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: 
Unfortunately the AM-GM inequality is the best you can do. If your data is $\{x,\frac{1}{x}\}$ the geometric mean will be $1$, yet you can make your arithmetic mean any value in $[1,+\infty)$ by choosing $x$ large enough.
Of course the answer is "no," not without more data. There's not a direct calculation.

In finance, what's interesting is that the arithmetic mean will always be a bit higher than the geometric for year on year stock returns. 

From 1929 to 2013, the average was 11.41% (S&P return) yet the geometric mean was 9.43% nearly 2% lower per year. An understanding of the math behind this difference is helpful when someone mentions the market's return over a particular period.