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?
 A: 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.
A: Hints: a) think about $$f = \chi_E - \chi_{\mathbb R \setminus E}$$
b) if $g$ is Lebesgue measurable, then so is $h\circ g$ for every continuous $h$ on $\mathbb R.$
A: By a composition being Lebesgue measurable I assume you mean $f \circ g$ is $(\mathcal{L}, \mathcal{B})$-measurable where $\mathcal{L}$ is the Lebesgue sigma algebra and $\mathcal{B}$ is the Borel sigma algebra.  We want to know whether $f$ is $(\mathcal{L}, \mathcal{B})$-measurable.
Let $g(y) = y^2$ and $f(x): \mathbb{R} \to \mathbb{R}$.  Then for any $B \in \mathcal{B}$ we necessarily have $f^{-1}(g^{-1}(B)) \in \mathcal{L}$ by assumption.  To show $f$ is $(\mathcal{L}, \mathcal{B})$-measurable we need that $f^{-1}(A) \in \mathcal{L}$ for any $A \in \mathcal{B}$.  But, if we let $B = g(A)$ for some $A \in \mathcal{B}$ then we have only that $A \subset g^{-1}(B)$ and know only that $f^{-1}(g^{-1}(B)) \in \mathcal{L}$; there is no guarentee that $f^{-1}(A) \in \mathcal{L}$.  In other words, there exist Borel sets $A$ for which we cannot be sure $f^{-1}(A) \in \mathcal{L}$, so $f$ is not necessarily $(\mathcal{L}, \mathcal{B})$-measurable.
If $g$ were injective on $\mathbb{R}$ like $g(y) = y^3$ then we would have $A = g^{-1}(g(A))$ and then $f$ would be $(\mathcal{L}, \mathcal{B})$-measurable.
