If $f:\mathbb{R}\to\mathbb{R}$ measurable with $f(x) \neq 0$, prove there exists measurable $g(x)$ with $\frac{1}{2}f(x) < g(x) < f(x)$

Question

The question I have been assigned is this:

If $f:\mathbb{R}\to\mathbb{R}$ measurable with $f(x) \neq 0, \forall x \in \mathbb{R}$, prove there exists measurable $g(x)$ such that $\frac{1}{2}f(x) < g(x) < f(x), \forall x\in \mathbb{R}$.

My question here is twofold:

Surely the question means $f(x) > 0, \forall x \in \mathbb{R}$ otherwise $f(x) = -1$ contradicts $\frac{1}{2}f(x) < f(x)$.

My other question is that $g(x) = \frac{3}{4}f(x)$ should work, shouldn’t it?

I am new to this topic so I feel it is likely I am wrong here, I’d just like to know how. Thank you.