when there are similar mappings?

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Pardon if this question seems trivial- I will ask only where I am stuck.

Suppose I have some probability space $\Omega = \{a,b,c,d\}$ where the $\sigma$-algebra $F$ is the collection of all subsets of $\Omega$. I'll define some random variable $Y$ such that: $$Y(a)=4, \space{} Y(b)=0, \space{}Y(c) = 4, \space{}Y(d)=2.$$

Now, I have no problem generating the different subsets- were it not for the definition of a random variable and the fact that there are two different values that yield the value $4$.

My question is two-fold:

  1. Is $Y$ even measurable?
  2. When generating $\sigma(Y)$, do I have to include both $a$ and $b$? I'm going to guess not, but my understanding is a little hazy.
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Any real valued function is measurable on a space if the sigma algebra contains all subsets. So Y is measurable. $\sigma (Y)$ consists of sets of the type $Y^{-1} (A)$. To write down these sets you have see which of the values 0,2,4 is in A. The answer is $\sigma (Y)=\{ \{.\},\Omega,\{b\},\{d\},\{a,c\},\{b,d\},\{a,b,c\}, \{a,c,d\}\}$