Proof for Linearity of Expectation Question 
It is mentioned that the second line of the proof is from the definition of the Union of Probabilities: 

I do not understand how that happened. I understand that the first line is the definition of Expectation expanded, but, why are the two random variables being operated by an AND? 
 A: It's actually in the first equality where you use (for the first time) that property. From lines 1 to 2 and 2 to 3 they just apply known properties of $\Sigma$ operator. And then from 3 to 4 we apply such a property of probability again. This last one happens because
$$\bigcup_j \{Y=j\}$$
(where as explained in the proof $j$ takes values in $R_Y$) equals the whole sample space, and then you can say for any $i\in R_X$
$$\{X=i\}=\{X=i\}\cap \left(\bigcup_j \{Y=j\} \right)=\bigcup_j \big(\{X=i\}\cap\{Y=j\}\big).$$
This explains why (going from 4 to 3)
$$P\big(\{X=i\}\big)=P\left(\bigcup_j \big(\{X=i\}\cap\{Y=j\}\big)\right)=\sum_j{P\big(\{X=i\}\cap\{Y=j\}\big)}.$$
(Second term of 4 comes from same argument interchanging $i$ and $j$.)
In addition, at the very beginning they omit the fact that by definition if we call $Z=X+Y$ then
$$\mathrm{E}(X+Y)=\mathrm{E}(Z)=\sum_k k\mathrm{P}(Z=k).$$
But $Z=X+Y=k$ happens iff $k=i+j \wedge X=i \wedge Y=j$. Since there might be many $(i,j)$ pairs in $R_X \times R_Y$ such that $i+j=k$, then we can just say that
$$\{X+Y=k\}=\bigcup_{i,j/i+j=k} \{X=i \cap Y=j\}$$
and so
$$k\mathrm{P}(Z=k)=\sum_i \sum_{j / i+j=k} (i+j)\mathrm{P}(X=i \cap Y=j\})$$.
But since letting $i$ and $j$ take all values in $R_X$ and $R_Y$ respectively accounts for all possible values $k=i+j$, the final formula reduces just to
$$\sum_k k\mathrm{P}(Z=k)=\sum_i \sum_j (i+j)\mathrm{P}(X=i \cap Y=j\}).$$
A: 
I do not understand how that happened. I understand that the first line is the definition of Expectation expanded, but, why are the two random variables being operated by an AND? 

You have two variables, $X,Y$, and wish to perform a weighed sum over all outcomes in the samples space, $\Omega$.   We then partition the sample space by the joint supports of the random variables; thus we are weighting by the probabilities for the conjunction.
$$\begin{align}\text{In full:}\qquad\\\mathsf E(X+Y) ~&= \sum_{\omega\in\Omega}(X(\omega)+Y(\omega)) \,\mathsf P\{\omega\} \\ &= \sum_{i\in X(\Omega)}\sum_{j\in Y(\Omega)}\mathop{\sum\qquad\quad}_{\omega:(X(\omega)=i )\wedge(Y(\omega)=j)}(X(\omega)+Y(\omega))\, \mathsf P\{\omega\} \\ &= \sum_{i\in X(\Omega)}\sum_{j\in Y(\Omega)}(i+j)\mathop{\sum\qquad\quad}_{\omega:(X(\omega)=i )\wedge(Y(\omega)=j)} \mathsf P\{\omega\} \\ &= \sum_{i\in X(\Omega)}\sum_{j\in Y(\Omega)} (i+j) \,\mathsf P\{\omega:X(\omega)=i~\wedge~ Y(\omega)=j\} \\[3ex]\text{In short:}\quad\\\mathsf E(X+Y) ~ &=\sum_{i}\sum_{j} (i+j)\,\mathsf P(X=i\cap Y=j)\end{align}$$

Alternatively, you can view it as $\{X+Y=u\} = \bigcup_{i}\{X=i\}\cap\{Y=u-i\}$
So $\sum_{u}(u)\mathsf P\{X+Y=u\}~{=\sum_{u}(u)\sum_{i}\mathsf P\{X=i\}\cap\{ Y=u-i\}\\=\sum_i\sum_j (i+j)\mathsf P\{X=i\}\cap\{Y=j\}}$
