A different proof for "If $f$ is Riemann integrable, Then so is $f^2$ "

Question

Theorem :

If $f$ is Riemann integrable on $[a,b]$, Then so is $f^2$.

There are many proofs for this theorem. But i don't want the proofs which use $U(f,P)$ and $L(f,P)$. My book gives other definitions :

$S(f,P^t)=\sum_{i=1}^n f(t_i)(x_i-x_{i-1})$

$\omega(f,[a,b])=\sup\{f(x):x\in[a,b]\}-\inf\{f(x):x\in[a,b]\}$

We say $f:[a,b] \to \mathbb R$ is Riemann integrable on $[a,b]$, if there exists $L$ such that :

For each $\epsilon \gt 0$ there exists $\delta \gt 0$ such that for each labeled partition $P^t$ such that $||P^t|| \lt \delta$ , We have :
$|S(f,P^t)-L|\lt \epsilon$

This definition is not related to $U(f,P)$ and $L(f,P)$ and The question wants us to proof the above theorem with this definition and the properties of $\omega(f,[a,b])$. I think this property helps :

$\omega(f^2,[a,b]) \le 2M\omega(f,[a,b])$ ( $M$ is the bound of $f$ )

My problem is that i can't rewrite this definition in a way that $f^2$ becomes Riemann integrable.

Edit :

We have two related theorems which i think may help :

1. A bounded function $f$ is Riemann integrable on $[a,b]$ if and only if for each $\epsilon \gt 0$ there exists $\delta \gt 0$ such that for each two partitionings $P_1^t$ and $P_2^t$ such that their norms are less than $\delta$ , $|S(f,P_1^t)-S(f,P_2^t)| \lt \epsilon$

2. Assume that $f$ is a bounded function defined on $[a,b]$. Assume that $P_1=\{z_i:0\le i\le n\}$ is a partitioning for $[a,b]$ and $P_2$ is an elegant partitioning for $P_1$ ( Meaning it has more points ) If $P_1^t$ and $P_2^t$ are two labeled partitionings for $[a,b]$ derived from $P_1$,$P_2$ , Then :

$|S(f,P_1^t)-S(f,P_2^t)| \le \sum_{i=1}^n \omega(f,[z_{i-1},z_i])(z_i-z_{i-1})$

Answer 1

$\newcommand{\eps}{\varepsilon}$Hints using your book's definitions:

1. For every $\eps > 0$ and for every partition $P = (z_{i})_{i=0}^{n}$ of $[a, b]$, there exist labelings $P_{1}^{t}$ and $P_{2}^{t}$ such that $$ \sum_{i=1}^{n} \omega(f, [z_{i-1}, z_{i}])(z_{i} - z_{i-1}) \leq |S(f, P_{1}^{t}) - S(f, P_{2}^{t})| + \eps. \tag{*} $$ Proof: For each subinterval $I = [z_{i-1}, z_{i}]$, pick the label $t_{i}$ of $P_{1}^{t}$ so that $$ 0 \leq \sup_{x \in I} f(x) - f(t_{i}) < \frac{\eps}{2(b-a)}, $$ and pick the label $t'_{i}$ of $P_{2}^{t}$ so that $$ 0 \leq f(t'_{i}) - \inf_{x \in I} f(x) < \frac{\eps}{2(b-a)}. $$ Rearranging gives $$ \sup_{x \in I} f(x) - \inf_{x \in I} f(x) < f(t_{i}) - f(t'_{i}) + \frac{\eps}{b - a}. $$ Summing over $i$ gives inequality (*).

2. For each interval $I$ contained in $[a, b]$, you have $\omega(f^{2}, I) \leq 2M\, \omega(f, I)$.

Can you take it from here?