Find the density function of the sum $(X,X+Y)$. Problem
I have to prove the following:
Let $X$ and $Y$ be independent continuous random variables with density function $f:\mathbb R\to\mathbb R$. \
Prove that $(X,X+Y)$ is a continuous bivariate random variable with density function $f_{X,X+Y}(x,z)=f(x)f(z-x)$.
My thoughts
Look at $$\mathbb{P}(X\leqslant x,X+Y\leqslant z)=\mathbb{P}(X\leqslant x,Y\leqslant z-x)=\int_{-\infty}^{z-x}\int_{-\infty}^x f_{X,Y}(u,v)\,du\,dv.$$
Since $X$ en $Y$ are independent and have the same density function, we have $f_{X,Y}(u,v)=f(u)f(v)$, so
$$=\int_{-\infty}^{z-x}\int_{-\infty}^x f(u)f(v)\,du\,dv=\int_{-\infty}^{z-x} f(v)\left(\int_{-\infty}^x f(u)\,du\right)dv=\int_{-\infty}^{z-x} f(v)\,dv\cdot \int_{-\infty}^x f(u)\,du$$
Now I feel like I am very close, but we can't use the FTC since $f$ is not necessarily continuous (only right continuous). This might be a stupid question, but I don't know how to proceed
 A: $$
\Pr(X\le x\ \&\ X+Y\le z) = \int_{-\infty}^x \int_{-\infty}^{z\,-\,u} f_{X,\,Y} (u,v) \, dv \, du = \int_{-\infty}^x \int_{-\infty}^{z-u}f(u)f(v) \, dv\,du
$$
For any particular value $u$ that the random variable $X$ can assume between $-\infty$ and the value $x$ being considered, the value of $X+Y = u+Y$ is $\le z$ precisely if $Y\le z-u.$
Next we have
\begin{align}
& \int_{-\infty}^x \int_{-\infty}^{z\,-\,u} f(u) f(v) \, dv \, du = \int_{-\infty}^x \int_{-\infty}^z f(u)f(v-u) \, dv\,du \\[10pt]
\text{because } & \int_{-\infty}^{z-u} f(v)\,dv = \int_{-\infty}^z f(v-u)\, du \text{ by routine substitution.} 
\end{align}
Then we will have
$$
\Pr(X\le x\ \&\ X+Y\le z) = \int_{-\infty}^x \int_{-\infty}^z f(u) f(v-u) \,dv\,du
$$
and that implies that the function being integrated on the right side is the density of $(X,\,X+Y).$
A: The continuity of $X+Y$ is obvious from that of $X$ and $Y$.  Let $g:\Bbb{R}^2\to\Bbb{R}$ be a bounded continuous function.
In the second step, make a change of variables $z = x+y$.  Thanks to Fubini's theorem, changing order of integration is allowed.
\begin{align}
\Bbb{E}[g(X,X+Y)] &= \int_{\Bbb{R}^2} g(x,x+y) f_X(x) f_Y(y) \,dx\,dy \\
&= \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} g(x,x+y) f_Y(y) \,dy \right) f_X(x) \,dx \\
&= \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} g(x,z) f_Y(z-x) \,dz \right) f_X(x) \,dx \\
&= \int_{\Bbb{R}^2} g(x,z) f_X(x) f_Y(z-x) \,dx \,dz
\end{align}
Therefore, $(X,X+Y)$ has density $f_{X,X+Y}(x,z) = f_X(x) f_Y(z-x)$.
In particular, when $f_X = f_Y = f$, $f_{X,X+Y}(x,z) = f(x) f(z-x)$.
