If $f^3(x)$ is integrable then $f(x)$ is integrable? 
Let $f:[a,b] \to \mathbb{R}$ be a function such that $f(x)$ is bounded and $f^3(x) \in \mathcal{R}$ [integrable] then does it follow that $f$ is integrable?

My attempt:
Does $f^3$ integrable imply $f$ integrable? .Heres an answer to the question but I have tried doing it differently.
Let us consider a partition $P$ on $[a,b]$ such that $x_o = a < x_1 < \cdots < x_n = b$ and an interval $[x_{r-1},x_r] $ and $U(P,f^3) - L(P,f^3) < \epsilon 3A^2$
$f(\alpha) - f(\beta) = \frac{ (f(\alpha) - f(\beta))(f^2(\alpha) + f^2(\beta) + f(\alpha).f(\beta))}{(f^2(\alpha) + f^2(\beta) + f(\alpha).f(\beta))} = \frac{(f^3(\alpha) - f^3(\beta))}{(f^2(\alpha) + f^2(\beta) + f(\alpha).f(\beta))} \le \frac{M_r-m_r}{(f^2(\alpha) + f^2(\beta) + f(\alpha).f(\beta))}$
where $M_r,m_r$ are the supremum and infimum of the function $f^3(x)$ in $[x_{r-1},x_r]$ and $\forall \alpha,\beta \in [x_{r-1},x_r]$.
Since $f(x)$ is bounded we can conclude that $f^2(x)$ is also bounded.
Case $1$: $f(x)$ is bounded below by $A$ where $A$ is negative then $\frac{1}{f^2(x)} \le \frac{1}{A^2}$ so we conclude that $\frac{1}{f^2(\alpha) + f^2(\beta) + f(\alpha).f(\beta)}$ is bounded by $\frac{1}{3A^2}$.
Case $2$:If $f(x)$ is bounded by $0$ then it is bounded by a smaller negative number $A$.Then $\frac{1}{f^2(x)} \le \frac{1}{A^2}$ so we conclude that $\frac{1}{f^2(\alpha) + f^2(\beta) + f(\alpha).f(\beta)}$ is bounded by $\frac{1}{3A^2}$.
Case $3$:If $A \le f^2(x)$ where $A > 0$ then it is bounded by a smaller negative number $A$.Then $\frac{1}{f^2(x)} \le \frac{1}{A^2}$ so we conclude that $\frac{1}{f^2(\alpha) + f^2(\beta) + f(\alpha).f(\beta)}$ is bounded by $\frac{1}{3A^2}$.
Thus, $U(P,f)-L(P,f) = \sum_{i=0}^{n-1}(\mbox{ sup}f(x)- \mbox{ inf}f(x))_{x \in [x_i,x_{i+1}]}(x_{i+1}-x_i) = \sum_{i=0}^{n-1}\frac{1}{3A^2}(M_{i+1}-m_{i+1})(x_{i+1}-x_i) = \frac{1}{3A^2} (U(P,f^3) - L(P,f^3)) < \epsilon $
Can someone go through this?
 A: There is a theorem

A function on a bounded interval [a,b] is Riemann integrable iff it is bounded and almost everywhere continuos.

So using this we obtain that $f^3(x)$ is bounded and almost everywhere continuous. Similarly let $\psi(x)=x^{1/3}$. Then we could say that $ (\psi\circ f^3)(x)=f(x)$ is almost everywhere continuous and bounded. Which means $f$ is integrable.
Note: Maybe you wanted a proof that contains "partitions" or  talks about partitions in a way but as you know they could get really cumbersome.
A: I have read your proof in which there are two places confusing me.
$1.\, |f^2(\alpha) + f^2(\beta) + f(\alpha).f(\beta)|$ may equals to $0$. How can you put it as a denominator?
$2.$ $f$ is bounded which means $\exists A>0$ s.t. $|f|\le A.$ However, you don't know if $|f|$ is bounded from below, i.e. whether there exists a $ B>0$ s.t. $|f|\ge B>0\,?$ If you don't have a bound from below for $|f|$, then you may not bound the term $1\over |f^2(\alpha) + f^2(\beta) + f(\alpha).f(\beta)|$.
