Question: Can we give an example to show that the measurability of the set $f^{-1}(\{a\})$ for every a does not imply that $f$ is a measurable function?

Answer: 
No it is not. Arithmetic mean gives you one equation. And there are two numbers to solve for. So there are infinite possibilities. For example, $$a=1,b=100$$ $$ a=0,b=101$$
They both have same A.M but widely different G.M
Since you want a estmated relationship between AM and GM, and there's no further constraint on the data, I can present a widely used formula for approximation in finance. Hope this will light you up a bit.

Let $A=\frac{1}{n}\sum\_{k=1}^n r\_k$ denote AM of a series of return $r\_1,r\_2...r\_k$, $G=[\prod\_{k=1}^n (1+r\_k)]^{1/n}-1$ denote GM of the returns, then $G\approx A-\frac{1}{2}V$, where $V$ is variance of these returns.

***Proof:***

We can take $1/n$ inside the square brackets, and $G=\prod\_{k=1}^n (1+r\_k)^{1/n}-1$ , use the Maclaurin series expansion for $(1+r\_k)^{1/n}$ up to degree 2 and ignore the remainder:
$$(1+r\_k)^{1/n}\approx 1+\frac{1}{n}r\_k+\frac{1-n}{2n^2}r\_k^2$$
Substitute the expansion into $G$, we have $G\approx\prod\_{k=1}^n(1+\frac{1}{n}r\_k+\frac{1-n}{2n^2}r\_k^2)-1$, expand the product and drop terms with degree 3 and above, that is, you can see the expansion as selecting one term in one bracket such that their product's degree is no more than 2, then we have another level of approximation of $G$:
$$G\approx\prod\_{k=1}^n(1+\frac{1}{n}r\_k+\frac{1-n}{2n^2}r\_k^2)-1\approx \frac{1}{n}\sum\_{k=1}^nr\_k+\frac{1}{n^2}\sum\_{k\neq l}^nr\_kr\_l+\frac{1-n}{2n^2}\sum\_{k=l}^nr\_k^2$$

As for $V$, we have $$V=\frac{1}{n}\sum\_{k=1}^n(r\_k-A)^2=\frac{1}{n}\sum\_{k=1}^nr\_k^2-A^2=\frac{n-1}{n^2}\sum\_{k=1}^nr\_k^2-\frac{2}{n^2}\sum\_{k\neq l}^nr\_kr\_l$$
Observe $V$ and the last two terms of $G$, here we are: $G\approx A-\frac{1}{2}V$.