Density of sum of two random variable I have a problem with a derivation of the density of the sum of nonnegative two random variables. 
Suppose the joint density of nonnegative random variables $X,Y$ is $f_{X,Y}(x,y)$, a brutal force way to compute the density of $X+Y$ would be 
$$  f_{X+Y}(a) = \frac{d}{da} \int_{x+y\leq a} f_{X,Y}(x,y) dy dx =  \frac{d}{da} \int_{\mathbb{R}_+} \int_{0}^{a-x} f_{X,Y}(x,y) dy dx .$$
If we allowed to exchange the sign of the  differentiation sign and the integral sign, we obtain 
$$f_{X+Y} (a) = \int_{0}^{\infty} f_{X,Y}(x,a-x)dx.$$ 
So my question is why we can exchange the differentiation sign and the integral sign here. I did not see a simple dominant function so that we could apply the dominant convergence theorem. However, the last expression seems to be always correct by Fubini's  theorem. 
Did I miss something obvious? 
 A: Moving the differentiation operator inside the integral is an application of the fundamental theorem of calculus.
$$\begin{align}\dfrac{\operatorname d ~~}{\operatorname d a}\int_b^c \dfrac{\operatorname d G(x,a)}{\operatorname d x\qquad\;}\operatorname d x ~=~& \dfrac{\operatorname d (G(c,a)-G(b,a))}{\operatorname d a\qquad\qquad\qquad\quad}\\[1ex] =~& \int_b^c \dfrac{\operatorname d^2 G(x,a)}{\operatorname d a~\operatorname d x\quad}\operatorname d x
\\[3ex] \dfrac{\operatorname d ~~}{\operatorname d a}\int_b^c g(x,a)\operatorname d x ~=~& \int_b^c \dfrac{\operatorname d g(x,a)}{\operatorname d a\qquad}\operatorname d x
\\[4ex]\therefore\qquad \dfrac{\operatorname d ~~}{\operatorname d a}\int_{0}^{\infty}\color{silver}{\left(\color{black}{ \int_{0}^{a-x} f(x,y)\operatorname d y}\right)}\operatorname d x~=~& \int_{0}^{\infty} \dfrac{\operatorname d ~~}{\operatorname d a}\color{silver}{\left(\color{black}{ \int_{0}^{a-x} f(x,y)\operatorname d y}\right)}\operatorname d x\end{align}$$
The next step follows from similar reasoning, except this time the derivation variable occurs in the bounds of the (inner) integral.
