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-
$$\frac{x\_1+x\_2+\cdots+x\_n}{n}\ge\sqrt[\large n]{x\_1\cdot x\_2\cdots x\_n}$$
For an arithmetic mean $a$ and for a geometric mean $g$, we can see

1. $\sqrt{x\cdot y}=a \leftrightarrow x\cdot y=a^2$
2. $\frac{x+y}{2}=b \leftrightarrow x+y=2b \leftrightarrow y=2b-x$

Substituting (2) into (1) we get

$$x\cdot(2b-x)=a^2 \leftrightarrow -x^2+2bx-a^2=0 \leftrightarrow x=-\frac{4b^2\pm\sqrt{4b^2-4a^2}}{2}$$

Thus not even a calculation of gm from am is possible, but for every positive $(a;b)$ am, gm pair we can find the $(x,y)$ real number pair, whose am and gm are the given values!