The title is basically the question:
Does there exist a non-measurable function $f$ where $\sqrt{f}$ is measurable?
I'm honestly not sure. My book does have a proposition that if $f$ is measurable, then $f^n$ where $n \in \mathbb{N}$ are measurable. However, it states this for naturals, not rationals. While no example is given, I'd assume the reason the statement is given this way is that $f^{q}$ where $q \in \mathbb{Q}$ need not be measurable. If it was true, I think they'd state the more general statement.
I know I am asking a different question than that proposition, but being that I don't believe the proposition would be true for a $q$ power, I'm not sure what to make of the question I'm being asked. I can't seem to think of a counterexample, but I'm also not convinced it's true. Can anyone shed some light on this?
Part 1: If you don't assume $f > 0$, then $f^q$ may not uniquely defined for $q\in \Bbb Q \setminus \Bbb N$. For instance, if $f(x)=0$ and $q=-1$, $f^q(x)$ is not defined. If $f(x)=-1$ and $q=1/2$, $f^q(x)$ can be $i$ or $-i$.
So, the real issue is what you meant by $\sqrt{f}$, when $f$ has negative values. You need to choose one branch of the complex square root and that leads to a more sophisticate environment. That is why your book (and other books) stay with $q\in \Bbb N$.
Part 2:"Does there exist a non-measurable function $f$ where $\sqrt{f}$ is measurable?" As we have seen in Part 1, the real issue is to assure that $\sqrt{f}$ is well-defined as function. If it is well-defined and it is a measurable function, then $f$ is measurable.
Proof: Let $g= \sqrt{f}$. Then $g$ is measurabe, So $f= g^2$ is measurable. $\square$
So, there exist not a non-measurable function $f$ where $\sqrt{f}$ is measurable.