2. 리만-스틸체스 적분 (iii): 계단함수·변수변환·미적분학의 기본정리The Riemann-Stieltjes Integral (iii): Step Functions, Change of Variable, and the FTC

Let $A := \sum_{n=1}^{\infty} a_n \in \mathbb{R}$. For each $k \in \mathbb{N}$, put

Since $|b_n| \ge 0$ for all $n \in \mathbb{N}$,

i.e. the sequence $\{S_k\}$ is monotonically increasing. Moreover, since $|b_n| \le a_n$ and $a_n \ge 0$, $T_k$ is also monotonically increasing and, since $\sum_{n=1}^{\infty} a_n$ converges, $T_k \le A$. Thus for each $k \in \mathbb{N}$,

i.e. $\{S_k\}$ is bounded above, and the set $\{S_k : k \in \mathbb{N}\}$ is nonempty and has $A$ as an upper bound. Thus by Remark 1.49,

exists in $\mathbb{R}$, and for any $\varepsilon > 0$, there exists a $K \in \mathbb{N}$ such that

Let $\varepsilon > 0$ be given. Take $K \in \mathbb{N}$ as above. Since $\{S_k\}$ is monotonically increasing, for all $k \ge K$,

Also since $S$ is an upper bound of $\{S_k : k \in \mathbb{N}\}$, $S_k \le S \lt S + \varepsilon$. Thus for all $k \ge K$,

Since $\varepsilon > 0$ is an arbitrary positive constant, $S_k \to S$ as $k \to \infty$, i.e. the sequence $\sum_{n=1}^{k} |b_n|$ converges to the real number $S$. Therefore the series $\sum_{n=1}^{\infty} b_n$ absolutely converges.

Since Lemma 2.48 only gives the infimum-over-refinements characterization for the upper integral, we first establish the corresponding fact for the lower integral. Let $P$ be a partition of $[a,b]$. We claim

(this mirrors the proof of Lemma 2.48). Obviously

since every refinement $P^*$ of $P$ is itself a partition of $[a,b]$, and $L(P^*,f,\alpha)$ is one of the values whose supremum defines the lower integral. On the other hand, let $P_1$ be an arbitrary partition of $[a,b]$. Then $P_1 \cup P$ is a refinement of $P$ and also a refinement of $P_1$, so by Theorem 2.15,

Since $P_1$ is an arbitrary partition of $[a,b]$,

Combining the two inequalities gives the claimed equality.

Now we prove (ii). Let $\varepsilon > 0$ be given. Since $\alpha'$ is Riemann integrable on $[a,b]$, there exists a partition $P = \{x_0, \ldots, x_n\}$ of $[a,b]$ such that

By the mean-value theorem, for each $i \in \{1, \ldots, n\}$, there exists $t_i \in [x_{i-1}, x_i]$ such that

As in the proof of (i), setting

we have, for all $s_i \in [x_{i-1}, x_i]$,

(this is exactly (2.52)). In particular, for all $s_i \in [x_{i-1}, x_i]$,

Taking the infimum over each $s_i \in [x_{i-1}, x_i]$ on the left,

(since $\Delta\alpha_i \ge 0$ and the $s_i$ vary independently, the infimum distributes across the terms). This value is at least the constant $L(P,f\alpha') - M\varepsilon$ on the right, so

Similarly,

Note that both inequalities also hold for any refinement $P^*$ of $P$, since

(by Theorem 2.15 and Corollary 2.16, refinement increases the lower sum and decreases the upper sum, so their difference only shrinks under refinement). That is, for any refinement $P^*$ of $P$,

Taking the supremum over refinements $P^*$ of $P$ and applying the lower-integral characterization shown above, we obtain

Since $\varepsilon > 0$ is an arbitrary positive constant, we conclude

By (2.53), for the partition $P = \{x_0,\ldots,x_n\}$ of $[a,b]$, for every choice $s_1 \in [x_0,x_1], \ldots, s_n \in [x_{n-1},x_n]$,

holds. Here the right side is a constant independent of the choice of $s_1,\ldots,s_n$, and the choice $s_i$ on each interval $[x_{i-1},x_i]$ is independent of the others. We use this to show that the supremum on the left distributes term by term.

Let $\varepsilon' > 0$ be given. For each $i = 1,\ldots,n$, put $M_i := \sup_{x \in [x_{i-1},x_i]} f(x)$. If $\Delta\alpha_i > 0$, then by the $\varepsilon$-characterization of Remark 1.49, there exists $s_i^* \in [x_{i-1},x_i]$ such that

and hence

If $\Delta\alpha_i = 0$, then for any choice of $s_i^* \in [x_{i-1},x_i]$,

trivially holds. In either case $f(s_i^*)\Delta\alpha_i \gt M_i\Delta\alpha_i - \varepsilon'/n$, so summing over $i = 1,\ldots,n$,

On the other hand, since $s_1^*,\ldots,s_n^*$ is a choice on each interval, (2.53) gives

Combining the two inequalities,

that is,

for every $\varepsilon' > 0$. Since $\varepsilon' > 0$ is arbitrary,

which is (2.54).

By (2.52), for the partition $P = \{x_0,\ldots,x_n\}$ of $[a,b]$ and every choice $s_1 \in [x_0,x_1],\ldots,s_n \in [x_{n-1},x_n]$,

holds. In particular,

Also, by definition $f(s_i) \le M_i = \sup_{x\in[x_{i-1},x_i]} f(x)$ for each $i$, and $\Delta\alpha_i \ge 0$, so

Thus for all $s_1,\ldots,s_n$,

holds. The right side is a constant independent of the choice of $s_1,\ldots,s_n$, and as in Exercise 2.51, each $s_i$ can be chosen independently on its own interval. For each $i = 1,\ldots,n$, put

Applying the same argument as in Exercise 2.51 (fix $\varepsilon' > 0$; for each $i$ with $\Delta x_i > 0$, use the $\varepsilon$-characterization of Remark 1.49 to choose $s_i^*$ with $f(s_i^*)\alpha'(s_i^*)\Delta x_i > M_i^{f\alpha'}\Delta x_i - \varepsilon'/n$, and any $s_i^*$ when $\Delta x_i = 0$; sum over $i = 1,\ldots,n$ and let $\varepsilon' \to 0$), we obtain

which is (2.55).

The hypotheses of Theorem 2.53 coincide with those of Lemma 2.49: $f$ is a bounded function on $[a,b]$, and $\alpha$ is a monotonically increasing differentiable function on $[a,b]$ whose derivative $\alpha'$ is Riemann integrable on $[a,b]$. Thus by Lemma 2.49 (i) and (ii),

Also, $f$ is a bounded function on $[a,b]$, and since $\alpha'$ is Riemann integrable on $[a,b]$, Exercise 2.10 makes $\alpha'$ bounded on $[a,b]$, so $f\alpha'$ is a bounded function on $[a,b]$ as well. Since the integrator $\alpha$ is monotonically increasing and so is the identity $x \mapsto x$, by Corollary 2.18 all four of the above values exist in $\mathbb{R}$. Thus by Definition 2.12, $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,b]$ if and only if

By the equalities shown above, this is equivalent to

which by Definition 2.6 is equivalent to $f\alpha'$ being Riemann integrable on $[a,b]$. Therefore $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,b]$ if and only if $f\alpha'$ is Riemann integrable on $[a,b]$.

By hypothesis, $f$ is a bounded function on $[a,b]$ and $\alpha$ is a monotonically increasing differentiable function on $[a,b]$ whose derivative $\alpha'$ is Riemann integrable on $[a,b]$, i.e. the hypotheses of Theorem 2.53 hold. Since we assumed that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,b]$ or $f\alpha'$ is Riemann integrable on $[a,b]$, by Theorem 2.53 both properties hold: $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,b]$ and $f\alpha'$ is Riemann integrable on $[a,b]$.

By Definition 2.12,

by Lemma 2.49(i),

and since $f\alpha'$ is Riemann integrable on $[a,b]$, by Definition 2.6,

Combining the three equalities, we obtain

Note. Definition 2.11 explicitly sets up $U(P,f,\alpha)$ and $L(P,f,\alpha)$ under the hypothesis that $f$ is a bounded function on $[a,b]$. In the counterexample below, $f$ is unbounded, so there can be an interval on which $M_i = \sup_{x\in[x_{i-1},x_i]} f(x)$ equals $+\infty$; on such an interval, if $\Delta\alpha_i = 0$, the product $M_i\Delta\alpha_i$ is, on its face, of the undefined form "$\infty \times 0$". Below we adopt the natural convention that, on any interval with $\Delta\alpha_i = 0$, we read $M_i\Delta\alpha_i := 0$ (likewise $m_i\Delta\alpha_i := 0$), regardless of whether $M_i$ is finite or $+\infty$ and whether $m_i$ is finite or $-\infty$, and compute $U(P,f,\alpha)$, $L(P,f,\alpha)$ accordingly. Definition 2.12 imposes no boundedness of its own and states integrability for "a function $f$ on $[a,b]$", so under this convention we can directly check the integrability condition of Definition 2.12 (the upper and lower integrals exist in $\mathbb{R}$ and are equal). This precisely targets the point at which the argument of Exercise 2.10 — that if $\alpha$ is strictly increasing, e.g. $\alpha(x) = x$, so that every $\Delta\alpha_i > 0$, then an unbounded $f$ forces $M_i\Delta\alpha_i$ to grow arbitrarily large on some interval, so the upper sum diverges — no longer applies once $\alpha$ is (locally) constant, so $\Delta\alpha_i = 0$. (If instead one reads the boundedness hypothesis of Definition 2.11 as a blanket precondition for the entire problem, then the phrase "Riemann-Stieltjes integrable $f$" already refers only to bounded $f$, and the claim is trivially true. The argument below instead follows the reading that extends the upper/lower sum formulas of Definition 2.11 to unbounded $f$ in the natural way described above.)

The claim is disproved. Let $[a,b] = [0,1]$ and

For all $x, y \in [0,1]$ with $x \lt y$, $\alpha(x) = 0 \le 0 = \alpha(y)$, so by Definition 1.43(i), $\alpha$ is monotonically increasing on $[0,1]$.

Since $f(x_k) = k \to \infty$ for $x_k := 1/k \to 0^+$ as $k \to \infty$, $f$ is unbounded on $[0,1]$.

Let $P = \{x_0, x_1, \ldots, x_n\}$ be an arbitrary partition of $[0,1]$. Since $\alpha$ is a constant function, for all $i \in \{1,\ldots,n\}$,

By the convention in the note above, $M_i\Delta\alpha_i = 0$ and $m_i\Delta\alpha_i = 0$ for all $i$ (for instance, in a partition with $x_1 > 0$ the first interval $[x_0,x_1] = [0,x_1]$ has $M_1 = \sup_{x\in[0,x_1]} f(x) = +\infty$, but this too is read as $0$ once multiplied by $\Delta\alpha_1 = 0$), so

Since $P$ was an arbitrary partition of $[0,1]$,

and both values exist in $\mathbb{R}$ and are equal. Thus by Definition 2.12, $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[0,1]$ and

But $f$ is unbounded on $[0,1]$. Therefore the claim "if $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,b]$, then $f$ is bounded" is false (under the convention in the note above), and it is disproved.

Let $x \in (x_0 - \delta, x_0) \cap [a,b]$. By Theorem 2.37(iii),

As already observed in the proof of Theorem 2.61,

Thus applying the linear property of the Riemann integral (Theorem 2.37(i), (ii)) again,

If $x \in (x_0 - \delta, x_0) \cap [a,b]$ and $t \in [x, x_0]$, then $t \in (x_0 - \delta, x_0 + \delta) \cap [a,b]$, so by (2.57),

Thus by Theorem 2.37(iv) (taking the interval of integration to be $[x,x_0]$, the integrator $\alpha(t) = t$, and $M = \varepsilon/2$),

Therefore

Since this holds for all $x \in (x_0 - \delta, x_0) \cap [a,b]$, (2.58) is proved.

The claim is true. Let $f$ be continuous on $[a,b]$. Define

(with $F(a) = 0$). For any $x \in [a,b]$, $f$ is also continuous on $[a,x]$, and the identity function $\alpha(t) = t$ is monotonically increasing on $[a,x]$, so by Theorem 2.26, $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,x]$. By Remark 2.13, this is equivalent to $f$ being Riemann integrable on $[a,x]$, so $F(x)$ is well-defined for every $x \in [a,b]$.

Since $f$ is continuous on $[a,b]$, it is in particular continuous at every $x_0 \in [a,b]$. Since $f$ is a Riemann integrable function on $[a,b]$ (as verified above with $x = b$), the hypotheses of Theorem 2.61 hold at every $x_0 \in [a,b]$. Thus by Theorem 2.61, for every $x_0 \in [a,b]$, $F$ is differentiable at $x_0$ and

(By Definition 1.17, when $x_0 = a$ or $x_0 = b$, $F'(x_0)$ means the right- or left-hand derivative, respectively; but the proof of Theorem 2.61 works throughout with intervals of the form $(x_0 - \delta, x_0 + \delta) \cap [a,b]$, $(x_0, x_0+\delta) \cap [a,b]$, and $(x_0-\delta, x_0) \cap [a,b]$, so it applies unchanged when $x_0$ is an endpoint and yields exactly the corresponding one-sided derivative.) Therefore $F$ is differentiable on $[a,b]$ and satisfies $F'(x) = f(x)$ for all $x \in [a,b]$, so such an $F$ exists, as claimed.