Problem:
Let $X$ and $Y$ be independent real random variables with distributions $\mu=f\lambda$ and $v=g\lambda$ respectively ($f$ and $g$ are densities w.r.t. $\lambda$ the Lebesgue measure). Then $Y\neq 0$ almost surely, so the quotient $Q:=X/Y$ may be formed. Show that its distribution has, with respect to $\lambda$, the density
$$q(z):=\int f(zy)g(y)|y| \lambda(dy)$$
My attempt:
$Q$ is defined a.s. and so there exist a measurable extension of $Q$ to the whole space (for example by setting $Q:=0$ on a null set). Every such extension will have the same distribution $Q(P)$ and so the distribution of $Q$ is well-defined. Also the function $(y,z)\mapsto f(zy)g(y)|y|$ is non-negative and $\mathcal{B}(\mathbb{R})\otimes \mathcal{B}(\mathbb{R})$ measurable, and so by Tonelli's theorem $q(z)$ is a non-negative measurable function. Hence $ q\lambda (A):= \int_A q(z) \lambda(dz)$ defines a Borel measure on $\mathbb{R}$ and we need to show that $Q(P)=q\lambda$. By the uniqueness theorem for measures, it is sufficient to show that both measures agree on sets of the form $(-\infty,t]$ with $t\in \mathbb{R}$. (those sets form a $\cap$-stable generator of $ \mathcal{B}(\mathbb{R})$).
Define $Z(\omega):=(X(\omega),Y(\omega))$. Then $Z$ is a $\mathcal{B}(\mathbb{R})\otimes \mathcal{B}(\mathbb{R})$ random variable, and because $X$ and $Y$ are independent we have $P_Z=P_X \otimes P_Y$. Let $E:=\{(x,y)\in \mathbb{R}^2: x\leq ty, y>0\}\in \mathcal{B}(\mathbb{R})\otimes \mathcal{B}(\mathbb{R})$ so that $\{X\leq tY, Y>0\}=Z^{-1}(E)$.
We have
$$P(Q\leq t,Y>0)=P(X\leq tY,Y>0)=P(Z^{-1}(E))=P_Z(E)=(P_X \otimes P_Y)(E)=\int 1_E d(P_X \otimes P_Y)$$
$$=\iint 1_E dP_X dP_Y=\iint 1_E f(x) \lambda(dx) dP_Y=\iint1_E f(x)g(y) \lambda(dx) \lambda(dy)$$
If $y>0$ then the change of variable formula tell us that $$\int1_E f(x)g(y) \lambda(dx)=\int 1_{\frac{1}{y}E} f(xy)g(y)|y| \lambda(dx)=\int 1_{\{x\leq t\}} f(xy)g(y)|y| \lambda(dx)$$
and if $y\leq0$ then $$\int1_E f(x)g(y) \lambda(dx)=0$$ so
$$\int1_E f(x)g(y) \lambda(dx)=\int 1_{\{x\leq t\}}1_{\{y>0\}} f(xy)g(y)|y| \lambda(dx)$$
when regarded as a function of $y$. Pugging back and changing the order of integration we get
$$P(Q\leq t,Y>0)=\int_{-\infty}^t \int_{0}^{\infty} f(xy)g(y)|y| \lambda(dy) \lambda(dx)$$
A similar argument, this time using $E:=\{(x,y)\in \mathbb{R}^2: x\geq ty, y<0\}$, gives
$$P(Q\leq t,Y<0)=P(X\geq tY,Y<0)=\int_{-\infty}^t \int_{-\infty}^{0} f(xy)g(y)|y| \lambda(dy) \lambda(dx)$$
Putting everything together we get
$$P(Q\leq t)=P(Q\leq t,Y>0)+P(Q\leq t,Y=0)+P(Q\leq t,Y<0)$$
$$=\int_{-\infty}^t \int_{0}^{\infty} f(xy)g(y)|y| \lambda(dy) \lambda(dx)+0+\int_{-\infty}^t \int_{-\infty}^{0} f(xy)g(y)|y| \lambda(dy) \lambda(dx) $$
$$=\int_{-\infty}^t \int f(xy)g(y)|y| \lambda(dy)\lambda(dx)=\int_{-\infty}^t q(z)dz$$
which is what we wanted to show.
Is this the right way to proceed?
Thanks lot for your help.