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Given independent random variables $X$, $Y$ with density functions $f$, $g$, what is the density function of the random variable $X+Y$?

I want to take the derivative of the cumulative distribution function:

$$\begin{array} FF_{X+Y}(a) & = P(X+Y\leq a)\\ & = P(X\leq a-Y)\\ & = P(\left\{\omega\in\Omega\colon X(\omega)\leq a-Y(\omega)\right\})\\ & = P(\left\{\omega\in\Omega\colon Y(\omega)\leq y,\, X(\omega)\leq a-y,\, y\in\mathbb{R}\right\})\\ & =^? \int_{\mathbb{R}} g(y)\int_{-\infty}^{a-y} f(x) \operatorname{d}x \operatorname{d}y \\ & = \int_{\mathbb{R}} g(y)F_X(a-y)\operatorname{d}y, \end{array}$$

but I am not sure about the marked equality. Intuitively, it seems true to me, but I am not sure about the implicit steps involved. The independency must have been used, and why, exactly, are you allowed write the two integrals? The set in the line above is not of a form that allows using the integration-property of the density functions.

Further, how do I proceed, and what is the final solution?

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    Look at the joint random variable $(X,Y)$ with pdf $h(x,y)$ then $P(X+Y \le a) = \iint_{x+y \le a} h(x,y)dxdy = \int_{-\infty}^\infty \int_{-\infty}^{a} h(u-y,y)dudy=\int_{-\infty}^\infty \int_{-\infty}^{a} f(u-y) g(y)dudy = \int_{-\infty}^{a} k(u)du$ where [$k(u) = \int_{-\infty}^\infty f(u-y) g(y)dy$](https://en.wikipedia.org/wiki/Convolution#Definition) which is the pdf of $X+Y$ – 2017-01-25

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