Proof of the linearity of expectation for continuous random variables All the demonstrations I found  of the linearity of the expectation in the continuous case start like this :
$$
\mathsf E(X+Y) = \iint (x+y) f_{X,Y}(x,y) \,dx \,dy
$$
I don't understand why the joint density of $X$ and $Y$ can be used here. Does it means that the density probability function of $X+Y$ is $f_{X,Y}(x,y)$ ?
 A: Let $Z=X+Y$.  Then we can first show the Law of Unconscious Statistician holds for the sum, and then the Linearity of Expectation. $$\begin{align}\mathsf E(X+Y) &= \mathsf E(Z)\\[1ex] &= \int_\Bbb R z\,f_{X+Y}(z)\,\mathrm d z\tag 1\\[1ex] &=\int_\Bbb R z\int_\Bbb R f_{X,X+Y}(x, z)\,\mathrm d x\,\mathrm d z\tag 2\\[1ex] &=\int_\Bbb Rz\int_\Bbb R f_{X,Y}(x, z-x)\,\mathrm d x\,\mathrm d z\tag 3\\[1ex] &=\int_\Bbb R\int_\Bbb R z\, f_{X,Y}(x, z-x)\,\mathrm d z\,\mathrm d x\tag 4\\[1ex]&=\int_\Bbb R\int_\Bbb R (x+y) f_{X,Y}(x,y)\,\mathrm d y\,\mathrm d x\tag 5\\[4ex]\hline\mathsf E(X+Y) &= \int_\Bbb R\int_\Bbb R x\, f_{X,Y}(x,y)\,\mathrm d y\,\mathrm d x+\int_\Bbb R\int_\Bbb R y\,f_{X,Y}(x,y)\,\mathrm d y\,\mathrm d x\tag 6\\[1ex] &= \int_\Bbb R x\int_\Bbb R f_{X,Y}(x,y)\,\mathrm d y\,\mathrm d x+\int_\Bbb R y\int_\Bbb R f_{X,Y}(x,y)\,\mathrm d x\,\mathrm d y\tag 7\\[1ex] &= \int_\Bbb R x\,f_{X}(x)\,\mathrm d x+\int_\Bbb R y\,f_{Y}(y)\,\mathrm d y\tag 8\\[1ex]\therefore\qquad\mathsf E(X+Y) &= \mathsf E(X)+\mathsf E(Y)\tag 9\end{align}$$
This is of course, for continuous random variables.   There are analogous proofs for discrete random variables, and so on.
A: This is an instance of what some call the law of the unconscious statistician (google it).
\begin{align}
\operatorname E(g(W)) & = \int_{\mathcal X} g(x) f_W(x)\, dx \\[8pt]
& = \int_{g(\mathcal X)} x f_{g(W)}(x)\, dx.
\end{align}
There is no need to find the second integral above; it can be evaluated by evaluating the first one.
This is the case where $W=(X,Y)$ and $g(W)=X+Y.$
A: Too long for a comment. I just do cw answer to expand Graham Kemp's answer.

All the demonstrations I found of the linearity of the expectation in the continuous case

It's funny you should say 'continuous case'.
I tried looking it up on proofwiki and saw that in the article Linearity of Expectation Function, the continuous part indeed doesn't give a justification!
But the discrete does! The justification is Expectation of Function of Joint Probability Mass Distribution. Here, the key idea of the proof here is to

*

*come up with this random variable $Z=g(X,Y)$,

*apply LOTUS to $Z$ and then

*discover the link (see next) between the joint distribution of $X$ and $Y$ to the distribution of $Z$.

This indeed works for continuous as with discrete. Still choose $g(x,y)=x+y$, to get what Graham Kemp did in h answer.
What is the link exactly? The preceding proofwiki article gives the link as Probability Mass Function of Function of Discrete Random Variable (hopefully you can come up with the continuous version). In fact Probability Mass Function of Function of Discrete Random Variable is used in the (discrete) LOTUS. See proofwiki's (discrete) LOTUS: Expectation of Function of Discrete Random Variable
