Proof of E(X + Y) explanation of step?

1) In general, do we have 1 sigma for each random variable?

Yes, you have to cover all combinations

so if n = m =5

you needs

(x(1), y(1) ) , (x(1), y(2) ) , (x(1), y(3) ) , (x(1), y(4) ) , (x(1), y(5) )

(x(2), y(1) ) , (x(2), y(2) ) , (x(2), y(3) ) , (x(2), y(4) ) , (x(2), y(5) )

(x(3), y(1) ) , (x(3), y(2) ) , (x(3), y(3) ) , (x(3), y(4) ) , (x(3), y(5) )

(x(4), y(1) ) , (x(4), y(2) ) , (x(4), y(3) ) , (x(4), y(4) ) , (x(4), y(5) )

(x(5), y(1) ) , (x(5), y(2) ) , (x(5), y(3) ) , (x(5), y(4) ) , (x(5), y(5) )

You can see you won't cover all combinations without

a sigma for each variable.

2) In the fourth line, we can take the x (which is summed over i) out of the summation operator over j. Is this always the case, because there's no j attached to the x term?

Yes,

x_sub_i is a constant as you vary j.

y_sub_j is not a constant as you vary j, but they

must have used some rule to justify

reversing the order of the summation statement.

I assumed there must be a rule that allows that.

3) In the fifth/last line, why does the Pj disappear, and we're left with the Pi probability mass function part thing?

See image

I'm not sure I don't a good job of explaining

but if you add the probabilities for P(x_i, y_j) for all value of y_j

it equals the probability of P(x_i)

remember, if you flip a coin and then roll a dice,

P(heads) = P(heads , roll a 1) + P(heads, 2) + P(head,3) + P(heads,4)

+ P(heads,5) + P(heads, 6)