# 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.

• If they haven't defined limits, what do they give for the definition of 'integrable'? That seems like the place to start. Commented Jul 19, 2017 at 19:10
• Very fair question; unfortunately, the answer is long winded. This is why I referred to Apostol's book which characterizes an ordinate set as "measurable" if it can be "enclosed by step functions". I think that it it pointless for me to elaborate further, since I believe that my original question can only be attacked by someone with access to Apostol's book. Commented Jul 21, 2017 at 20:05
• Your question helps me a lot. I read Apostol's Calculus in 2018, then I had the some question on the proof of theorem 1.11 and couldn't understand why the author asserted like that. Later, I gave up this book. Yesterday I picked up this book again, rethink this question :) Commented Nov 16, 2021 at 8:56

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 $a(Q^\prime)=I=a(Q)$.

• wow, right on. I voted to award 2 points because I couldn't find a flaw in your logic and because I feel that this analysis "belongs in the book". Apparently, I don't have enough reputation points to alter your answer's "score". I ASK THAT A SITE ADMIN AWARD 2 POINTS. Commented Mar 1, 2018 at 18:11
• How did you decide that $S' \subseteq Q'$, why isn't it possible that $S' \not\subseteq Q'$? By definition, $s(x) \leq f(x)$, and this is how we infer that $S \subseteq Q$. However, if the ordinate set $f(x)$ outline is removed, then possible that $S' \not\subseteq Q'$.
– John
Commented May 7, 2019 at 15:29
• @John The reasoning is that when you remove the outline of $Q$, you are removing the outline of $S$ if it's overlapping the one of $Q$. So if we define $S' = S - G$, being $G$ the graph of $f$, and $Q' = \{(x, y) | a \leq x \leq b, 0 \leq y < f(x) \}$, it's clear that if $x \in S'$ then $x \in Q$. Or more generally, in set theory $S \subseteq Q$ implies $S - G \subseteq Q - G$ for all $S, Q, G$. Commented Apr 21, 2021 at 15:39
• @John Partial alternative to the response of Daniel Muñoz Parsapoormoghadam : $~S' \subseteq S \subset Q.~$ Assume $~S'~$ not $~\subseteq Q'.~$ Then $~\exists ~(x_0,y_0) \in S'~$ such that $~ (x_0,y_0) \in Q - Q'.~$ This implies that $~y_0 < s(x_0)~$ and $~y_0 = f(x_0).~$ This implies that $~f(x_0) < s(x_0),~$ which contradicts the specification of $~s(x).$ Commented Jan 17 at 14:47