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


*

*dealing with the possibility that $f(x)$ may be equal to zero anywhere in $[a,b]$

*visualizing the set of of all step functions $s$ such that $s(x) \lt  f(x)$ throughout $[a,b]$

*visualizing the set of of all step functions $t$ such that $t(x) \gt f(x)$ throughout $[a,b]$.


I request a proof of the assertion.  
Note:
Although I was impressed with the approach taken in 
Proving area equal to zero of a continuos function.,
it involves a continuous function, uniform continuity, and limits.  Theorem 1.11's premise merely requires that f be integrable.  Further, the concepts of uniform continuity and limits are after page 75 in the referenced book.
 A: I got stuck on this one too. To get unstuck, I considered the simpler case of rectangles:
Let $$R=\{(x,y)|a\leq x\leq b, 0\leq y\leq h \} $$
$$R^\prime=\{(x,y)|a\leq x\leq b, 0\leq y< h \}$$
$$L=\{(x,y)|a\leq x\leq b, y = h \}$$
$L$ is a line segment, which we decided earlier in the book was measurable with $a(L)=0$. $R$ is measurable by axiom 5, with $a(R)=h(b-a)$. 
 $R^\prime=R-L$, and is therefore measurable by axiom 3 with $a(R^\prime)=a(R-L)=a(R)-a(L)=h(b-a)$. You can come to a similar conclusion working with the exhaustion property, but it is a bit more longwinded. If we, for want of a better term, define the mapping of $R$ to $R^\prime$ as “removing the top edge”, we can conclude that removing the top edge of a rectangle does not affect its area. This naturally extends to step regions by considering each sub-interval of its partition separately then applying axiom 2: if you remove the top edge of a step region, you do not affect its area.
To bring this result into the proof of theorem 1.10 and 1.11, we map each $S$ to an $S^\prime$ and each $T$ to a $T^\prime$ by removing the top edge. $S^\prime$ and $T^\prime$ then satisfy $S^\prime\subseteq Q^\prime\subseteq T^\prime$ with $a(S^\prime) = a(S)$ and $a(T^\prime)=a(T)$. Following the logic of the theorem 1.10 proof, $Q^\prime$ is measurable with $a(Q^\prime)=I=a(Q)$.
