How to proof that the ratio of two RANDOM variables is still a RANDOM variable? I know that there is a similar post on this on the following link:
How to prove the ratio of two random variables is also a random variable
However, the answers were not very clear.
So, how to prove that $Q = \frac{X}{Y}$ is a random variable given that $X$ and $Y$ are random variables.
If the solution is using the product of two random variables, how can i prove it? and also, how can i prove that the inverse of a random variable is still random variable.
Sorry if this question is too basic.

Let $Y$ be a random variable then it's easy to show that for each measurable function $f$ also $f\circ Y$ is a random variable due to $(f\circ Y)^{-1}(A) = Y^{-1}\left(f^{-1}(A)\right)$.
Because $f$ is measurable so is $B=f^{-1}(A)$ for each measurable $A$ and because $Y$ is a RV so is $Y^{-1}(B) = Y^{-1}\left(f^{-1}(A)\right)$.
Take $f: \Bbb R \setminus\{0\} \to  \Bbb R \setminus\{0\}, f(x) = \frac{1}{x}$ which is measurable because it's continuous gives you that $\frac{1}{Y}$ is a RV.
To show that the product of two RV is a RV is the same argument for the continuous hence measurable function $u(x,y) = xy$ and the fact that $(XY)^{-1} = \binom{X}{Y}^{-1}\circ u^{-1}$ and $\binom{X}{Y}^{-1}(A \times B) = \binom{X^{-1}(A)}{Y^{-1}(B)}$.
Consider that $\{A \times B \;|\; A,B\in\mathcal{B}(\Bbb R)\}$ is a generator of $\mathcal{B}(\Bbb R^2)$ which is stable under intersections so it's enough to show it for these kinds of sets…
