The complete question is
What is the significance of proving the integrability of monotonic functions when it was proved the integrability of a general one?
Doesn't the integrability of a general function implies that of a monotonic one?
I'm reading Calculus I, which says:
Theorem 1.12: If $g$ is monotonic on a closed interval [a,b], then $g$ is integrable on $[a,b]$.
But by Theorem 1.9 we stablished the existence of the integral of a general function $f$ bounded on a closed interval iff $\underline{\mathbf{I}}(f) = \bar{\mathbf{I}}(f)$.
I don't think the proof of Theorem 1.12 seems to care about the character of $g$ as a monotonic function. It looks like one can stuck that proof with Theorem 1.19 and it would work equally well.
Certainly a general function $f$ whose domain is $[a,b]$ cannot be said to be monotonic, but, I think, one always can choose an arbitrary partition $P$ of $[a,b]$ with $n$ subintervals $[x_{k-1}, x_k]$ where $f$ is either increasing or decreasing and therefore monotonic on $[x_{k-1}, x_k]$.
In the way I'm (mis)understanding this, I would show the integrability of a general function $f$ on a closed interval (Theorem 1.19 does that), and then show that a set of monotonic functions is a subset of a set of general ones, which implies they are integrable too.
Well, I haven't dedicaded further thoughts to the last proposal, but I haven't been able to see the difference between the two cases either.
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