从定义到判别

上一讲把 Riemann 积分定义成「下积分与上积分相等时的公共值」，但没有回答什么时候相等。这一讲补上三件事：

1. 对任何有界函数，下积分都不超过上积分——所以可积性的定义不会自相矛盾。
2. 给出一个不可积的例子（Dirichlet 函数），说明有界性远远不够。
3. 给出可积的充要判据：对每个 $\varepsilon>0$，能找到分割使上下和之差小于 $\varepsilon$。

判据本身只涉及有限个矩形，但要把它用在连续函数上，还需要一个新工具：一致连续。整讲的逻辑是

$$\text{连续} \xrightarrow{\text{紧性}} \text{一致连续} \xrightarrow{\text{上下和}} \text{可积}.$$

时间轴

回顾：分割、上下和、加细

设 $f$ 在 $[a,b]$ 上有界，$P=\{a=x_0<x_1<\cdots<x_n=b\}$ 是分割，记 $\Delta x_i=x_i-x_{i-1}$，并令

$$M_i=\sup_{[x_{i-1},x_i]}f, \qquad m_i=\inf_{[x_{i-1},x_i]}f .$$

则

$$U(f,P)=\sum_{i=1}^{n}M_i\Delta x_i, \qquad L(f,P)=\sum_{i=1}^{n}m_i\Delta x_i, \qquad L(f,P)\le U(f,P).$$

加细引理：若 $P_2$ 是 $P_1$ 的加细，则

$$L(f,P_1)\le L(f,P_2)\le U(f,P_2)\le U(f,P_1).$$

上积分是所有上和的下确界，下积分是所有下和的上确界：

$$\overline{\int_a^b f}=\inf_P U(f,P), \qquad \underline{\int_a^b f}=\sup_P L(f,P).$$

下积分永远不超过上积分

命题。 对任意有界函数 $f$，

$$\underline{\int_a^b f}\le \overline{\int_a^b f}.$$

证明。 先证一个更强的结论：任意两个分割的下和不超过上和的任意组合，即对任意分割 $P,P^*$，

$$L(f,P)\le U(f,P^*).$$

取 $P$ 与 $P^*$ 分点的并集，得到公共加细 $\bar P$。它同时是两者的加细，所以由加细引理

$$L(f,P)\le L(f,\bar P)\le U(f,\bar P)\le U(f,P^*).$$

于是 $L(f,P)\le U(f,P^*)$ 对一切 $P,P^*$ 成立。

现在固定 $P$。上式说明 $L(f,P)$ 是集合 $\{U(f,P^*):P^*\}$ 的一个下界，因此不超过它的下确界：

$$L(f,P)\le\inf_{P^*}U(f,P^*)=\overline{\int_a^b f}.$$

这说明 $\overline{\int_a^b f}$ 是集合 $\{L(f,P):P\}$ 的一个上界，因此不小于其上确界：

$$\sup_P L(f,P)=\underline{\int_a^b f}\le \overline{\int_a^b f}.$$

$\square$

这个证明的技巧值得记住：先证「任意下和 ≤ 任意上和」，再用上下确界的定义各取一次。加细引理在这里承担了全部工作。

一个不可积的函数：Dirichlet 函数

定义在 $[0,1]$ 上：

$$f(x)= \begin{cases} 1, & x\in\mathbb Q,\\ 0, & x\notin\mathbb Q . \end{cases}$$

它显然有界（$0\le f\le1$），但不可积。

理由。 取任意分割 $P$。任何一个非退化的子区间 $[x_{i-1},x_i]$ 中，既有有理数（稠密性），又有无理数（稠密性），所以

$$M_i=1, \qquad m_i=0 .$$

于是

$$U(f,P)=\sum_{i=1}^{n}1\cdot\Delta x_i=1, \qquad L(f,P)=\sum_{i=1}^{n}0\cdot\Delta x_i=0 .$$

这对所有分割都成立。因此

$$\overline{\int_0^1 f}=1, \qquad \underline{\int_0^1 f}=0,$$

两者不相等，函数不可积。

这个例子说明：有界、甚至处处有定义，都不够。可积性要求函数在小区间上的「起伏」能整体压小，而 Dirichlet 函数在每个区间上的起伏都是 $1$。

可积的充要判据

定理。 设 $f$ 在 $[a,b]$ 上有界。则 $f$ Riemann 可积，当且仅当对每个 $\varepsilon>0$，存在分割 $P$，使

$$U(f,P)-L(f,P)<\varepsilon .$$

证明（必要性）。 设 $f$ 可积，公共值为 $I$。由

$$I=\underline{\int_a^b f}=\sup_P L(f,P),$$

给定 $\varepsilon>0$，存在分割 $P_1$，使

$$L(f,P_1)>I-\frac{\varepsilon}{2}.$$

（否则所有下和都 $\le I-\varepsilon/2$，上确界就小于 $I$。）同理，由 $I=\inf_P U(f,P)$，存在分割 $P_2$，使

$$U(f,P_2)<I+\frac{\varepsilon}{2}.$$

取公共加细 $P$，它同时加细 $P_1,P_2$。于是

$$L(f,P)\ge L(f,P_1)>I-\frac{\varepsilon}{2}, \qquad U(f,P)\le U(f,P_2)<I+\frac{\varepsilon}{2}.$$

两式相减：

$$U(f,P)-L(f,P)<\left(I+\frac{\varepsilon}{2}\right)-\left(I-\frac{\varepsilon}{2}\right)=\varepsilon .$$

证明（充分性）。 设对每个 $\varepsilon>0$ 都有分割 $P$ 使 $U(f,P)-L(f,P)<\varepsilon$。对任意分割恒有

$$L(f,P)\le\underline{\int_a^b f}\le\overline{\int_a^b f}\le U(f,P),$$

所以

$$0\le\overline{\int_a^b f}-\underline{\int_a^b f}\le U(f,P)-L(f,P)<\varepsilon .$$

$\varepsilon$ 任意，故两个积分相等，$f$ 可积。$\square$

判据的意义在于：可积性只需要一列分割，不需要对所有分割验证。只要上下和能任意接近，上下确界就被夹住。

一致连续

定义。 设 $f$ 定义在集合 $E$ 上。称 $f$ 在 $E$ 上一致连续，若对每个 $\varepsilon>0$，存在 $\delta>0$，使得对所有 $x,y\in E$，只要 $|x-y|<\delta$，就有

$$|f(x)-f(y)|<\varepsilon .$$

与普通连续的区别只在一个词：$\delta$ 不能依赖 $x$。普通连续是「对每个 $x_0$，给定 $\varepsilon$，存在 $\delta(\varepsilon,x_0)$」；一致连续要求同一个 $\delta$ 对所有点都管用。

几何上：给定 $y$ 轴上的误差带 $(-\varepsilon,\varepsilon)$，一致连续保证存在一个统一的水平宽度 $\delta$，使得宽度小于 $\delta$ 的任何一段图像都落在高度小于 $2\varepsilon$ 的带子里。

反例一：$x^2$ 在 $\mathbb R$ 上

设 $f(x)=x^2$。若一致连续，则存在 $\delta>0$，使 $|x-y|<\delta$ 时 $|x^2-y^2|<\varepsilon$。取 $y=x+\delta/2$，则

$$|f(x)-f(y)|=\left|x^2-\left(x+\frac{\delta}{2}\right)^2\right| =\delta x+\frac{\delta^2}{4}.$$

当 $x\to+\infty$ 时这个量无界，不可能对所有 $x$ 都小于固定的 $\varepsilon$。所以 $x^2$ 在 $\mathbb R$ 上连续但不一致连续。

原因不在函数本身，而在定义域无界：图像越往右越陡，同样的水平宽度对应越来越大的竖直变化。

反例二：$1/x$ 在 $(0,1]$ 上

设 $f(x)=1/x$，定义域 $(0,1]$。取

$$x_n=\frac1n, \qquad y_n=\frac1{2n}.$$

则

$$|x_n-y_n|=\frac1{2n}\to0, \qquad |f(x_n)-f(y_n)|=|n-2n|=n\to\infty .$$

所以无论 $\delta$ 多小，总有一对距离小于 $\delta$ 的点，函数值相差超过给定 $\varepsilon$。$1/x$ 在 $(0,1]$ 上不一致连续。

原因不在无界，而在定义域不闭：越靠近 $0$，函数越陡，且没有定义在 $0$ 处。

两个反例都指向同一件事：一致连续性依赖于定义域的紧性。

紧区间上连续蕴含一致连续

定理。 设 $f$ 在闭区间 $[a,b]$ 上连续，则 $f$ 在 $[a,b]$ 上一致连续。

证明（反证 + 序列紧性）。 假设不一致连续。否定定义：存在 $\varepsilon>0$，使得对每个 $\delta>0$，都能找到 $x,y\in[a,b]$ 满足

$$|x-y|<\delta \quad\text{但}\quad |f(x)-f(y)|\ge\varepsilon .$$

取 $\delta=1/n$（$n=1,2,3,\dots$），得到两列点 $x_n,y_n\in[a,b]$，满足

$$|x_n-y_n|<\frac1n, \qquad |f(x_n)-f(y_n)|\ge\varepsilon .$$

由 Bolzano–Weierstrass 定理（有界闭区间上的有界数列有收敛子列），存在子列 $x_{n_k}\to x$，且 $x\in[a,b]$（闭区间包含极限点）。

再看对应的 $y_{n_k}$。由三角不等式

$$|y_{n_k}-x|\le|y_{n_k}-x_{n_k}|+|x_{n_k}-x| <\frac{1}{n_k}+|x_{n_k}-x| .$$

右边两项都趋于 $0$，故 $y_{n_k}\to x$。

由于 $f$ 在 $x$ 处连续，

$$f(x_{n_k})\to f(x), \qquad f(y_{n_k})\to f(x).$$

于是

$$|f(x_{n_k})-f(y_{n_k})|\to0,$$

这与它对每个 $k$ 都 $\ge\varepsilon>0$ 矛盾。所以假设不成立，$f$ 一致连续。$\square$

这个证明是紧性论证的典型样本：否定一致性，得到两条距离趋于零但函数值距离不趋于零的点列；用 B–W 抽出收敛子列；两条子列收敛到同一点；连续性迫使函数值也靠拢，矛盾。

连续函数可积

定理。 设 $f$ 在 $[a,b]$ 上连续，则 $f$ Riemann 可积。

证明。 由判据，只需对每个 $\varepsilon>0$ 构造一个上下和之差小于 $\varepsilon$ 的分割。

设 $a<b$（$a=b$ 时积分显然为零）。由一致连续性，取 $\delta>0$，使 $|x-y|<\delta$ 时

$$|f(x)-f(y)|<\frac{\varepsilon}{b-a}.$$

取分割 $P$，使每段长度 $\Delta x_i<\delta$。

在每段 $[x_{i-1},x_i]$ 上，$f$ 连续，由极值定理，$M_i$ 与 $m_i$ 分别在两个点 $y_i,y_i^*$ 取到。这两点的距离不超过该段长度，小于 $\delta$，所以

$$M_i-m_i=f(y_i)-f(y_i^*)<\frac{\varepsilon}{b-a}.$$

于是

$$U(f,P)-L(f,P) =\sum_{i=1}^{n}(M_i-m_i)\Delta x_i <\sum_{i=1}^{n}\frac{\varepsilon}{b-a}\Delta x_i =\frac{\varepsilon}{b-a}\sum_{i=1}^{n}\Delta x_i =\varepsilon .$$

最后一步用了 $\sum\Delta x_i=b-a$。由判据，$f$ 可积。$\square$

注意证明中三件事各有其位：一致连续提供统一的 $\delta$；极值定理保证 sup 与 inf 被取到（否则只能取近似点，得不到严格不等式）；分割把这些局部估计加起来，$\sum\Delta x_i=b-a$ 把常数消掉。

积分的基本性质

课程最后只陈述了四条，证明留到后面：

1. 数乘：$f$ 可积，$c\in\mathbb R$，则 $cf$ 可积，且 $\int_a^b cf=c\int_a^b f$。
2. 加法：$f,g$ 可积，则 $f+g$ 可积，且 $\int_a^b(f+g)=\int_a^b f+\int_a^b g$。
3. 单调性：$f,g$ 可积且 $f\le g$，则 $\int_a^b f\le\int_a^b g$。
4. 区间可加：$f$ 在 $[a,b]$ 上可积，$c\in[a,b]$，则 $f$ 在 $[a,c]$ 与 $[c,b]$ 上都可积，且

$$\int_a^b f=\int_a^c f+\int_c^b f .$$

第四条用面积理解最直接：以 $c$ 为界把图像切成两段，两块有向面积相加等于整块。

本节结论

• 对任意有界函数，$\underline{\int}\le\overline{\int}$；证明靠公共加细与确界定义。
• Dirichlet 函数在每个区间上 sup 为 $1$、inf 为 $0$，故 $U\equiv1$、$L\equiv0$，不可积。
• $f$ 可积 $\iff$ 对每个 $\varepsilon>0$ 存在分割使 $U-L<\varepsilon$。
• 一致连续 = 同一个 $\delta$ 对所有点有效；$x^2$ 在 $\mathbb R$、$1/x$ 在 $(0,1]$ 都是连续但非一致连续。
• 紧区间上连续 $\Rightarrow$ 一致连续（B–W + 反证）。
• 紧区间上连续 $\Rightarrow$ 可积（一致连续 + 极值定理 + 上下和估计）。
• 积分满足线性、单调与区间可加（暂未证明）。

理解检验

1. 证明：对任意两个分割 $P,P^*$，有 $L(f,P)\le U(f,P^*)$，并由此推出下积分不超过上积分。

参考答案：取 $P$ 与 $P^*$ 分点的并集得到公共加细 $\bar P$。由加细引理，$L(f,P)\le L(f,\bar P)$ 且 $U(f,\bar P)\le U(f,P^*)$；又对同一分割总有 $L(f,\bar P)\le U(f,\bar P)$。三者串起来得 $L(f,P)\le U(f,P^*)$。固定 $P$，右端对一切 $P^*$ 成立，故 $L(f,P)\le\inf_{P^*}U(f,P^*)=\overline{\int}f$。再对一切 $P$ 取上确界：$\underline{\int}f=\sup_P L(f,P)\le\overline{\int}f$。

2. 证明 Dirichlet 函数在 $[0,1]$ 上不可积。

参考答案：任取分割 $P$。每个非退化子区间既含有理数又含无理数（两种数都稠密），所以 $M_i=1$、$m_i=0$。于是 $U(f,P)=\sum1\cdot\Delta x_i=1$，$L(f,P)=\sum0\cdot\Delta x_i=0$，对所有分割都成立。因此上积分是常值 $1$ 的下确界，等于 $1$；下积分是常值 $0$ 的上确界，等于 $0$。上下积分不相等，故不可积。

3. 陈述并证明可积判据（$U-L<\varepsilon$ 的充要条件）的充分性部分。

参考答案：设对每个 $\varepsilon>0$ 存在分割 $P$ 使 $U(f,P)-L(f,P)<\varepsilon$。对任意分割恒有 $L(f,P)\le\underline{\int}f\le\overline{\int}f\le U(f,P)$，因此 $0\le\overline{\int}f-\underline{\int}f\le U(f,P)-L(f,P)<\varepsilon$。由于这对任意 $\varepsilon>0$ 成立，两个积分必须相等，故 $f$ 可积。

4. 说明 $1/x$ 在 $(0,1]$ 上连续但不一致连续，并指出定义域缺少哪条性质。

参考答案：取 $x_n=1/n$、$y_n=1/(2n)$，都在 $(0,1]$ 内。则 $|x_n-y_n|=1/(2n)\to0$，而 $|f(x_n)-f(y_n)|=|n-2n|=n\to\infty$。所以不存在统一的 $\delta$：任意小的水平距离都可能对应巨大的竖直差。定义域 $(0,1]$ 有界但不闭（缺端点 $0$），因此不紧致；紧性是连续升级为一致连续的关键条件。

5. 用 Bolzano–Weierstrass 定理证明：$[a,b]$ 上的连续函数一致连续。

参考答案：反设不一致连续，则存在 $\varepsilon>0$，使对每个 $n$ 都有 $x_n,y_n\in[a,b]$ 满足 $|x_n-y_n|<1/n$ 且 $|f(x_n)-f(y_n)|\ge\varepsilon$。由 B–W，存在子列 $x_{n_k}\to x\in[a,b]$。由三角不等式 $|y_{n_k}-x|\le|y_{n_k}-x_{n_k}|+|x_{n_k}-x|<1/n_k+|x_{n_k}-x|\to0$，故 $y_{n_k}\to x$。连续性给出 $f(x_{n_k})\to f(x)$ 与 $f(y_{n_k})\to f(x)$，于是 $|f(x_{n_k})-f(y_{n_k})|\to0$，与它始终 $\ge\varepsilon$ 矛盾。

6. 设 $f$ 在 $[a,b]$ 上连续。用一致连续与判据证明 $f$ 可积，并指出证明中极值定理用在哪里。

参考答案：给定 $\varepsilon>0$，由一致连续取 $\delta>0$ 使 $|x-y|<\delta\Rightarrow|f(x)-f(y)|<\varepsilon/(b-a)$。取分割使每段 $\Delta x_i<\delta$。在每个子区间上，极值定理保证 $M_i,m_i$ 被某两点取到，而这两点距离小于 $\delta$，所以 $M_i-m_i<\varepsilon/(b-a)$。于是 $U-L=\sum(M_i-m_i)\Delta x_i<\frac{\varepsilon}{b-a}\sum\Delta x_i=\varepsilon$。由判据 $f$ 可积。极值定理用在「sup 与 inf 确实被取到」，否则只能取到近似点，只能得到 $\le$ 而非严格不等式；改用 $\varepsilon/(2(b-a))$ 也可以绕开，但极值定理让论证更直接。

与前后课的关系

这一讲把可积性从定义变成了可用的工具：判据把无穷多个分割压缩成一个 $\varepsilon$ 论证，而紧性把「逐点连续」升级成「一致连续」。两条线索在后文反复出现。

下一讲进入微积分基本定理：定义变上限积分 $F(x)=\int_a^x f$，证明 $F$ 可导且 $F'=f$（当 $f$ 连续），再用它计算具体积分，并讨论反常积分。再往后，积分与导数的交换、函数项级数的逐项积分，都要回到本讲的一致连续与一致收敛思想。

From definition to criterion

The previous lecture defined the Riemann integral as the common value of the lower and upper integrals, but said nothing about when they agree. This lecture supplies three things:

1. For every bounded function the lower integral is at most the upper one, so the definition is consistent.
2. A non-integrable example (Dirichlet's function) shows that boundedness is far from enough.
3. An if-and-only-if criterion: for every $\varepsilon>0$ there is a partition whose upper and lower sums differ by less than $\varepsilon$.

The criterion only involves finitely many rectangles, but applying it to continuous functions needs a new tool: uniform continuity. The logic of the lecture is

$$\text{continuous} \xrightarrow{\text{compactness}} \text{uniformly continuous} \xrightarrow{\text{upper/lower sums}} \text{integrable}.$$

Timeline

Review: partitions, sums, refinements

Let $f$ be bounded on $[a,b]$ and $P=\{a=x_0<x_1<\cdots<x_n=b\}$ a partition, with $\Delta x_i=x_i-x_{i-1}$ and

$$M_i=\sup_{[x_{i-1},x_i]}f, \qquad m_i=\inf_{[x_{i-1},x_i]}f .$$

Then

$$U(f,P)=\sum_{i=1}^{n}M_i\Delta x_i, \qquad L(f,P)=\sum_{i=1}^{n}m_i\Delta x_i, \qquad L(f,P)\le U(f,P).$$

Refinement lemma: if $P_2$ refines $P_1$, then

$$L(f,P_1)\le L(f,P_2)\le U(f,P_2)\le U(f,P_1).$$

The upper integral is the infimum of all upper sums, the lower integral the supremum of all lower sums:

$$\overline{\int_a^b f}=\inf_P U(f,P), \qquad \underline{\int_a^b f}=\sup_P L(f,P).$$

The lower integral never exceeds the upper

Proposition. For every bounded $f$,

$$\underline{\int_a^b f}\le \overline{\int_a^b f}.$$

Proof. First prove a stronger statement: for any two partitions $P,P^*$,

$$L(f,P)\le U(f,P^*).$$

Take the union of the division points of $P$ and $P^*$, giving the common refinement $\bar P$. It refines both, so the refinement lemma gives

$$L(f,P)\le L(f,\bar P)\le U(f,\bar P)\le U(f,P^*).$$

Hence $L(f,P)\le U(f,P^*)$ for all $P,P^*$.

Fix $P$. The inequality says $L(f,P)$ is a lower bound for the set $\{U(f,P^*):P^*\}$, so it does not exceed the infimum:

$$L(f,P)\le\inf_{P^*}U(f,P^*)=\overline{\int_a^b f}.$$

Thus $\overline{\int_a^b f}$ is an upper bound for $\{L(f,P):P\}$, and therefore at least the supremum:

$$\sup_P L(f,P)=\underline{\int_a^b f}\le \overline{\int_a^b f}.$$

$\square$

The technique is worth remembering: first show "any lower sum ≤ any upper sum", then apply the definitions of sup and inf once each. The refinement lemma does all the work.

A non-integrable function: Dirichlet's

Define on $[0,1]$:

$$f(x)= \begin{cases} 1, & x\in\mathbb Q,\\ 0, & x\notin\mathbb Q . \end{cases}$$

It is plainly bounded ($0\le f\le1$), yet not integrable.

Reason. Take any partition $P$. Every nondegenerate subinterval $[x_{i-1},x_i]$ contains rationals and irrationals (both are dense), so

$$M_i=1, \qquad m_i=0 .$$

Therefore

$$U(f,P)=\sum_{i=1}^{n}1\cdot\Delta x_i=1, \qquad L(f,P)=\sum_{i=1}^{n}0\cdot\Delta x_i=0 .$$

This holds for all partitions. Hence

$$\overline{\int_0^1 f}=1, \qquad \underline{\int_0^1 f}=0,$$

so the function is not integrable.

The example shows that boundedness is not the issue. Integrability requires that the oscillation over small intervals can be made small in aggregate; here the oscillation is $1$ on every interval.

The integrability criterion

Theorem. Let $f$ be bounded on $[a,b]$. Then $f$ is Riemann integrable if and only if for every $\varepsilon>0$ there is a partition $P$ with

$$U(f,P)-L(f,P)<\varepsilon .$$

Proof (necessity). Suppose $f$ is integrable with common value $I$. Since

$$I=\underline{\int_a^b f}=\sup_P L(f,P),$$

given $\varepsilon>0$ there is $P_1$ with

$$L(f,P_1)>I-\frac{\varepsilon}{2}.$$

(Otherwise every lower sum would be $\le I-\varepsilon/2$ and the supremum would be smaller than $I$.) Likewise, since $I=\inf_P U(f,P)$, there is $P_2$ with

$$U(f,P_2)<I+\frac{\varepsilon}{2}.$$

Take the common refinement $P$, which refines both. Then

$$L(f,P)\ge L(f,P_1)>I-\frac{\varepsilon}{2}, \qquad U(f,P)\le U(f,P_2)<I+\frac{\varepsilon}{2}.$$

Subtracting,

$$U(f,P)-L(f,P)<\left(I+\frac{\varepsilon}{2}\right)-\left(I-\frac{\varepsilon}{2}\right)=\varepsilon .$$

Proof (sufficiency). Suppose that for every $\varepsilon>0$ some $P$ has $U(f,P)-L(f,P)<\varepsilon$. For any partition,

$$L(f,P)\le\underline{\int_a^b f}\le\overline{\int_a^b f}\le U(f,P),$$

so

$$0\le\overline{\int_a^b f}-\underline{\int_a^b f}\le U(f,P)-L(f,P)<\varepsilon .$$

Since $\varepsilon$ is arbitrary, the two integrals agree and $f$ is integrable. $\square$

The point of the criterion: integrability needs one partition for each $\varepsilon$, not verification over all partitions. If upper and lower sums can be brought arbitrarily close, the sup and inf are squeezed together.

Uniform continuity

Definition. Let $f$ be defined on a set $E$. Say $f$ is uniformly continuous on $E$ if for every $\varepsilon>0$ there is $\delta>0$ such that for all $x,y\in E$,

$$|x-y|<\delta\implies|f(x)-f(y)|<\varepsilon .$$

The only difference from ordinary continuity is one word: $\delta$ may not depend on the point. Ordinary continuity says "for each $x_0$ and each $\varepsilon$ there is $\delta(\varepsilon,x_0)$"; uniform continuity requires one $\delta$ that works everywhere.

Geometrically: given an error band of height $\varepsilon$ on the $y$-axis, there is a uniform horizontal width $\delta$ such that any piece of the graph of width less than $\delta$ stays within a band of height $2\varepsilon$.

Counterexample 1: $x^2$ on $\mathbb R$

If $x^2$ were uniformly continuous, there would be $\delta>0$ with $|x^2-y^2|<\varepsilon$ whenever $|x-y|<\delta$. Take $y=x+\delta/2$:

$$|f(x)-f(y)|=\left|x^2-\left(x+\frac{\delta}{2}\right)^2\right| =\delta x+\frac{\delta^2}{4},$$

which is unbounded as $x\to+\infty$. So no single $\delta$ works. The function is continuous but not uniformly continuous on $\mathbb R$.

The cause is not the function but the unbounded domain: the graph steepens without limit, so a fixed horizontal width produces ever larger vertical change.

Counterexample 2: $1/x$ on $(0,1]$

Let $f(x)=1/x$ on $(0,1]$ and take

$$x_n=\frac1n, \qquad y_n=\frac1{2n}.$$

Then

$$|x_n-y_n|=\frac1{2n}\to0, \qquad |f(x_n)-f(y_n)|=|n-2n|=n\to\infty .$$

So no matter how small $\delta$ is, some pair of points closer than $\delta$ has images more than a given $\varepsilon$ apart. Hence $1/x$ is not uniformly continuous on $(0,1]$.

The cause is not unboundedness but the domain not being closed: near $0$ the function steepens without bound, and the endpoint $0$ is missing.

Both counterexamples point to the same thing: uniform continuity depends on compactness of the domain.

Continuity on a compact interval gives uniform continuity

Theorem. If $f$ is continuous on $[a,b]$, then $f$ is uniformly continuous on $[a,b]$.

Proof (contradiction plus sequential compactness). Suppose not. Negating the definition gives $\varepsilon>0$ such that for every $\delta>0$ there are $x,y\in[a,b]$ with

$$|x-y|<\delta \quad\text{and}\quad |f(x)-f(y)|\ge\varepsilon .$$

Take $\delta=1/n$ for $n=1,2,3,\dots$ to obtain sequences $x_n,y_n\in[a,b]$ with

$$|x_n-y_n|<\frac1n, \qquad |f(x_n)-f(y_n)|\ge\varepsilon .$$

By Bolzano–Weierstrass (a bounded sequence in a closed bounded interval has a convergent subsequence), some subsequence $x_{n_k}\to x$, and $x\in[a,b]$ because the interval is closed.

Consider $y_{n_k}$. The triangle inequality gives

$$|y_{n_k}-x|\le|y_{n_k}-x_{n_k}|+|x_{n_k}-x| <\frac{1}{n_k}+|x_{n_k}-x| .$$

Both terms tend to $0$, so $y_{n_k}\to x$.

Continuity of $f$ at $x$ gives

$$f(x_{n_k})\to f(x), \qquad f(y_{n_k})\to f(x).$$

Hence

$$|f(x_{n_k})-f(y_{n_k})|\to0,$$

contradicting the fact that it is $\ge\varepsilon>0$ for every $k$. Therefore $f$ is uniformly continuous. $\square$

This is a model compactness argument: negate uniformity, obtain two sequences whose points converge together but whose images stay apart; use Bolzano–Weierstrass; continuity forces the images together; contradiction.

Continuous functions are integrable

Theorem. If $f$ is continuous on $[a,b]$, then $f$ is Riemann integrable.

Proof. By the criterion it suffices, for each $\varepsilon>0$, to construct a partition with $U-L<\varepsilon$.

Assume $a<b$ (if $a=b$ the integral is zero). Uniform continuity gives $\delta>0$ such that $|x-y|<\delta$ implies

$$|f(x)-f(y)|<\frac{\varepsilon}{b-a}.$$

Choose a partition $P$ with every $\Delta x_i<\delta$.

On each $[x_{i-1},x_i]$, the function is continuous, so by the extreme value theorem $M_i$ and $m_i$ are attained at points $y_i,y_i^*$. Those points lie at distance less than $\delta$, hence

$$M_i-m_i=f(y_i)-f(y_i^*)<\frac{\varepsilon}{b-a}.$$

Therefore

$$U(f,P)-L(f,P) =\sum_{i=1}^{n}(M_i-m_i)\Delta x_i <\sum_{i=1}^{n}\frac{\varepsilon}{b-a}\Delta x_i =\frac{\varepsilon}{b-a}\sum_{i=1}^{n}\Delta x_i =\varepsilon .$$

The last step uses $\sum\Delta x_i=b-a$. The criterion gives integrability. $\square$

Each ingredient has a job: uniform continuity supplies a uniform $\delta$; the extreme value theorem guarantees that the sup and inf are actually attained (otherwise only nearby points are available and the inequality is not strict); the partition sums the local estimates, and $\sum\Delta x_i=b-a$ cancels the constant.

Basic properties of the integral

Four properties were stated, with proofs deferred:

1. Scalar multiples: if $f$ is integrable and $c\in\mathbb R$, then $cf$ is integrable and $\int_a^b cf=c\int_a^b f$.
2. Sums: if $f,g$ are integrable, so is $f+g$, and $\int_a^b(f+g)=\int_a^b f+\int_a^b g$.
3. Monotonicity: if $f,g$ are integrable and $f\le g$, then $\int_a^b f\le\int_a^b g$.
4. Additivity over intervals: if $f$ is integrable on $[a,b]$ and $c\in[a,b]$, then $f$ is integrable on $[a,c]$ and on $[c,b]$, and

$$\int_a^b f=\int_a^c f+\int_c^b f .$$

The last is clearest in terms of area: cutting the graph at $c$ splits the signed area into two parts whose sum is the whole.

What this lecture establishes

• For every bounded function, $\underline{\int}\le\overline{\int}$; the proof uses a common refinement and the definitions of sup and inf.
• Dirichlet's function has $M_i=1$ and $m_i=0$ on every interval, so $U\equiv1$ and $L\equiv0$; it is not integrable.
• $f$ is integrable iff for every $\varepsilon>0$ some partition has $U-L<\varepsilon$.
• Uniform continuity means one $\delta$ works for all points; $x^2$ on $\mathbb R$ and $1/x$ on $(0,1]$ are continuous but not uniformly continuous.
• Continuity on a compact interval implies uniform continuity (Bolzano–Weierstrass plus contradiction).
• Continuity on a compact interval implies integrability (uniform continuity, extreme value theorem, upper/lower sum estimate).
• The integral is linear, monotone and additive over subintervals (proofs deferred).

Comprehension check

1. Prove that $L(f,P)\le U(f,P^*)$ for any two partitions, and deduce that the lower integral is at most the upper one.

Reference answer: Let $\bar P$ be the common refinement obtained from the union of the division points. The refinement lemma gives $L(f,P)\le L(f,\bar P)$ and $U(f,\bar P)\le U(f,P^*)$, while for any single partition $L(f,\bar P)\le U(f,\bar P)$. Chaining them gives $L(f,P)\le U(f,P^*)$. Fix $P$; the right side holds for every $P^*$, so $L(f,P)\le\inf_{P^*}U(f,P^*)=\overline{\int}f$. Taking the supremum over all $P$ gives $\underline{\int}f=\sup_P L(f,P)\le\overline{\int}f$.

2. Prove that Dirichlet's function is not Riemann integrable on $[0,1]$.

Reference answer: Take any partition $P$. Every nondegenerate subinterval contains both rationals and irrationals (both sets are dense), so $M_i=1$ and $m_i=0$. Hence $U(f,P)=\sum1\cdot\Delta x_i=1$ and $L(f,P)=\sum0\cdot\Delta x_i=0$ for every partition. The upper integral is therefore the infimum of the constant $1$, namely $1$, and the lower integral is the supremum of the constant $0$, namely $0$. They differ, so the function is not integrable.

3. State the integrability criterion and prove its sufficiency.

Reference answer: A bounded $f$ on $[a,b]$ is Riemann integrable iff for every $\varepsilon>0$ there is a partition with $U(f,P)-L(f,P)<\varepsilon$. For sufficiency, suppose such a partition exists for each $\varepsilon$. For any partition, $L(f,P)\le\underline{\int}f\le\overline{\int}f\le U(f,P)$, hence $0\le\overline{\int}f-\underline{\int}f\le U(f,P)-L(f,P)<\varepsilon$. Since this holds for every $\varepsilon>0$, the two integrals are equal and $f$ is integrable.

4. Show that $1/x$ is continuous but not uniformly continuous on $(0,1]$, and identify the missing property of the domain.

Reference answer: Take $x_n=1/n$ and $y_n=1/(2n)$, both in $(0,1]$. Then $|x_n-y_n|=1/(2n)\to0$, while $|f(x_n)-f(y_n)|=|n-2n|=n\to\infty$. So no single $\delta$ works: arbitrarily small horizontal distances can produce arbitrarily large vertical differences. The domain $(0,1]$ is bounded but not closed (the endpoint $0$ is missing), hence not compact; compactness is exactly what upgrades continuity to uniform continuity.

5. Use Bolzano–Weierstrass to prove that a continuous function on $[a,b]$ is uniformly continuous.

Reference answer: Suppose not. Then there is $\varepsilon>0$ such that for every $n$ there are $x_n,y_n\in[a,b]$ with $|x_n-y_n|<1/n$ and $|f(x_n)-f(y_n)|\ge\varepsilon$. By Bolzano–Weierstrass, a subsequence $x_{n_k}$ converges to some $x\in[a,b]$. The triangle inequality gives $|y_{n_k}-x|\le|y_{n_k}-x_{n_k}|+|x_{n_k}-x|<1/n_k+|x_{n_k}-x|\to0$, so $y_{n_k}\to x$. Continuity yields $f(x_{n_k})\to f(x)$ and $f(y_{n_k})\to f(x)$, hence $|f(x_{n_k})-f(y_{n_k})|\to0$, contradicting the lower bound $\varepsilon$.

6. For continuous $f$ on $[a,b]$, prove integrability from uniform continuity and the criterion, and say where the extreme value theorem enters.

Reference answer: Given $\varepsilon>0$, uniform continuity gives $\delta>0$ with $|x-y|<\delta\Rightarrow|f(x)-f(y)|<\varepsilon/(b-a)$. Choose a partition with every $\Delta x_i<\delta$. On each subinterval, the extreme value theorem guarantees that $M_i$ and $m_i$ are attained at points less than $\delta$ apart, so $M_i-m_i<\varepsilon/(b-a)$. Then $U-L=\sum(M_i-m_i)\Delta x_i<\frac{\varepsilon}{b-a}\sum\Delta x_i=\varepsilon$, and the criterion gives integrability. The extreme value theorem is what makes the sup and inf attained, producing a strict inequality; using $\varepsilon/(2(b-a))$ would also work, but the extreme value theorem makes the argument direct.

Relation to neighbouring lectures

This lecture turns integrability from a definition into a usable tool: the criterion compresses infinitely many partitions into one $\varepsilon$-argument, and compactness upgrades pointwise continuity to uniform continuity. Both ideas recur constantly.

The next lecture introduces the fundamental theorem of calculus: the function $F(x)=\int_a^x f$ is differentiable with $F'=f$ when $f$ is continuous; this is then used to compute concrete integrals, and improper integrals are discussed. Later, exchanging limits with integrals, and term-by-term integration of series, return to the uniform continuity and uniform convergence ideas from this lecture.