2. 리만-스틸체스 적분 (i): 정의와 적분가능성 판정The Riemann-Stieltjes Integral (i): Definitions and the Integrability Criterion

Let $P = \{x_0, x_1, \ldots, x_n\}$ be an arbitrary partition of $[0,1]$. As in Example 2.8, since $x_0 = 0$ and $x_n = 1 > 0$, there exists $i_0 \in \{1, \ldots, n\}$ such that

Since $P$ is a partition of $[0,1]$, $x_{i_0-1} \ge x_0 = 0$, and hence $x_{i_0-1} = 0$. Thus $[x_{i_0-1}, x_{i_0}] = [0, x_{i_0}]$ and $\Delta x_{i_0} = x_{i_0} > 0$.

We show that $M_{i_0} = \sup_{0 \le x \le x_{i_0}} f(x) = \infty$. Choose $k \in \mathbb{N}$ with $k \ge 1/\sqrt{x_{i_0}}$. Then $0 \lt 1/k^2 \le x_{i_0}$, so $1/k^2 \in [0, x_{i_0}]$ and

Thus $M_{i_0} \ge k$ for every such $k$, and letting $k \to \infty$ gives $M_{i_0} = \infty$.

Since $f \ge 0$, we have $M_i \ge 0$ for all $i$, and hence

Since this holds for every partition $P$ of $[0,1]$,

i.e. $\overline{\int_0^1} f dx$ does not exist in $\mathbb{R}$. By Definition 2.6, Riemann-integrability requires both $\overline{\int_a^b} f dx$ and $\underline{\int_a^b} f dx$ to exist in $\mathbb{R}$, so $f$ is not Riemann-integrable on $[0,1]$.

Assume that $f$ is unbounded on $[a,b]$, i.e. there is no $M > 0$ such that

First observe that this means $f$ is not bounded above on $[a,b]$ or not bounded below on $[a,b]$. Indeed, if $f$ were both bounded above and bounded below, i.e. $f(x) \le M_1$ and $f(x) \ge M_2$ for all $x \in [a,b]$ for some $M_1, M_2 \in \mathbb{R}$, then setting $M := \max\{|M_1|,|M_2|\} + 1 > 0$ would give $|f(x)| \lt M$ for all $x \in [a,b]$, contradicting the assumption.

• (Case 1) $f$ is not bounded above on $[a,b]$. Let $P = \{x_0, \ldots, x_n\}$ be an arbitrary partition of $[a,b]$. We show that there exists $i_0 \in \{1, \ldots, n\}$ such that
  $$M_{i_0} := \sup_{x_{i_0-1} \le x \le x_{i_0}} f(x) = \infty.$$

  Suppose not, i.e. $M_i \lt \infty$ for all $i \in \{1, \ldots, n\}$. Then $M := \max\{M_1, \ldots, M_n\}$ is a finite real number. Since $[a,b] = \bigcup_{i=1}^{n} [x_{i-1}, x_i]$, for any $x \in [a,b]$ there exists $i$ with $x \in [x_{i-1}, x_i]$, so $f(x) \le M_i \le M$. This means $f$ is bounded above by $M$ on $[a,b]$, contradicting the assumption. Hence $M_{i_0} = \infty$ for some $i_0$.

  Moreover, if $x_{i_0-1} = x_{i_0}$, then $[x_{i_0-1}, x_{i_0}] = \{x_{i_0-1}\}$, so $M_{i_0} = f(x_{i_0-1}) \in \mathbb{R}$, contradicting $M_{i_0} = \infty$. Hence $x_{i_0-1} \lt x_{i_0}$, i.e. $\Delta x_{i_0} > 0$. Therefore

  $$U(P,f) = \sum_{i=1}^{n} M_i \Delta x_i \ge M_{i_0} \Delta x_{i_0} = \infty.$$

  Since this holds for every partition $P$ of $[a,b]$,

  $$\overline{\int_a^b} f dx := \inf U(P,f) = \infty,$$

  i.e. $\overline{\int_a^b} f dx$ does not exist in $\mathbb{R}$. By Definition 2.6, $f$ is not Riemann-integrable on $[a,b]$.

• (Case 2) $f$ is not bounded below on $[a,b]$. We use a symmetric argument to Case 1. Let $P = \{x_0, \ldots, x_n\}$ be an arbitrary partition of $[a,b]$. If $m_i > -\infty$ for all $i$, then $m := \min\{m_1, \ldots, m_n\}$ is a finite real number, and by the same reasoning as above $f$ would be bounded below by $m$ on $[a,b]$, contradicting the assumption. Hence there exists $i_0 \in \{1, \ldots, n\}$ such that
  $$m_{i_0} := \inf_{x_{i_0-1} \le x \le x_{i_0}} f(x) = -\infty.$$

  Also, if $x_{i_0-1} = x_{i_0}$, then $m_{i_0} = f(x_{i_0-1}) \in \mathbb{R}$, a contradiction, so $\Delta x_{i_0} > 0$. Therefore

  $$L(P,f) = \sum_{i=1}^{n} m_i \Delta x_i \le m_{i_0} \Delta x_{i_0} = -\infty.$$

  Since this holds for every partition $P$ of $[a,b]$,

  $$\underline{\int_a^b} f dx := \sup L(P,f) = -\infty,$$

  i.e. $\underline{\int_a^b} f dx$ does not exist in $\mathbb{R}$. By Definition 2.6, $f$ is not Riemann-integrable on $[a,b]$.

In either case, $f$ is not Riemann-integrable on $[a,b]$; thus if $f$ is unbounded on $[a,b]$, then $f$ is not Riemann-integrable on $[a,b]$.

Since $P$ is a partition of $[a,b]$, we can put

for some $n \in \mathbb{N}$.

• Case 1) $P^*$ contains only one point more than $P$. Let $x^*$ be the extra point. Then there exists $i_0 \in \{1, \ldots, n\}$ such that
  $$x_{i_0-1} \le x^* \le x_{i_0}.$$

  Thus

  $$P^* = \{x_0, x_1, \ldots, x_{i_0-1}, x^*, x_{i_0}, \ldots, x_n\}.$$

  Recall

  $$M_{i_0} = \sup_{x_{i_0-1} \le x \le x_{i_0}} f(x)$$

  and put

  $$W_1 = \sup_{x_{i_0-1} \le x \le x^*} f(x) \qquad \text{and} \qquad W_2 = \sup_{x^* \le x \le x_{i_0}} f(x).$$

  Since $[x_{i_0-1}, x^*] \subset [x_{i_0-1}, x_{i_0}]$ and $[x^*, x_{i_0}] \subset [x_{i_0-1}, x_{i_0}]$,

  $$W_1 \le M_{i_0} \qquad \text{and} \qquad W_2 \le M_{i_0}.$$

  Therefore, recalling the definition of $U(P,f,\alpha)$ and $\Delta\alpha_i \ge 0$, we have

  $$\begin{aligned} &U(P,f,\alpha) \\ &= \sum_{i=1}^{n} M_i \Delta\alpha_i \\ &= \left( \sum_{i=1}^{i_0-1} M_i \Delta\alpha_i \right) + M_{i_0} \Delta\alpha_{i_0} + \left( \sum_{i=i_0+1}^{n} M_i \Delta\alpha_i \right) \\ &= \left( \sum_{i=1}^{i_0-1} M_i \Delta\alpha_i \right) + M_{i_0} \left( \alpha(x_{i_0}) - \alpha(x^*) + \alpha(x^*) - \alpha(x_{i_0-1}) \right) + \left( \sum_{i=i_0+1}^{n} M_i \Delta\alpha_i \right) \\ &\ge \left( \sum_{i=1}^{i_0-1} M_i \Delta\alpha_i \right) + W_2 \left( \alpha(x_{i_0}) - \alpha(x^*) \right) + W_1 \left( \alpha(x^*) - \alpha(x_{i_0-1}) \right) + \left( \sum_{i=i_0+1}^{n} M_i \Delta\alpha_i \right) \\ &= U(P^*,f,\alpha). \end{aligned}$$
• Case 2) (General case). Since $P^*$ is a refinement of $P$, there exists $k \in \{0, 1, \cdots\}$ such that $P^*$ contains $k$-points more than $P$. If $k = 0$, then $P = P^*$, so the desired inequality holds. If $k = 1$, the desired inequality is proved in Case 1. So we may assume that $k \ge 2$. Put
  $$P^* - P = \{x_1^*, \ldots, x_k^*\}, \qquad P_0^* = P,$$

  and

  $$P_j^* := P \cup \{x_1^*, \ldots, x_j^*\} \quad (j = 1, \ldots, k).$$

  Then

  $$P = P_0^* \subset P_1^* \subset \cdots \subset P_k^* = P^*$$

  and $P_i^*$ contains only one more point than $P_{i-1}^*$ for all $i = 1, \ldots, k$. Therefore by Case 1,

  $$U(P_i^*,f,\alpha) \le U(P_{i-1}^*,f,\alpha) \qquad \forall i \in \{1, \ldots, k\}.$$

  In particular, we have

  $$U(P^*,f,\alpha) = U(P_k^*,f,\alpha) \le \cdots \le U(P_0^*,f,\alpha) = U(P,f,\alpha).$$

This is exactly $U(P^*,f,\alpha) \le U(P,f,\alpha)$, which proves Corollary 2.16.

(1) False (a counterexample). Let $D = (0,1]$ and $f(x) = 1/x$. First, $f$ is continuous on $D$: let $c \in D$ and $\varepsilon > 0$ be given. Set

If $x \in D$ satisfies $|x-c| \lt \delta$, then $x > c - \delta \ge c/2 > 0$, so $xc > c^2/2$, and hence

Since $c \in D$ was arbitrary, $f$ is continuous on $D$.

However, $f$ is not uniformly continuous on $D$. Let $\varepsilon = 1$. For any $\delta > 0$, set

Then $0 \lt x \le 1/2 \le 1$ and $0 \lt y \lt x \le 1$, so $x, y \in D$. Moreover

while

(since $x \le 1/2$ gives $1/x \ge 2$). Thus no $\delta > 0$ can satisfy the condition for uniform continuity at $\varepsilon = 1$, so $f$ is not uniformly continuous on $D$. Therefore (1) is false.

(2) True. Assume that $f$ is uniformly continuous on $D$. Let $c \in D$; it suffices to show that $f$ is continuous at $c$. Let $\varepsilon > 0$ be given. Since $f$ is uniformly continuous on $D$, there exists $\delta > 0$ such that

for all $x, y \in D$ satisfying $|x-y| \lt \delta$. In particular, taking $y = c$,

for all $x \in D$ satisfying $|x-c| \lt \delta$. This means $f$ is continuous at $c$. Since $c \in D$ was arbitrary, $f$ is continuous on $D$. Therefore (2) is true.

Let $\varepsilon > 0$ be given. Since $f$ is continuous on $K$, for each $c \in K$ there exists $\delta_c > 0$ such that

for all $x \in (c - \delta_c, c + \delta_c) \cap K$. Since each $(c - \delta_c/2, c + \delta_c/2)$ is open and

(as each $c \in K$ belongs to $(c-\delta_c/2,c+\delta_c/2)$), the collection $\{(c - \delta_c/2, c + \delta_c/2) : c \in K\}$ is an open cover of $K$. Since $K$ is compact, there exists a finite subcover

such that

Set

Let $x, y \in K$ satisfy $|x - y| \lt \delta$. It is sufficient to show that

By the subcover condition above, there exists $i_0 \in \{1, \ldots, n\}$ such that

Then since

we have

i.e.

Since $x, y \in K$, we have $x, y \in (c_{i_0} - \delta_{c_{i_0}}, c_{i_0} + \delta_{c_{i_0}}) \cap K$, and hence

Since $x, y \in K$ with $|x-y| \lt \delta$ were arbitrary, $f$ is uniformly continuous on $K$.