What's wrong of my proof of that $f^2$ is Riemann integrable implies $f$ is Riemann integrable?

Question

Suppose $f$ is a bounded real function on $[a,b]$ and $f^2$ is Riemann integrable on $[a,b]$.

$f^2$ is Riemann integrable does not imply $f$ is Riemann integrable. As an example, $f = -1$ when $x$ is irrational; $f = 1$ when $x$ is rational.

However, if $f^3$ is Riemann integrable, then $f$ is Riemann integrable because: if $f^3$ is Riemann integrable on $[a,b]$, then $[f(a)]^3\le f^3 \le [f(b)]^3$. Since $g(u) = u^\frac13$ is obiviously on $[[f(a)]^3, [f(b)]^3], g[f^3(x)] = f(x)$ is Riemann integrable on $[a,b]$.

What if I use $\psi(u) = u^\frac12$ to replace the $g(u)$ above, then we can prove that $f^2$ is Riemann integrable implies $f$ is Riemann integrable, doesn't it? But this is wrong. Then, what's wrong with the later proof?

Answer 1

$u^{1/2}$ is not onto, unlike $u^{1/3}$. You already gave a counterexample. How can you reconstruct $-1$ from $1^{1/2}=1$? You cannot just randomly choose different root for $u^{1/2}$. Because such choice won't be (Riemann) integrable in general. And secondly picking $-1$ instead of $1$ for $1^{1/2}$ will reconstruct $-1$, but now you lose $1$!

When we talk about $u^{1/2}$ we only consider non-negative root, because this choice leads to good properties (continuous, integrable, etc.). Unlike $u^{1/3}$ which is a well defined homeomorphism, and no choice is involved here (the equation $x^3=u$ has a single solution).

So in reality, $\big(f^{2}\big)^{1/2}\neq f$. This is true only when $f\geq 0$ to begin with, in fact $\big(f^{2}\big)^{1/2}=|f|$ (the absolute value). While $\big(f^{3}\big)^{1/3}= f$ is always true.

Note that you did show something though. That if $f^2$ is Riemann integrable then so is $|f|$.

Answer 2

The difference between $g$ and $\psi$ is that it is true that $f=g\circ f^3$, but it is not always true that $f=\psi\circ f^2$. For example, whenever $f(x)$ is negative, you have $\psi(f^2(x)) = -f(x)$.