Squeeze theorem for riemann integrable functions

Question

$\text{Squeeze theorem :}$ Let $f: [a,b]\rightarrow \mathbb{R}$. Then $f\in \mathcal{R}[a,b]$ if and only if for every $\epsilon>0$ there exists functions $\alpha_\epsilon$ and $\omega_\epsilon$ in $\mathcal{R}[a,b]$ with $$\alpha_\epsilon\le f(x)\le\omega_\epsilon\space \text{for all}\space x\in[a,b]$$ and such that $$\int\limits_a^b(\omega_\epsilon-\alpha_\epsilon)<\epsilon$$

Question : What is the negation of the squeeze theorem? Would the following work?

$f\notin \mathcal{R}[a,b]$ if and only if there exists an $\epsilon>0$ such that for all functions $\alpha_\epsilon$ and $\omega_\epsilon$ in $\mathcal{R}[a,b]$ with $\alpha_\epsilon\le f(x)\le\omega_\epsilon\space \text{for all}\space x\in[a,b]$ we have $\int\limits_a^b(\omega_\epsilon-\alpha_\epsilon)\ge\epsilon$

Answer 1

Denote by $P$ the proposition "$f\in\mathcal{R}[a,b]$", and by $Q$ the proposition "for every $\epsilon>0$ there exists... such that..."
The squeeze theorem states that $(P\Rightarrow Q)\land(Q\Rightarrow P)$. Hence the nagation of it is $(\lnot(P\Rightarrow Q))\lor(\lnot(Q\Rightarrow P))$.
We see $\lnot(P\Rightarrow Q)$ may be stated as:

There exists a function $f\in\mathcal{R}[a,b]$ such that $\lnot Q$ is true, i.e., such that the following holds:

There exists some $\epsilon>0$ such that for any function $\alpha_\epsilon,\omega_\epsilon\in\mathcal{R}[a,b]$ satisfying $\alpha_\epsilon\leq f\leq\omega_\epsilon$, we have $$\int_a^b(\omega_\epsilon-\alpha_\epsilon)dx\geq\epsilon$$

On the other hand, $\lnot(Q\Rightarrow P)$ is

There exists a function $f\not\in\mathcal{R}[a,b]$ such that $Q$ is true, i.e., such that the following holds:

For every $\epsilon>0$ there exists a pair of functions $\alpha_\epsilon,\omega_\epsilon\in\mathcal{R}[a,b]$ satisfying $\alpha_\epsilon\leq f\leq\omega_\epsilon$ such that $$\int_a^b(\omega_\epsilon-\alpha_\epsilon)dx<\epsilon$$

The negation of the squeeze theorem is of the form "$A$ or $B$", where $A,B$ is the two propositions stated above.