Consider
I understand everything except of the first line of the note. In particular $$ \int_\Omega 1_{X(\omega ) + Y (\omega )\in A} \;d \mathbb P(\omega ) =\iint \mathbf{1}_A (x+y) \; \;d(\mu, \nu) (x, y) $$ I don't quite understand what is $d(\mu,\nu)$. I assume that the author is using the change of variables formula but I can't quite see how are the measures related. In particular, I thought I had to use the change of variables formula \begin{align*} \int_{}^{} g \, \mathrm{d}\nu = \int g \circ f \, \mathrm{d} \mu . \end{align*} where $\nu = \mu (f^{-1})$.
Question: Rigorously, how do you justify $$ \int_\Omega 1_{X(\omega ) + Y (\omega )\in A} \;d \mathbb P(\omega ) =\iint \mathbf{1}_A (x+y) \; \;d(\mu, \nu) (x, y) $$
In the formula (changing the names for the measures, since $X$, $Y$ have laws $\mu$ and $\nu$) $$\int g d\lambda = \int g \circ f d\eta,$$ set $$g(x,y) = 1_A(x+y),$$ $$f(\omega) = (X(\omega), Y(\omega)),$$ $$\eta(S) = \mathbb P(S),$$ for events $S$ and finally $$\lambda(U) = \mathbb P(f^{-1}(U))$$ for measurable $U \subset \mathbb R^2$. Note that when $X$ and $Y$ are independent, then this is just the product measure, as $$\lambda(U_1 \times U_2) = \mathbb P(X \in U_1, Y \in U_2) = \mu(U_1) \times \nu(U_2) $$ which is what the author means by $d(\mu, \nu)$. Once everything is substituted, you get $$\int gdv = \int 1_{A}(x + y) d\lambda = \int 1_{A}(x + y) d(\mu, \nu)$$ and $$ \int g \circ f d \eta = \int 1_{\{X(\omega) + Y(\omega) \in A\}} d\mathbb P $$ as desired.