$f \ RI \ on \ [a,b]$
$inf_{[a,b]}(f) = m_1 \leq f \leq m_2 = Sup_{[a,b]}(f)$
Then,
$$m_1(b - a) \leq \int_{a}^{b}f(x)dx \leq m_2(b - a)$$
I think I just need some clarification on this. Why is this not an obvious result? I feel like this is trivial and there is nothing to show. We know that the definition of the upper and lower sums are the infimum and supremum times the size of the interval.
By defininig the inf and sup as $m_1$ and $m_2$, I feel like $m_1(b - a)$ and $m_2(b - a)$ are literally $L(f,P)$ and $U(f,P)$.
Then, since $U(f) = Inf_{P} (U(f,P))$ and $I(f) = Sup_{P} (I(f,P))$,
it is a known result that $U(f) \leq U(f,P)$ and $I(f) \geq I(f,P)$
Since we also know that for the integral to exist, $\int_{a}^{b} f(x) dx = U(f) = I(f)$. So,
$$m_1(b - a) = I(f,P) \leq I(f) = \int_{a}^{b}f(x)dx = U(f) \leq U(f,P)) = m_2(b - a)$$
Is this actually the entire proof or am I missing something? It's basically a one-step problem right now, so I'm not entirely confident I'm doing it right. Can anyone confirm\deny this for me? Thanks!
You wrote that “$m_1(b - a)$ and $m_2(b - a)$ are literally $L(f,P)$ and $U(f,P)$”, but this doesn't make sense unless you tell is which partition $P$ is. It turns out that what you wrote is actually true if $P=\{a,b\}$. And it follows from this that, indeed\begin{align}m_1(b-a)&=L(f,\{a,b\})\\&\leqslant\int_a^bf(x)\,\mathrm dx\\&\leqslant U(f,\{a,b\})\\&=m_2(b-a).\end{align}