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?

Why not carefully write down the "proof" applied to your specific counterexample and see which steps work amd which steps do not? — Jason DeVito - on hiatus, May 15 '23 at 11:57
Just because $f^3$ is integrable on $[a,b]$, doesn't necessarily mean that $a^3 \leq f^3 \leq b^3$. — stange, May 15 '23 at 11:58
If someone writes a good answer here we effectively get a dupe candidate. Basically Cameron Williams' comments there form the answer to this question. — Sarvesh Ravichandran Iyer, May 15 '23 at 12:05
@JasonDeVito One problem I found may be $\psi(u) = u^\frac12$ . However, I don't know how to describe the problem exactly. Could you please offer me a statement? Thanks very much! — Kenton Pendragon Su, May 16 '23 at 09:33
@SarveshRavichandranIyer Sorry, I made such a big mistake. I've corrected it. — Kenton Pendragon Su, May 16 '23 at 09:35
Maybe I understand. $\psi(u) = [f(x)^2]^\frac12$ does not equals $ f $; it's just seems like to be the same; but actually $f < 0$ will happen while $\psi(u) = [f(x)^2]^\frac12 < 0$ won't. — Kenton Pendragon Su, May 16 '23 at 09:46

freakish · Accepted Answer · 2023-05-16T10:02:13.013

$u^{1/2}$ 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 $|f|$.

Thanks very much. – Kenton Pendragon Su May 16 '23 at 09:49 — Kenton Pendragon Su, May 16 '23 at 09:49

score 1 · Answer 2 · answered May 16 '23 at 09:50

1

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 $\psi(f^2(x)) = -f(x)$.

answered May 16 '23 at 09:50

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

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