Linearity of Expectation Derivation I'm trying to understand how to prove linearity of expectation. The explanation I'm reading is:
$$E[X+aY] = \int_{(x,y)}(x+ay)p_{X,Y}dxdy$$
$$= \int_{(x,y)}xp_{X,Y}(x,y)dxdy + a\int_{(x,y)}yp_{X,Y}(x,y)dxdy$$
$$= \int_{(x)}xp_{X}(x)dx + a\int_{(y)}yp_{Y}(y)dy$$
I'm having trouble intuitively understanding how to go from the first to the second step. Mathematically why is it that I only need to worry about the PDF of $X$, and $Y$ for their respective parts rather than the joint $X,Y$ PDF?
 A: Using the notation as in your question:$$\int_{(y)}p_{X,Y}(x,y)dy=p_X(x)$$
and $$\int_{(x)}p_{X,Y}(x,y)dx=p_Y(y)$$
So we have $$\int_{(x,y)}xp_{X,Y}(x,y)dxdy=\int_{(x)}x\left(\int_{(y)}p_{X,Y}(x,y)dy\right)dx=\int_{(x)}xp_X(x)dx$$
and similarly for the other integral. Does this answer your question?
A: If it is your aim to prove linearity of expectation then you better go this route:$$\mathbb E [X+aY]=\int X(\omega)+aY(\omega)\mathsf P(d\omega)=$$$$\int X(\omega)\mathsf P(d\omega)+a\int Y(\omega)\mathsf P(d\omega)=\mathbb EX+a\mathbb EY$$Note that this is valid also if there is no PDF. In fact the distributions that are involved have no added value here and are left aside. 

edit 
Concerning your effort:
Note that $\mathbb Ef(X,Y)=\int_{(x,y)}f(x,y)p_{X,Y}(x,y)dxdy$ for every measurable function $f:\mathbb R^2\to\mathbb R$. 
If you take the function $f$ prescribed by $(x,y)\mapsto x$ then you find $$\mathbb EX=\mathbb Ef(X,Y)=\int_{(x,y)} f(x,y)p_{X,Y}(x,y)dxdy=\int_{(x,y)} xp_{X,Y}(x,y)dxdy$$
So in order to prove that $\mathbb E[X+aY]=\mathbb EX+a\mathbb EY$ it is already enough to observe that:$$\int_{(x,y)} [x+ay]p_{X,Y}(x,y)dxdy=\int_{(x,y)} xp_{X,Y}(x,y)dxdy+a\int_{(x,y)} yp_{X,Y}(x,y)dxdy=\mathbb EX+a\mathbb EY$$
Also we have equalities like $\int_{(x,y)} xp_{X,Y}(x,y)dxdy=\int_{(x)}xp_X(x)dx$ but in this context there is no need to get them involved.

The connection between $\mathsf P$ in my answer and $P_{X,Y}$ on $\mathbb R^2$ (which has been assumed to have a PDF $p_{X,Y}$ in your effort) is: $$P_{X,Y}(B)=\mathsf P(\{\omega\in\Omega\mid \langle X(\omega),Y(\omega)\rangle\in B\})=\mathsf P(\{\langle X,Y\rangle\in B\})=\mathsf P(\langle X,Y\rangle\in B)$$
The second and third equality only involve practicized abbreviations.
