If $X$ and $Y$ are continuously distributed independent random variables with respective densities $f$ and $g$ then the density of $X+Y$ is
$$h(z) = \int_{-\infty}^{\infty} f(x)g(z-x)\ \mathrm {dx}.$$
I found that if $Z=X+Y$ and if $z \in \Bbb R$ then
$$\Bbb {P} (Z \leq z) = \int_{-\infty}^{z} \int_{-\infty}^{\infty} f(x)g(w-x)\ \mathrm {dx}\ \mathrm {dw}.$$ i.e.
$$F_Z (z) = \int_{-\infty}^{z} \int_{-\infty}^{\infty} f(x)g(w-x)\ \mathrm {dx}\ \mathrm {dw}.$$ where $F_Z$ is the CDF of $Z$. Can I differentiate it? I want to apply Fundamental Theorem of Integral Calculus here. For that we need to ensure that for all $w \in \Bbb R$ the function $h$ given by
$$h(w)= \int_{-\infty}^{\infty} f(x) g(w-x)\ \mathrm {dx}$$ should be continuous. Right? But how can I say that? Please help me in this regard.
Thank you very much.
For $B\in\mathscr{B}(\mathbb{R})$ a borel set. We have \begin{align} \mathbb{P}_{X+Y}(B)=\mathbb{P}(X+Y \in B)&\stackrel{\text{Defn.}}{=}\int_\Omega {1}_B(X+Y) ~d\mathbb{P} \\ &\stackrel{\text{dist.}}{=}\int_{\mathbb{R}^2} {1}_B(x+y)~\mathbb{P}\left((X,Y) \in d(x,y)\right) \\ &\stackrel{X\perp Y}{=} \int_{\mathbb{R}^2} {1}_B(x+y)~\mathbb{P}_X\otimes \mathbb{P}_Y(d x,d y) \\ &\stackrel{\text{Fubini}}{=}\iint {1}_B(x+y)~d \mathbb{P}_X(x)d\mathbb{P}_Y(y)\\ &\stackrel{\text{density}}{=}\iint {1}_B(x+y)f_X(x)g_Y(y)~dx~dy\\ &\stackrel{z=x+y}{=}\iint {1}_B(z) f(z-y)g(y)~dydz\\ &=\int_B \int f(z-y)g(y)~dy ~dz &\end{align}
So $X+Y$ has the density $\int f(z-y)g(y)~dy$. Here $\mathbb{P}$ is the probability measure, $\mathbb{P}_X$ is the image measure under $X$ and $f_X$ the density.