Linearity of the expectation value for simple conception of random variables In secondary education, random variables aren't rigorously defined. Contrary to tertiary education, where they are usually maps $X\colon \;\Omega \rightarrow \mathbb{R}$ with $\Omega$ the sample space of the probability space, they are just seen as placeholders for some undetermined real quantity which assumes certain values with a well-defined probability.
If random variables are defined rigorously, the proof of the linearity of the expectation value is trivial, it follows (with $a,b$ real) from the linearity of the integral (or of summation):
$$\mathbf{E}[a X+ b Y] = \sum_{\omega \in \Omega} \mathbf{P}[\omega]\cdot\bigl(a X(\omega)+ b Y(\omega)\bigr) = \\ =  a\sum_{\omega \in \Omega} \mathbf{P}[\omega]\cdot X(\omega) + b\sum_{\omega \in \Omega}  \mathbf{P}[\omega] \cdot Y(\omega) = a\,\mathbf{E}[X] + b\,\mathbf{E}[Y]\,.  $$
But what would be the best way to prove/explain, that the expectation value is linear if one hasn't defined random variables rigorously? After all, in the non-rigorous "definition", $X$ and $Y$ are just known by their possibly different probability distributions on the same sample space, $\mathbb{R}$.
 A: I guess in the non-rigorous approach you'd still have definitions like
$$
E(X) = \int xP(x,y,z) dxdydz
$$
and 
$$
E(Y) = \int y P(x,y,z)dxdydz
$$
for, say, a setting with continuous RVs X,Y,Z. So $E(aX+bY) = aE(X) + bE(Y)$ would still follow from the linearity of these integrals. (Or sums in the discrete case).
We can also get to the formula $$ E(aX+bY) = \int (ax+by)P(x,y)dxdy$$ starting from $Z = aX+bY$ and the formula for the expectation value for $Z$ $$E(aX+bY) = \int zP_Z(z) dz.$$
Here, we need to use the formula for the PDF of the sum of two RVs, $P_Z(z) = \int P_{aX,bY}(\xi, z-\xi)d\xi$ where $P_{aX,bY}$ is the joint PDF for $aX$ and $bY.$ Then we transform from $(aX,bY)$ to $(X,Y)$ and get $P_{X,Y}(x,y) = abP_{aX,bY}(ax,by)$  so that $$ P_Z(z) = \int \frac{1}{ab} P_{X,Y}\left(\frac{\xi}{a},\frac{z-\xi}{b}\right)d\zeta.$$
Finally, plug this into the formula for expectation and get
$$
E(aX+bY = \int zdz\int d\xi \frac{1}{ab}P_{X,Y}\left(\frac{\xi}{a},\frac{z-\xi}{b}\right)= \int zdz\int dx \frac{1}{b}P_{X,Y}\left(x,\frac{z-ax}{b}\right).
$$
Where we substituted $x = \xi/a.$ Now, change the order of integration and sub $y = (z-ax)/b$ for $z$ to get $$ E(aX+bY) = \int dx \int zdz\frac{1}{b}P_{X,Y}\left(\xi,\frac{z-a\xi}{b}\right)= \int dx \int dy (ax+by)P_{X,Y}(x,y)$$ and again, the linearity of the integral shows this is equal to $aE(X) + bE(Y).$
