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.


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.

  • $\begingroup$ 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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