Definition of $\sigma(X) = \{ X^{-1}(B): B \in \mathcal{B} \}$ Definition of $\sigma(X) = \{ X^{-1}(B): B \in \mathcal{B} \}$. 
Could someone explain why $\sigma(X)$ does not contain the union of the sets $\{X =1 \}$ and $\{X = 0\}$ but $\sigma(Y)$ does?
 A: Does it help to view it like this?
$\sigma(X) = \big\{ \mathcal X\times\{1,2,3,4,5,6\} : \mathcal X\in\mathcal P(\{0,1\})\big\}$
$$\begin{array}{rl:l}\sigma(X) = \Big\{ & \{\}, & \varnothing \\ & \{0\}\times\{1,2,3,4,5,6\}, & \{X=0\} \\ & \{1\}\times\{1,2,3,4,5,6\}, & \{X=1\}
\\ & \{0,1\}\times\{1,2,3,4,5,6\} & \Omega \textsf{ aka }\{X=0\}\cup \{X=1\} \\ \Big\} & \end{array}$$
$\sigma(Y) = \big\{ \{0,1\}\times\mathcal Y : \mathcal Y\in\mathcal P(\{1,2,3,4,5,6\})\big\}$
$$\begin{array}{rl:l}\sigma(Y) = \Big\{ & \{\}, & \varnothing \\ & \{0,1\}\times\{1\}, & \{Y=1\} \\ & \{0,1\}\times\{2\}, & \{Y=2\}
\\ & \{0,1\}\times\{3\} & \{Y=3\} \\ & \vdots & \vdots \\ & \{0,1\}\times\{1,2\} & \{Y=1\}\cup\{Y=2\} \\ & \vdots & \vdots \\ & \{0,1\}\times\{5,6\} & \{Y=5\}\cup\{Y=6\} \\ & \{0,1\}\times\{1,2,3\} & \big\{Y\in\{1,2,3\}\big\} \\ & \vdots & \vdots \\ & \vdots & \vdots \\ & \{0,1\}\times\{1,2,3,4,5,6\} & \Omega \\ \Big\} \end{array}$$
So $\sigma(X)$ does contain the event $\{X=0\}\cup\{X=1\}$ because, since $0$ and $1$ are the only evaluations of $X$, that union is in fact $\Omega$.   
$\sigma(Y)$ contains many more unions of the atoms ($\{Y=y\}$) because the power set of dice rolls is larger than that of coin flips.
$$\lvert \mathcal P\{1,2,3,4,5,6\}\rvert = 2^6 \\ \lvert\mathcal P\{0,1\}\rvert = 2^2$$ 
