Measure Theory: $X$, $Y$ independent, proof that $P_{X+Y} = P_{X} * P_{Y}$ I am studying measure theory, and I bumped in the the following: Let $X$ and $Y$ be two independent random variables with distributions $P_X$ and $P_Y$ respectively. A lot of sources conclude (or use in) the convolution theorem that $P_{X+Y} = P_{X} * P_{Y}$. However, I do not see how this can be proved using measure theory. 
The definition concerning convolution I am given from my lecture notes is the following:
$\textit{Let } f,g: \bar{\mathbb{R}} \to [0,\infty) \textit{ be Lebesgue measurable. The convolution product of $f$ and $g$ is defined by}$ $$h(x) = \int_{\bar{\mathbb{R}}} f(x-y)g(y) \; \mathrm{d}m(y), $$
$\forall x \in \bar{\mathbb{R}}\textit{ for which the integrand is Lebesgue integrable.}$ $\textit{The function $h$ is called the convolution of $f$ and $g$, denoted by $h=f*g$}.$
Can anyone explain to me how  $P_{X+Y} = P_{X} * P_{Y}$?
 A: Let $Z:=(X,Y)$, then we have that
$$
\Pr[X+Y\leqslant c]:=\Pr(\{\omega \in \Omega :X(\omega )+Y(\omega )\leqslant c\})\\
=\Pr(\{\omega \in \Omega :x+y\leqslant c\,\land\,(X(\omega ), Y(\omega ))=(x,y)\})\\
=\Pr(\{\omega \in \Omega :x+y\leqslant c\,\land\,Z(\omega)=(x,y)\})\tag1
$$
Because $X$ and $Y$ are independent then $F_Z=F_X\cdot F_Y$, that is, the distribution of $Z$ is the product of the distributions of $X$ and $Y$; and from $\mathrm{(1)} $ we find that $\Pr[X+Y\leqslant c]=\Pr[Z\in S_c]$ for $S_c:=\{(x,y)\in \Bbb R ^2:x+y\leqslant c\}$, so
$$
\Pr[X+Y\leqslant c]=\iint \mathbf{1}_{S_c}(x,y)\,\mathrm d F_X(x) \,\mathrm d F_Y(y)\tag2
$$
Now, the convolution of two measures $\mu $ and $\nu $ is defined by
$$
\mu *\nu (A):=\iint \mathbf{1}_{A}(x+y)\,\mathrm d \mu(x)  \,\mathrm d \nu(y) \tag3
$$
In our case
$$
\,\mathrm d F_X*\,\mathrm d F_Y(A):=\iint \mathbf{1}_{A}(x+y)\,\mathrm d F_X(x) \,\mathrm d F_Y(y)\tag4
$$
Equating both definitions we need to show that
$$
\mathbf{1}_{S_c}(x,y)=\mathbf{1}_{(-\infty ,c]}(x+y)\tag5
$$
what is not hard to see using the definition of $S_c$.
