Suppose $f$ is a bounded real function on $[a, b]$, and $f^2 \in \mathscr{R}$ on $[a, b]$. Does it follow that $f \in \mathscr{R}$? Does the answer change if we assume that $f^3 \in \mathscr{R}$?
Question: (1) Claim: If $f^3 \in \mathscr{R}$, then $f \in \mathscr{R}$. Proof: let $g:\Bbb{R} \to \Bbb{R}$, defined by $g(x)=x^3$. By 4.11 example, $g$ is continuous, because it’s a polynomial, to be more specific monomial. Now we’ll show it’s bijective. Let $y \in \Bbb{R}^+$. Then by theorem 1.21(existence of nth roots of positive reals), $\exists ! x \in \Bbb{R}^+ \subset \Bbb{R}$, such that $g(x)=x^3=y$. If $y=0 \in \Bbb{R}$, then it is clear that $\exists ! 0 \in \Bbb{R}$, s.t $0^3=0$. Let $y \in \Bbb{R}^-$, $y \lt 0$. So $-y \gt 0$. Again by theorem 1.21, $\exists ! x \in \Bbb{R}^+ \subset \Bbb{R}$, such that $x^3=-y \Leftrightarrow -(x)^3 =y$. $-(x)^3 = -1 \cdot x \cdot (x)^2 = (-x) \cdot (x)^2= (-x) \cdot (-x)^2= (-x)^3$. Thus $(-x)^3=y$, $ -x \in \Bbb{R}^- $. By axioms of field $-x$ is unique for each $x$. In other word, $\forall y \in \Bbb{R}^-, \exists ! -x \in \Bbb{R}^-$ s.t $(-x)^3=y$. Thus, function $g$ is continuous and bijective. By theorem 4.17, it’s inverse, $g^{-1}(y)=(y)^{\frac{1}{3}}$, is continuous on $\Bbb{R}$. Since $f^3 \in \mathscr{R}$, $f^3$ is bounded. Take $\phi (t)= (t)^{\frac {1}{3}}$. By theorem 6.11, $f \in \mathscr{R}$. Is this proof correct? proof would be much easier if we could show cube root function is continuous on $\Bbb{R}$, perhaps by using $\epsilon – \delta$ definition. Yeah I don’t known how to prove using $\epsilon – \delta$ definition. Is there any formula for $x^{\frac {1}{3}}-c^ {\frac {1}{3}}$?
(2) Is if $f^2 \in \mathscr{R}$, then $f \in \mathscr{R}$ ? I know it’s false implication. There are trivial functions for which above implication don’t hold. I was wondering, what if we use the same mechanism as we did in the proof of que(1). Since $f$ is bounded on $[a,b]$, $m \leq f \leq M$. So $0 \leq m^{\ast}\leq f^{2} \leq M^{\ast} \lt \infty$. If we take $\phi (t)= (t)^{1/2}$ on $[m^{\ast}, M^{\ast}]$, then by theorem 6.11 $f$ should be Riemann integrable. I must be missing something obvious.
Best Answer
For the first part, your proof would have been clearer if you didn't use the notation $f(x)=x^3$ (since $f$ is already used to denote a function with $f^3$ being Riemann-integrable). But yes, $\phi:\Bbb{R}\to\Bbb{R}$, $\phi(t)=t^{1/3}$ (defined the usual way for negative reals) is continuous, hence $f=\phi\circ f^3$ is also Riemann-integrable.
If you take $\psi:[0,\infty)\to[0,\infty)$, $\psi(t)=t^{1/2}$, then all you can conclude is that $\psi\circ f^2=|f|$. So, the correct statement is that "if $f:[a,b]\to\Bbb{R}$ is a function such that $f^2$ is Riemann-integrable then $|f|$ is Riemann-integrable". Alternatively, you can say "if $f:[a,b]\to\Bbb{R}$ is a function such that $f^2$ is Riemann-integrable AND $f$ is non-negative then $f$ is Riemann-integrable".