## zero area of a graph (request proof of theorem in book by Tom M. Apostol)

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.