Problem about jointly continuous and linearity of expectation. When defining jointly continuous, we  are saying: if there exists a function $f(x,y)$ satisfying that 
$$
P\{X\in A, Y\in B\}=\int_B\int_a f(x,y)dxdy
$$
then $X,Y$ are called jointly continuous.
So not every pair of random variable $X$ and $Y$ are jointly continuous, because such $f(x,y)$ may not exists, right? 
Then in Ross's book Introduction to Probability Models, there is something I'm confused:
He proved the formula
$$
E[X+Y]=E[X]+E[Y]
$$
for jointly continuous random variables, but then use it for arbitrary pair of random variables.
When $X,Y$ is jointly continuous, it is easy to prove, since
$$
E[g(X,Y)]=\int_{\mathbb{R}}\int_{\mathbb{R}}g(x,y)f(x,y)dxdy
$$
Then let $g(X,Y)=X+Y$.
To be clear, my question is :
(1) Isn't every pair of random variable $X$ and $Y$ are jointly continuous, right?
(2) The formula $E[X+Y]=E[X]+E[Y]$ is valid for any pair $X, Y$ even if they are not independent, right?
(3) Ross in his book proved $E[X+Y]=E[X]+E[Y]$ under the assumption $X$ and $Y$ is jointly continuous. It is quite easy. How to prove general case?
 A: 
Is there $X$ and $Y$ such that they are both continuous, but not jointly continuous? It is what I want to ask in the question (1).

Sure: take $X=Y$, where $X$ is any random variable whose distribution has a density. Then, $\mathbb P((X,Y)\in D)=1$, where $D=\{(x,x)\mid x\in\mathbb R\}$ has Lebesgue measure zero, hence the distribution of $(X,Y)$ has no density.

The formula $E[X+Y]=E[X]+E[Y]$ is valid for any pair $X, Y$ even if they are not independent, right?

Right.

How to prove [it in the] general case?

One may first prove that for every random variable $Z$ and every measurable function $u$ such that $u(Z)$ is integrable,
$$
\mathbb E(u(Z))=\int u(z)\,\mathrm d\mathbb P_{Z}(z).
$$
Assume this key relation is known and apply it to $Z=(X,Y)$ and $u:(x,y)\mapsto x+y$. This yields
$$
\mathbb E(X+Y)=\int (x+y)\,\mathrm d\mathbb P_{(X,Y)}(x,y)=(*).
$$
By linearity of the integrals with respect to the measure $\mathbb P_{(X,Y)}$,
$$
(*)=\int x\,\mathrm d\mathbb P_{X,Y}(x,y)+\int y\,\mathrm d\mathbb P_{(X,Y)}(x,y).
$$
Apply the key relation backwards to the functions $v:(x,y)\mapsto x$ and $w:(x,y)\mapsto y$. This yields
$$
(*)=\mathbb E(v(X,Y))+\mathbb E(w(X,Y))=\mathbb E(X)+\mathbb E(Y).
$$
