Quotient of measurable functions is measurable

I need to show that if $f,g : X \to \mathbb{R}$ are measurables with respect to the $\sigma$-algebra $S$ of $X$, and $g(x) \neq 0, \forall x \in X$, then $f/g: X \to \mathbb{R}$ is measurable. So far what i got is: I need to show that
{$x \in X: (f/g)(x) \leq c$} $\in S$.

1. First Proof: {$x \in X: (f/g)(x) < c$} $\in S$

   Without loss of generality I can write this set as

$$\{x \in X: f(x) < c \cdot g(x)\} \implies \exists q \in \mathbb{Q}: \{x \in X: f(x) < q < c \cdot g(x)\} \implies \bigcup_q [ \{x \in X: f(x) < q\} \cap \{x \in X: q < c \cdot g(x)\} ] $$ Here's where i'm stuck. I know that $\{x \in X: f(x) < q \} \in S$ but, how do i prove that $\{x \in X: q < c \cdot g(x)\} \in S$ ?

I'm thinking the rest of the proof isn't that complicated, but if someone has more ideas, better!

Solution 1:

If $g$ is a measurable function then $\forall c\in \mathbb{R}, c\ne 0$, we have $c*g$ measurable. Moreover ${x \in X: g(x)> \frac{q}{c}}$ which is measurable.