Context
I'm trying to wrap my head around the measure-theoretic definition of a random variable. I put forward this example in the hopes that someone can verify that my understanding is correct, or else indicate how it is incorrect.
I have done a non-exhaustive search on this site. I have found 1 helpful, but it still does not give me full comprehension.
Problem
In the figure below, I offer a circular wheel with uniform density. The experiment is to spin the wheel and record what color is aligned with the rightmost vertice of the black triangle.
By $\Omega$, I denote the sample space, which I enumerate with three outcomes as
$$\Omega = \left\{\text{red}, \text{blue}, \text{green} \right\}.$$
By $\mathcal{F}$, I denote the $\sigma$-algebra, which I enumerate with $2^3$ events as
$$\mathcal{F} = \left\{\varnothing, \left\{\text{red}\right\}, \left\{\text{blue}\right\}, \left\{\text{green}\right\}, \left\{\text{red}, \text{green}\right\}, \left\{\text{red}, \text{blue}\right\}, \left\{\text{green}, \text{blue}\right\}, \left\{\text{red}, \text{green}, \text{blue}\right\} \right\}.$$
By $P$, I denote the probability measure, which I enumerate
\begin{align}
P(\varnothing)
&=
0,
\\
P(\left\{\text{red}\right\})
&=
1/6,
\\
P(\left\{\text{blue}\right\})
&=
3/6,
\\
P(\left\{\text{green}\right\})
&=
2/6,
\\
P(\left\{\text{red}, \text{blue}\right\})
&=
4/6,
\\
P(\left\{\text{red}, \text{green}\right\})
&=
3/6,
\\
P(\left\{\text{green}, \text{blue}\right\})
&=
5/6,~\text{and}
\\
P(\left\{\text{red}, \text{green}, \text{blue}\right\} )
&=
6/6.
\end{align}
By the triplet $(\Omega ,{\mathcal {F}},P)$, I denote the probability space.
By $E$, I denote the set $\left\{1,2,3\right\}.$
By $\mathcal{E}$, I denote a $\sigma$-algebra that I give as
$$\mathcal{E} = \left\{\varnothing,\{1\},\{2,3\},\{2\},\{1,3\},\{3\},\{1,2\}, \{1,2,3\}\right\}.$$
By the tuple $(E ,{\mathcal {E}})$, I denote the measurable space.
By the measurable function $ X\colon \Omega \to E$, I denote an $(E,{\mathcal {E}})$-valued random variable that I define as
\begin{align}
X(\omega) =
\begin{cases}
1,&~\text{if}~\omega = \text{red};
\\
2,&~\text{if}~\omega = \text{blue};~\text{or}
\\
3,&~\text{if}~\omega = \text{green}.
\end{cases}
\end{align}
For every subset $B\in {\mathcal {E}}$, I denote its preimage as $X^{-1}(B)$, where $X^{-1}(B)=\{\omega :X(\omega )\in B\} $.
I want to check that for every subset in $\mathcal {E} $, its preimage $X^{-1}(B)$ is in $\mathcal {F}$.
\begin{align}
X^{-1}(\varnothing)
&= \{\omega : X(\omega)\in \varnothing\} = \varnothing
\\
X^{-1}(\{1\})
&= \{\omega : X(\omega)\in \{1\}\} = \{\text{red} \}
\\
X^{-1}(\{2\})
&= \{\omega : X(\omega)\in \{2\}\} = \{\text{blue} \}
\\
X^{-1}(\{3\})
&= \{\omega : X(\omega)\in \{3\}\} = \{\text{green} \}
\\
X^{-1}(\{1,2\})
&= \{\omega : X(\omega)\in \{1,2\}\} = \{\text{red},\text{blue}\}
\\
X^{-1}(\{1,3\})
&= \{\omega : X(\omega)\in \{1,3\}\} = \{\text{red},\text{green} \}
\\
X^{-1}(\{2,3\})
&= \{\omega : X(\omega)\in \{2,3\}\} = \{\text{blue},\text{green} \}
\\
X^{-1}(\{1,2,3\})
&= \{\omega : X(\omega)\in \{1,2,3\}\} = \{\text{red},\text{blue},\text{green} \}
\end{align}
Biibliography
Best Answer
I'm pretty sure that my example is verified as correct. Since I could not find a lucid example on-line, I am leaving this question up for pedagogical purposes.