2. 리만-스틸체스 적분 (ii): 적분가능 함수와 연산The Riemann-Stieltjes Integral (ii): Integrable Functions and Algebraic Properties

Let $\varepsilon > 0$ be given. Since $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$, by Theorem 2.19 there exists a partition $P = \{x_0, x_1, \ldots, x_n\}$ of $[a, b]$ such that

For each $i = 1, \ldots, n$, put

We first show

For all $x \in [x_{i-1}, x_i]$, $f(x) \ge m_i$, so $-f(x) \le -m_i$, and thus $-m_i$ is an upper bound of $\{-f(x) : x_{i-1} \le x \le x_i\}$. Let $\varepsilon' > 0$ be given. Since $m_i = \inf_{[x_{i-1}, x_i]} f$, by Remark 1.49 there exists a $y \in [x_{i-1}, x_i]$ such that

which means $-f(y) > -m_i - \varepsilon'$. Since $\varepsilon' > 0$ is arbitrary, by the definition of the supremum (Remark 1.49),

Similarly, since $M_i = \sup_{[x_{i-1}, x_i]} f$, for any $\varepsilon' > 0$ there exists a $y' \in [x_{i-1}, x_i]$ such that $f(y') > M_i - \varepsilon'$, which means $-f(y') < -M_i + \varepsilon'$. On the other hand, $f(x) \le M_i$ for all $x \in [x_{i-1}, x_i]$, so $-f(x) \ge -M_i$ and thus $-M_i$ is a lower bound of $\{-f(x) : x_{i-1} \le x \le x_i\}$. Combining the two facts shows

This proves (*). Therefore by (*),

and

Thus

By Theorem 2.19, $-f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$. (As a remark, repeating the same argument over every partition $P$ of $[a, b]$ gives

and similarly $\underline{\int_a^b} (-f) d\alpha = -\overline{\int_a^b} f d\alpha = -\int_a^b f d\alpha$, so we also obtain

)

Put $g := -f$. Since $f$ is monotonically decreasing on $[a, b]$, by Definition 1.43(iii), $f(x) \ge f(y)$ whenever $x, y \in [a, b]$ and $x < y$. Thus $g(x) = -f(x) \le -f(y) = g(y)$, and by Definition 1.43(i), $g = -f$ is monotonically increasing on $[a, b]$. Since $\alpha$ is monotonically increasing and continuous on $[a, b]$, applying Theorem 2.27 to $g$ shows that $g$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$. Now applying Exercise 2.28 to $g$ (which is integrable with respect to $\alpha$), $-g$ is also Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$. But

so $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$.

Put

where $k$ is the number of the points $c_1, \ldots, c_k$ listed in Theorem 2.32 (they include every point of discontinuity of $f$) and $M = \sup_{x \in [a,b]} |f(x)|$. Since $f$ is bounded and $k$ is finite, $C$ is a finite constant, and crucially $C$ is a fixed value depending only on $f$, $\alpha$, and $k$ — not on the choice of $\varepsilon$ nor on the partition $P$ constructed to obtain (2.31). What (2.31) says is that no matter which value of the free parameter $\varepsilon > 0$ is chosen in the proof above, the partition $P$ built from that $\varepsilon$ satisfies

Now, to verify the criterion of Theorem 2.19, let an arbitrary $\varepsilon' > 0$ be given (this is the target tolerance appearing in (2.18) of Theorem 2.19, a quantity distinct from the parameter $\varepsilon$ used in the proof above). Put

(since $C \ge 0$, $C + 1 > 0$, so this $\varepsilon$ is a well-defined positive number even when $C = 0$). Substituting this $\varepsilon$ into the construction in the proof of Theorem 2.32, by (2.31) we obtain a partition $P$ such that

where the last inequality holds since $\frac{C}{C+1} < 1$. That is, for any given $\varepsilon' > 0$, choosing $\varepsilon$ as above (depending on $\varepsilon'$) and running the construction of Theorem 2.32 yields a partition $P$ satisfying

as required by (2.18) of Theorem 2.19. Since $\varepsilon' > 0$ was arbitrary, the criterion of Theorem 2.19 is satisfied, and therefore $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$. In short, the fact that $C$ is independent of $\varepsilon$ lets us invert the single inequality $U - L \le \varepsilon C$ to recover an arbitrarily small $U - L$ for any target $\varepsilon'$, by taking $\varepsilon = \varepsilon'/(C+1)$; this is exactly why (2.31) completes the proof of Theorem 2.32.

Let $\varepsilon > 0$ be given. Put $M := \sup_{x \in [a,b]} |f(x)|$. If $M = 0$, then $f \equiv 0$, so $f$ is trivially Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$. So we may assume $M > 0$. Let $p$ be the (by hypothesis, unique) limit point that $A$ has in $[a, b]$.

We choose $\eta > 0$ as follows.

• Case a) $p \notin A$. Then $p \in [a, b] \setminus A$, and since $f$ is continuous on $[a, b] \setminus A$, $f$ is in particular continuous at $p$. Thus there exists $\delta > 0$ such that
  $$|f(x) - f(p)| < \frac{\varepsilon}{2} \qquad \forall x \in (p - \delta, p + \delta) \cap [a, b].$$

  Put $\eta := \delta/2$ and $I := [p - \eta, p + \eta] \cap [a, b]$. Then $I \subset (p - \delta, p + \delta) \cap [a, b]$, so for all $x, y \in I$,

  $$f(x) - f(y) \le |f(x) - f(p)| + |f(p) - f(y)| < \varepsilon,$$

  and thus

  $$M_I - m_I \le \varepsilon, \qquad \text{where} \quad M_I := \sup_{x \in I} f(x), \ m_I := \inf_{x \in I} f(x). \tag{a}$$
• Case b) $p \in A$. Then since $\alpha$ is continuous on $A$, $\alpha$ is in particular continuous at $p$. Thus there exists $\delta > 0$ such that
  $$|\alpha(x) - \alpha(p)| < \frac{\varepsilon}{2} \qquad \forall x \in (p - \delta, p + \delta) \cap [a, b].$$

  Put $\eta := \delta/2$ and $I := [p - \eta, p + \eta] \cap [a, b]$. Then $I \subset (p - \delta, p + \delta) \cap [a, b]$, so

  $$\alpha(\sup I) - \alpha(\inf I) \le |\alpha(\sup I) - \alpha(p)| + |\alpha(p) - \alpha(\inf I)| < \varepsilon. \tag{b}$$

  Moreover, since $|f| \le M$ on $[a,b]$, obviously

  $$M_I - m_I \le 2M, \qquad \text{where} \quad M_I := \sup_{x \in I} f(x), \ m_I := \inf_{x \in I} f(x).$$

In either case, $I := [p - \eta, p + \eta] \cap [a, b]$ for the $\eta > 0$ chosen above. Next we show that $A \setminus (p - \eta, p + \eta)$ is a finite set. The set

is closed, being $[a, b]$ with the open interval $(p - \eta, p + \eta)$ removed, and bounded, being a subset of $[a, b]$. Thus by the Heine-Borel theorem (Remark 2.23), $K'$ is compact. Now take an arbitrary $q \in K'$. Since $q \in K'$ we have $q \notin (p - \eta, p + \eta)$, while $p \in (p - \eta, p + \eta)$, so $q \neq p$; and since $q \in K' \subset [a, b]$ while $p$ is the only limit point that $A$ has in $[a, b]$, $q$ is not a limit point of $A$. Hence, by the definition of a limit point, there is an open interval $V_q$ containing $q$ such that

Then $\{V_q : q \in K'\}$ is an open cover of $K'$, so by the compactness of $K'$ (Remark 2.23) it has a finite subcover: there exist $q_1, \ldots, q_m \in K'$ with $K' \subset \bigcup_{r=1}^{m} V_{q_r}$. Since $A \subset [a, b]$, we have $A \setminus (p - \eta, p + \eta) = A \cap K'$, and therefore

That is, $A \setminus (p - \eta, p + \eta)$ has at most $m$ elements and hence is finite. (Only the definition of compactness in Remark 2.23 — every open cover admits a finite subcover — is used here; no theorem outside the lecture notes is invoked.)

Define

(if $p - \eta \le a$, we do not consider $J_1$ — there is no left piece of $[a,b] \setminus \mathring{I}$ in this case — and similarly we do not consider $J_2$ if $p + \eta \ge b$). When $J_1$ exists, $A \cap J_1 \subset A \setminus (p - \eta, p + \eta)$ is finite by the finiteness shown above. Since $f$ is continuous on $[a, b] \setminus A$, $f$ is continuous at every point of $J_1 \setminus A$, and since $\alpha$ is continuous on $A$, $\alpha$ is continuous at every point of $A \cap J_1$. Moreover $f$ is bounded on $J_1$ and $\alpha$ is monotonically increasing on $J_1$, being restrictions of the bounded $f$ and the monotonically increasing $\alpha$ on $[a, b]$. Hence, taking $A \cap J_1$ as the exceptional set, all the hypotheses of Theorem 2.32 are met on $J_1$. If $A \cap J_1 = \emptyset$, then $f$ is continuous on all of $J_1$, so by Theorem 2.26, $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $J_1$; if $A \cap J_1 \neq \emptyset$, applying Theorem 2.32 with $J_1$ in place of $[a,b]$ and $A \cap J_1$ in place of $\{c_1, \ldots, c_k\}$ gives the same conclusion. In either case $f$ is integrable with respect to $\alpha$ on $J_1$, so by Theorem 2.19 there exists a partition $P_1$ of $J_1$ such that

Similarly, when $J_2$ exists, $A \cap J_2$ is finite (by the same argument), so by Theorem 2.26 or Theorem 2.32, $f$ is integrable with respect to $\alpha$ on $J_2$, and by Theorem 2.19 there exists a partition $P_2$ of $J_2$ such that

Now form a partition $P$ of $[a, b]$ by combining $P_1$ (if it exists), the two endpoints $\max\{a, p-\eta\}$ and $\min\{b, p+\eta\}$ of $I$, and $P_2$ (if it exists). Since $J_1$, $I$, $J_2$ cover $[a, b]$ consecutively without overlap, $P$ is indeed a partition of $[a, b]$, and $U(P, f, \alpha)$ and $L(P, f, \alpha)$ decompose as the sum of the contributions over $J_1$, $I$, and $J_2$ respectively:

where $\Delta\alpha_I := \alpha(\min\{b, p+\eta\}) - \alpha(\max\{a, p-\eta\})$, and the corresponding term is taken to be $0$ if $J_1$ or $J_2$ does not exist. Thus

In Case a), $\Delta\alpha_I \le \alpha(b) - \alpha(a)$ and $M_I - m_I \le \varepsilon$ by (a), so $(M_I - m_I)\Delta\alpha_I \le \varepsilon(\alpha(b) - \alpha(a))$. In Case b), $M_I - m_I \le 2M$ and $\Delta\alpha_I < \varepsilon$ by (b), so $(M_I - m_I)\Delta\alpha_I < 2M\varepsilon$. In either case,

and thus

where $C := 2 + \alpha(b) - \alpha(a) + 2M \ge 0$ is a fixed constant independent of $\varepsilon$ and of the choice of partition $P$. This is exactly the situation of Exercise 2.33: given any $\varepsilon' > 0$, starting with $\varepsilon := \varepsilon'/(C+1)$ and repeating the entire construction above (i.e. the choice of $\eta$, $I$, $P_1$, $P_2$, and $P$) yields a partition $P$ of $[a, b]$ such that

Since $\varepsilon' > 0$ is arbitrary, by Theorem 2.19, $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$.

Assume $A$ is nonempty and bounded both above and below, so that all the suprema and infima below exist in $\mathbb{R}$ (Remark 1.49). We first prove (i). Put $s := \sup A$. For all $a \in A$, $a \le s$, so $a + b \le s + b$, and thus $s + b$ is an upper bound of $A + b$. Let $\varepsilon > 0$ be given. Since $s = \sup A$, by Remark 1.49 there exists a $y \in A$ such that $s - \varepsilon < y$, and thus $y + b \in A + b$ and $(s + b) - \varepsilon < y + b$. Since $\varepsilon > 0$ is arbitrary, by the $\varepsilon$-characterization of the supremum (Remark 1.49),

Second, we prove (ii). Put $s := \sup A$. Since $c > 0$, $a \le s$ implies $ca \le cs$ for all $a \in A$, so $cs$ is an upper bound of $cA$. Let $\varepsilon > 0$ be given. Since $\varepsilon/c > 0$, by Remark 1.49 there exists a $y \in A$ such that $s - \varepsilon/c < y$. Since $c > 0$, multiplying both sides by $c$ does not reverse the inequality, so we obtain $cs - \varepsilon < cy$, and $cy \in cA$. Therefore, by the $\varepsilon$-characterization of the supremum,

Third, we prove (iii). Put $m := \inf A$. Since $d < 0$, $a \ge m$ implies $da \le dm$ for all $a \in A$ (multiplying by a negative number reverses the inequality), so $dm$ is an upper bound of $dA$. Let $\varepsilon > 0$ be given. Since $-d > 0$, $\varepsilon/(-d) > 0$, and by Remark 1.49 there exists a $y \in A$ such that $y < m + \varepsilon/(-d)$. Since $d < 0$, multiplying both sides by $d$ reverses the inequality, giving

and $dy \in dA$. Thus $dm - \varepsilon < dy$, so by the $\varepsilon$-characterization of the supremum,

Fourth, we prove (iv). Put $m := \inf A$. For all $a \in A$, $a \ge m$, so $a + b \ge m + b$, and thus $m + b$ is a lower bound of $A + b$. Let $\varepsilon > 0$ be given. By Remark 1.49 there exists a $y \in A$ such that $y < m + \varepsilon$, and thus $y + b \in A + b$ and $y + b < (m + b) + \varepsilon$. Therefore, by the $\varepsilon$-characterization of the infimum,

Fifth, we prove (v). Put $m := \inf A$. Since $c > 0$, $a \ge m$ implies $ca \ge cm$, so $cm$ is a lower bound of $cA$. Let $\varepsilon > 0$ be given. Since $\varepsilon/c > 0$, by Remark 1.49 there exists a $y \in A$ such that $y < m + \varepsilon/c$, and multiplying both sides by $c$ gives $cy < cm + \varepsilon$, with $cy \in cA$. Therefore,

Finally, we prove (vi). Put $s := \sup A$. Since $d < 0$, $a \le s$ implies $da \ge ds$ (multiplying by a negative number reverses the inequality), so $ds$ is a lower bound of $dA$. Let $\varepsilon > 0$ be given. Since $-d > 0$, $\varepsilon/(-d) > 0$, and by Remark 1.49 there exists a $y \in A$ such that $s - \varepsilon/(-d) < y$. Since $d < 0$, multiplying both sides by $d$ reverses the inequality, giving

and $dy \in dA$. Thus $dy < ds + \varepsilon$, so by the $\varepsilon$-characterization of the infimum,

We first prove (iii). Let $\varepsilon > 0$ be given. Since $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$, by Theorem 2.19 there exists a partition $P$ of $[a, b]$ such that

Put $P' := P \cup \{c\}$. Then $P'$ is a refinement of $P$, so by Theorem 2.20(i),

Put $P' = \{x_0, x_1, \ldots, x_n\}$; since $c \in P'$, there exists $j$ such that $c = x_j$. As $c \in (a, b)$ we have $x_0 = a < c < b = x_n$, hence $j \in \{1, \ldots, n-1\}$ (so that $P_1$ and $P_2$ below each contain at least two points and are therefore genuine partitions of $[a, c]$ and $[c, b]$). Define

Then $P_1$ is a partition of $[a, c]$ and $P_2$ is a partition of $[c, b]$. Since $U(P', f, \alpha)$ and $L(P', f, \alpha)$ are sums over the subintervals of $P'$, and the subintervals of $P'$ split exactly, at the boundary $x_j = c$, into those forming $P_1$ and those forming $P_2$, we have

Thus

By (2.8), both terms on the left-hand side are nonnegative, so each is individually less than $\varepsilon$:

Since $\varepsilon > 0$ is arbitrary, applying Theorem 2.19 on $[a, c]$ and on $[c, b]$ respectively shows that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, c]$ and on $[c, b]$.

Now we prove the identity. Since $f$ is integrable with respect to $\alpha$ on $[a, c]$, $[c, b]$, and $[a, b]$, by (2.15),

and adding these two inequalities gives

Also, applying (2.15) directly to $P'$ and $[a, b]$,

Combining the two chains of inequalities above,

Since $\varepsilon > 0$ is arbitrary,

Next, since the proof of (iv) requires (v), we prove (v) first. For any partition $P = \{x_0, \ldots, x_n\}$ of $[a, b]$ and each $i = 1, \ldots, n$, since $f_1(x) \le f_2(x)$ for all $x \in [x_{i-1}, x_i]$,

(since $\inf_{[x_{i-1},x_i]} f_1$ is a lower bound of $f_1$ on $[x_{i-1}, x_i]$, and $f_1(x) \le f_2(x)$ for all $x \in [x_{i-1}, x_i]$, it is also a lower bound of $f_2$ on $[x_{i-1}, x_i]$, hence at most $\inf_{[x_{i-1},x_i]} f_2$). That is, $m_i^{f_1} \le m_i^{f_2}$. Since $\Delta\alpha_i \ge 0$,

and this holds for every partition $P$ of $[a, b]$. Thus the supremum of the left-hand side is at most the supremum of the right-hand side, so

Since both $f_1$ and $f_2$ are integrable with respect to $\alpha$ on $[a, b]$,

Finally, we prove (iv). Since $\sup_{x \in [a,b]} |f(x)| \le M$,

Consider the constant function $g \equiv M$. For any partition $P = \{x_0, \ldots, x_n\}$ of $[a, b]$, since $\sup_{[x_{i-1},x_i]} g = \inf_{[x_{i-1},x_i]} g = M$ for every $i$,

(the sum telescopes, so $\sum_{i=1}^{n} \Delta\alpha_i = \alpha(b) - \alpha(a)$). That is, $U(P, g, \alpha) = L(P, g, \alpha) = M(\alpha(b) - \alpha(a))$ for every partition $P$, so by Theorem 2.19, $g$ is integrable with respect to $\alpha$ on $[a, b]$, and computing the upper integral $\inf_P U(P, g, \alpha)$ gives (writing $\int_a^b M d\alpha$ for the integral of $g$)

In the same way,

Now, since $-M \le f(x)$ for all $x \in [a, b]$, by (v),

Also, since $f(x) \le M$ for all $x \in [a, b]$, by (v),

Therefore

and since $\alpha(b) - \alpha(a) \ge 0$, this is exactly

Applying Theorem 2.37(ii) to $f_1$ and the constant $c_1$, $c_1 f_1$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ and

Similarly, applying Theorem 2.37(ii) to $f_2$ and the constant $c_2$, $c_2 f_2$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ and

Now put $g_1 := c_1 f_1$ and $g_2 := c_2 f_2$. Since both $g_1$ and $g_2$ are Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$, by Theorem 2.37(i), $g_1 + g_2 = c_1 f_1 + c_2 f_2$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ and

We first prove (i). Since $\alpha_1$ and $\alpha_2$ are monotonically increasing on $[a, b]$, if $x, y \in [a, b]$ and $x < y$, then $\alpha_1(x) \le \alpha_1(y)$ and $\alpha_2(x) \le \alpha_2(y)$, so $(\alpha_1 + \alpha_2)(x) \le (\alpha_1 + \alpha_2)(y)$. That is, $\alpha_1 + \alpha_2$ is also monotonically increasing on $[a, b]$. For a partition $P = \{x_0, \ldots, x_n\}$ of $[a, b]$ and each $i = 1, \ldots, n$,

Noting that $M_i := \sup_{[x_{i-1},x_i]} f$ and $m_i := \inf_{[x_{i-1},x_i]} f$ depend only on $f$, not on $\alpha$,

and similarly

These two identities hold for every partition $P$ of $[a, b]$.

Let $\varepsilon > 0$ be given. Since $f$ is integrable with respect to $\alpha_1$ on $[a, b]$, by Theorem 2.19 there exists a partition $P_1$ of $[a, b]$ such that

Similarly, since $f$ is integrable with respect to $\alpha_2$ on $[a, b]$, there exists a partition $P_2$ of $[a, b]$ such that

Let $P^*$ be the common refinement of $P_1$ and $P_2$ (Definition 2.14). Since $P^*$ is a refinement of $P_1$, by Theorem 2.15 and Corollary 2.16,

and since $P^*$ is a refinement of $P_2$, similarly

Therefore, by the identity shown above,

By Theorem 2.19, $f$ is Riemann-Stieltjes integrable with respect to $\alpha_1 + \alpha_2$ on $[a, b]$.

Now we prove the value identity. Since the identity above holds for every partition of $[a, b]$, not just $P^*$, for any partitions $P_1$, $P_2$ of $[a, b]$ and their common refinement $P^*$ (following the exact pattern of the proof of Theorem 2.37(i)),

and

Therefore

and

Combining these two inequalities,

Next we prove (ii). Since $c > 0$ and $\alpha$ is monotonically increasing on $[a, b]$, if $x, y \in [a, b]$ and $x < y$, then $\alpha(x) \le \alpha(y)$, and since $c > 0$, $c\alpha(x) \le c\alpha(y)$. That is, $c\alpha$ is also monotonically increasing on $[a, b]$. For a partition $P$ of $[a, b]$ and each $i$,

so since $c > 0$ can be factored out of the sum,

and similarly $L(P, f, c\alpha) = cL(P, f, \alpha)$.

Let $\varepsilon > 0$ be given. Since $f$ is integrable with respect to $\alpha$ on $[a, b]$, applying Theorem 2.19 with tolerance $\varepsilon/c > 0$, there exists a partition $P$ of $[a, b]$ such that

Thus

By Theorem 2.19, $f$ is Riemann-Stieltjes integrable with respect to $c\alpha$ on $[a, b]$.

We show the value identity. Since $c > 0$, by Exercise 2.36(v),

and by Exercise 2.36(ii),

Since both values equal $c\int_a^b f d\alpha$,

This statement is false. Let $[a, b] := [0, 1]$ and $\alpha(x) := x$. Then $\alpha$ is continuous and monotonically increasing on $[0, 1]$. Put $f \equiv 1$. Then $f$ is bounded, and for any partition $P = \{x_0, \ldots, x_n\}$ of $[0, 1]$, since $\sup_{[x_{i-1},x_i]} f = \inf_{[x_{i-1},x_i]} f = 1$ for every $i$,

That is, $U = L$ for every partition, so by Theorem 2.19, $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[0, 1]$.

Now consider the function from Example 2.7,

Since the values of $g$ are only $0$ or $1$, $g$ is bounded on $[0, 1]$. As already shown in Example 2.7, for every partition $P$ of $[0, 1]$,

so $U(P, g) - L(P, g) = 1$, which means that there is no partition $P$ satisfying $U(P,g) - L(P,g) < \varepsilon$ for any $\varepsilon \in (0, 1)$. Thus by Theorem 2.19, $g$ is not Riemann-integrable on $[0, 1]$, and by Remark 2.13 (i.e. since Riemann-Stieltjes integrability with respect to $\alpha(x) = x$ is equivalent to Riemann integrability), $g$ is also not Riemann-Stieltjes integrable with respect to $\alpha$ on $[0, 1]$.

But

so $fg = g$, and hence $fg$ is not Riemann-Stieltjes integrable with respect to $\alpha$ on $[0, 1]$. That is, even when $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ and $g$ is merely bounded on $[a, b]$, $fg$ need not be Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$.