measure theory - Lebesgue Measurable function raised to a power.

I would like to prove or give a counterexample to the following. Thanks for any help in advance.

Let $f$ : R → R be a function.

a) Suppose $f^{2}$ is Lebesgue measurable. Does it follow that $f$ is Lebesgue measurable?

b) Suppose $f^{3}$ is Lebesgue measurable. Does it follow that $f$ is Lebesgue measurable?

Answer

Consider the function

$
f(x) =\begin{cases}
&\frac{1}{|x|} & x>1 \\
&1 & |x| \le 1.
\end{cases}
$

It's easy to see that this is not integrable where as its square and its cube is.

Edit: Sorry I miss-read integrable instead of measurable. To answer the actual question...consider

$
f(x) =\begin{cases}
&-1 & x\in \{\text{Some non-measurable set}\} \\
&1 &\text{otherwise.}
\end{cases}
$

Clearly $f^2$ is measurable where as $f$ is not.

For the second part:
Note that $x^\frac{1}{3}$ is monotonic increasing hence measurable and since the composition of measurable functions is measurable we can deduce $f^3$ measurable implies $f$ measurable.