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: 
You can use the A.M. - G.M. inequality which is as follows-
$$\frac{x\_1+x\_2+\cdots+x\_n}{n}\ge\sqrt[\large n]{x\_1\cdot x\_2\cdots x\_n}$$
If you could, hey, why do you think people would have defined two different notions of mean?

However you can compute the arithmetic mean from the geometric mean of *other* numbers. Namely the arithmetic mean of $x\_1,\ldots,x\_n$ is the natural logarithm of the geometric mean of $e^{x\_1},\ldots,e^{x\_n}$. Not sure why anybody would like to compute it this way though.