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There is a small detail in the proof of Theorem 2.1 that I'm having trouble reasoning with. Hoping someone can provide some insight!

Theorem 2.1: Assume $f$ and $g$ are integrable and satisfy $f \leq g \text{ on } [a, b]$. Then the region $S$ between their graphs is measurable and its area $a(S)$ is given by the integral

$$a(S) = \int_{a}^{b}{[g(x) - f(x)]dx}\label{2.1}\tag{2.1}$$

Proof: Assume first that $f$ and $g$ are nonnegative Let $F$ and $G$ denote the following sets:

$$F = \{(x,y) \;|\; a \leq x \leq b,\; 0 \leq y < f(x)\}, \quad G = \{(x,y)\;|\;a \leq x \leq b, \; 0 \leq y \leq g(x)\}$$

That is $G$ is the ordinate set of $g$ and $F$ is the ordinate set of $f$ minus the graph of $f$. The region $S$ between the graphs of $f$ and $g$ is the difference $S = G - F$. By theorems 1.10 (integral of non-negative $f$ on $[a,b]$ is the area of the ordinate set) and 1.11 (graph of $f(x)$ is measurable and area zero), both $F$ and $G$ are measurable. Since $F \subseteq G$, the difference $S = G - F$ is also measurable (difference property of area) and we have

$$a(S) = a(G) - a(F) = \int_{a}^{b}{g(x)dx} - \int_{a}^{b}{f(x)dx} = \int_{a}^{b}{[g(x) - f(x)]dx}$$

This proves (2.1) when $f$ and $g$ are non-negative.

Apostol then shows the general case by translating $f$ and $g$ so that they are non-negative.

My question why is it necessary to have $F$ be the ordinate set of $f$, minus the graph of $f$? My intuition tells me it shouldn't matter because the graph $f(x)$ has area = 0, but if we're being rigorous, why do we need to consider $F$ this way?

My thought is that if we didn't and we let $F$ and $G$ be defined as

$$F = \{(x,y) \;|\; a \leq x \leq b, \; 0 \leq y \leq f(x)\}, \quad G = \{(x,y)\;|\;a \leq x \leq b, \; 0 \leq y \leq g(x)\}$$

Then $S = G - F$ would be defined as

$$S = \{(x,y) \; |\; a \leq x \leq b, f(x) < y \leq g(x)\}$$

But this is still measurable right? $G$ and $F$ are still measurable, so the difference is measurable. It just seems like a detail that is unnecessary but the fact that Apostol decided to do it this way makes me believe it is important for some reason.

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The set referred to in the theorem is the set $$ \{(x,y) \in \mathbb R^2| a \le x \le b \wedge f(x) \le y \le g(x)\}; $$ thus, the theorem statement concerns the measure of this set. Of course, the measure of the similar set $$ \{(x,y) \in \mathbb R^2| a \le x \le b \wedge f(x) < y \le g(x)\} $$ is exactly the same. In order to calculte its measure, one would modify the proof as indicated by you below the original proof.

Note: Since the measure of both sets is equal, the measure of their difference, the graph of $f$ on $[a,b]$, is a nullset of $\mathbb R^2$, ie. a set of measure zero.

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  • $\begingroup$ ok, glad I'm thinking about this correctly! Your answer also helps because previously I wasn't sure why the theorem suggests defining $F$ this way, but now I realize when you think about what $G - F$ should be, it is a lot more natural to define it as $$\{(x,y) \; | \; a \leq x \leq b, \; f(x) \leq y \leq g(x)\}$$ than $$\{(x,y) \; | \; a \leq x \leq b, \; f(x) < y \leq g(x)\}$$ when setting up the proof $\endgroup$ – Jake Kirsch Feb 8 at 12:14
  • $\begingroup$ Well, glad to be of help! Good luck with your further mathematical adventures! $\endgroup$ – AlgebraicsAnonymous Feb 8 at 12:21

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