Expected value is affine in distribution How to show that $E_{P_X}[X]$ is affine in $P_X$. That is for two distributions $P_{X,1}$ and $P_{X,2}$ we have that
\begin{align*}
E_{\alpha P_{X,1}+(1-\alpha) P_{X,2}}[X]=\alpha E_{ P_{X,1}}[X]+(1-\alpha) E_{ P_{X,2}}[X]
\end{align*}
for $\alpha \in [0,1]$.
I was thinking that we have to work with Lebesgue integral and show
\begin{align*}
\int X \ d(\alpha P_{X,1}+(1-\alpha) P_{X,2})(x)= \alpha \int X \ d( P_{X,1})(x)+(1-\alpha)\int X \ d( P_{X,2})(x)
\end{align*}
My problem is that $dP_X(x)$ is just notation. I don't think I can do what I did above, right? 
For example above is obviously true for $X$ that have densities. But how to show it for any distribution. 
 A: If you want to formalize your answer, your argument should pass through the Radon-Nikodym's theorem. 
So, if for $ \alpha >0 $, the measure $\lambda = \alpha \mu$ means the measure which $\lambda(A) = \alpha \mu(A)$, then the Radon-Nikodym's theorem assure what we expect, i.e., $d\lambda = \alpha d\mu$.
I suggest you take a look at the reference I stated above. If you don't understand, I can improve my answer.
Well, first of all you should understand what $dP$ means. When you write $\int_{\Omega} X dP$ you are saying that you want to determine the expected value of the random variable $X$ with respect the distribution $P$. When $X$ is the indicator function of some event $A$, $\int_{\Omega} X dP = EX = P(A)$.
So, second you must convince yourself that $$\int_{\Omega} X d(P_1+P_2) = \int_{\Omega} X dP_1 + \int_{\Omega} X dP_2 $$
This is clear if you let $X= \mathbb{1}_A$, then $$\int_{\Omega} X d(P_1+P_2) = (P_1+P_2)(A)=P_1(A)+P_2(A)=\int_{\Omega} X dP_1 + \int_{\Omega} X dP_2$$
Once you prove the result for indicators functions, the result extends to others measurable functions via standards convergence arguments (prove for simple functions and then approximate $X$ by increasing sequence of simple functions and use Monotone Convergence Theorem) 
Now, you have to convince yourself that $d(\alpha P) = \alpha dP$. Here you have to read and understand the concept of absolute continuous measures and the apply the Radon-Nikodym's theorem. But, another way is to repeat the argument I described above. Observe that for $X= \mathbb{1}_A$ you have 
$$\int_{\Omega} X d(\alpha P) = \alpha P(A)=\alpha\int_{\Omega} X dP$$
