Is the proof of Theorem 1.10 in Apostol's Calculus, Volume 1 incomplete?

Question

Theorem 1.10 in Apostol's Calculus, Volume 1 states that if $f$ is nonnegative and integrable on $[a, b]$, and $Q$ is the ordinate set of $f$ over $[a, b]$, then $Q$ is measurable and its area is equal to $\int_a^b f(x)\,dx$.

The ordinate set of a nonnegative function $f$ over $[a, b]$ is defined as $\{(x, y) : a \le x \le b,\;0 \le y \le f(x)\}$.

The proof uses what Apostol calls the Exhaustion Property, which is the following axiom:

Let $Q$ be a set that can be enclosed between two step regions $S$ and $T$, so that

(1.1) $S \subseteq Q \subseteq T$.

If there is one and only one number $c$ which satisfies the inequalities $a(S) \le c \le a(T)$ for all step regions $S$ and $T$ satisfying (1.1), then $Q$ is measurable and $a(Q) = c$.

$a$ is an area function; a step region is the union of finitely many side-by-side rectangles whose bases are on the $x$-axis; a rectangle is a set congruent to a set of the form $\{(x, y) : 0 \le x \le h,\;0 \le x \le k\}$ for some nonnegative real numbers $h$ and $k$.

Here is Apostol's proof:

Let $S$ and $T$ be two step regions satisfying $S \subseteq Q \subseteq T$. Then there are two step functions $s$ and $t$ satisfying $s \le f \le t$ on $[a, b]$, such that $a(S) = \int_a^b s(x)\,dx$ and $a(T) = \int_a^b t(x)\,dx$. Since $f$ is integrable on $[a, b]$, the number $I = \int_a^b f(x)\,dx$ is the only number satisfying the inequalities $\int_a^b s(x)\,dx \le I \le \int_a^b t(x)\,dx$ for all step functions $s$ and $t$ with $s \le f \le t$. Therefore this is also the only number satisfying $a(S) \le I \le a(T)$ for all step regions $S$ and $T$ with $S \subseteq Q \subseteq T$. By the exhaustion property, this proves that $Q$ is measurable and that $a(Q) = I$.

If I understand the Exhaustion Property correctly, the following need to be proved:

1. There exist step regions $S$ and $T$ such that $S \subseteq Q \subseteq T$.
2. For all step regions $S$ and $T$ satisfying $S \subseteq Q \subseteq T$, $a(S) \le \int_a^b f(x)\,dx \le a(T)$.
3. If $c$ is a number such that $a(S) \le c \le a(T)$ for all step regions $S$ and $T$ such that $S \subseteq Q \subseteq T$, then $c = \int_a^b f(x)\,dx$.

It doesn't look like Apostol proves (1). Is it actually unnecessary to do that?

It's easy to do - we can let $S = \{(x, y) : a \le x \le b,\;y = 0\}$. Since $f$ is integrable, it's bounded, so there's a $u$ such that $f(x) \le u\;\forall x \in [a, b]$. We can let $T = \{(x, y) : a \le x \le b,\;0 \le y \le u\}$.

A proof of (2) is sketched out in the second and third sentences. Apostol doesn't seem to prove (3), as he doesn't show that $\int_a^b s(x)\,dx \le c \le \int_a^b t(x)\,dx$ for all step functions $s$ and $t$ on $[a, b]$ such that $s \le f \le t$, which is the only way I can see to prove (3). Am I right, or am I misreading the proof?

If my thought about proving (3) is correct, then it doesn't seem to be possible to prove (3) within Apostol's framework, because it doesn't seem to be true that given a step function $s$ such that $s \le f$, there always exists a step region $S$ such that $S \subseteq Q$ and $\int_a^b s(x)\,dx \le a(S)$. We would need this in order to show that $\int_a^b s(x)\,dx \le c$.

Here's an example:

Define $f: [0, 1] \to \mathbb{R}$ by $f(x) = 0$ if $x \in \{0, 1\}$ and $f(x) = 1$ otherwise. Clearly, $f$ is nonnegative. Since $f$ is a step function, it is integrable. Let $s = f$. Then $s$ is a step function such that $s \le f$; $\int_0^1 s(x)\,dx = 1$. Let $Q$ be the ordinate set of $f$ over $[0, 1]$. Suppose for the sake of contradiction that there exists a step region $S$ such that $S \subseteq Q$ and $\int_0^1 s(x)\,dx \le a(S)$. If the left side of $S$ were to the right of the line $x = 0$, then $a(S) < 1 = \int_0^1 s(x)\,dx$, which isn't the case. So, the left side of $S$ must be along the line $x = 0$. Similarly, the right side of $S$ must be along the line $x = 1$. Because $S$ is a subset of $Q$, the height of each rectangle in $S$ is no more than 1. If the height of some rectangle were less than 1, then $a(S) < 1$, which is impossible, as we just saw. So, the height of each rectangle must be 1. This implies that $(0, 1)$ is the point at the upper left-hand corner of the leftmost rectangle in $S$. This point isn't in $Q$, so we have a contradiction of the fact that $S \subseteq Q$. Thus, there is no step region $S$ such that $S \subseteq Q$ and $\int_0^1 s(x)\,dx \le a(S)$.

Does this show that it is impossible to prove (3) within Apostol's framework?

Answer 1

I did have a look at Apostol's Calculus Vol 1 and found that there is no issue with the proof given by Apostol (this is expected because Apostol's books are best of the lot).

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The doubts raised by OP about Apostol's proof can be explained without much effort. I begin with a simple theorem which will help us here.

Theorem: If $s$ is a non-negative step function defined on $[a, b]$ then there is a step region $S$ with $a(S) = \int_{a}^{b}s(x)\,dx$ and conversely if $S$ is a step region between ordinates $x = a, x = b$ then there is a step function $s$ on $[a, b]$ such that $a(S) = \int_{a}^{b}s(x)\,dx$.

I prove the first part and the converse is left as an easy exercise. Let $$P = \{x_{0}, x_{1}, x_{2},\dots, x_{n}\}$$ be a partition of $[a, b]$ such that $s(x) = s_{k}$ for all $x \in (x_{k - 1}, x_{k})$. We now construct another step function $\sigma(x)$ such that $$\sigma(x_{0}) = s_{1}, \sigma(x_{n}) = s_{n}, \sigma(x_{k}) = \max(s_{k}, s_{k + 1}), k = 1, 2, \dots, n - 1$$ and $\sigma(x) = s_{k}$ if $x \in (x_{k - 1}, x_{k})$ for $k = 1, 2, \dots, n$. It is now clear that if $S$ is the ordinate set of $\sigma$ then $S$ is a step region and clearly $$a(S) = \int_{a}^{b}\sigma(x)\,dx = \int_{a}^{b}s(x)\,dx$$

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Let $f$ be non-negative and integrable on $[a, b]$. By this we mean that there is a unique number $I$ such that for any step functions $s, t$ with $s\leq f\leq t$ we have $$\int_{a}^{b}s(x)\,dx \leq I\leq \int_{a}^{b}t(x)\,dx, I = \int_{a}^{b}f(x)\,dx$$ Our objective here is to prove that the ordinate set $Q$ of $f$ is measurable and $a(Q) = I =\int_{a}^{b}f(x)\,dx$.

The proof begins with step regions $S, T$ such that $S\subseteq Q\subseteq T$ and you doubt whether such regions exist. Yes they do. If $s, t$ are step functions with $s\leq f\leq t$ then as mentioned in theorem in the beginning of this answer we can find step regions $S, T$ whose areas match the integrals of $s, t$ respectively. Note that such regions $S, T$ will satisfy $S\subseteq Q\subseteq T$ because $s\leq f\leq t$. Also note that by construction $a(S) \leq I \leq a(T)$.

Next we show that if $S, T$ are step regions with $S\subseteq Q\subseteq T$ then $a(S)\leq I \leq a(T)$. By the theorem mentioned in the beginning we can find step functions $s, t$ such that integrals of $s, t$ match areas of $S, T$ and clearly because of $S\subseteq Q\subseteq T$ we will have $s(x)\leq f(x)\leq t(x)$ for all $x \in [a, b]$ and hence by integrability of $f$ we have $$a(S) = \int_{a}^{b}s(x)\,dx \leq I \leq \int_{a}^{b}t(x)\,dx = a(T)$$ Thus we have shown that for any step regions $S, T$ with $S\subseteq Q\subseteq T$ we have $a(S)\leq I\leq a(T)$. The third point raised by OP deals with showing that $I$ is the only number which satisfies this property.

Assume that there is some number $c$ which satisfies $a(S) \leq c \leq a(T)$ for all step regions $S, T$ with $S\subseteq Q\subseteq T$. Then we have step functions $s, t$ whose integrals equal areas of $S, T$ and $s\leq f\leq t$ and then we have $\int_{a}^{b}s(x)\,dx \leq c \leq \int_{a}^{b}t(x)\,dx$. But since $f$ is integrable we know that $I$ is the unique number which lies between such two integrals. Hence we must have $c = I$. It now follows that $Q$ is measurable and $a(Q) = I$.

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As I had mentioned in my comments, there is a correspondence between step regions and step functions and that is the key to the link between area under the graph of a function and the integral of that function. Like the way the integral of the given function is sandwiched between integrals of suitable step functions, the area under graph of the function is also sandwiched between areas of corresponding step regions.