This question references "Calculus", vol 1, 2nd ed. 1966, by Tom M. Apostol. Theorem 1.11, page 75, says that if non-negative function f is integrable on [a,b] then its graph [ i.e. {(x,y) : a <= x <= b, y=f(x)} ] is "measurable" and has zero area. Just prior to this theorem, theorem 1.10 states that if f is integrable on [a,b] and Q is its ordinate set on [a,b] then Q is measurable and a(Q) = int[a,b] f(x)dx.

I followed the given proof to theorem 1.10 okay. Unfortunately, the book asserted that the same analysis that proved theorem 1.10 also applies to Q' = {(x,y) : a <= x <= b, y < f(x)}. The proof to theorem 1.11 is supposed to be a consequence of the assertion. I see no routine way of proving the assertion, especially because a "measurable" set must be "enclosable in step regions".

I am having trouble:
1. dealing with the possibility that f(x) may = 0 anywhere in [a,b]
2. visualizing the set of of all step functions s such that s(x) < f(x) throughout [a,b]
3. visualizing the set of of all step functions t such that t(x) > f(x) throughout [a,b].

I request a proof of the assertion.