测度论：Measure Theory (5) —— The Deficiency of Riemann Integral

We introduced measure and Lebesgue Integral due to the deficiency of Riemann Integral. In general, there are at least three problematic issues about Riemann Integral:

1. Riemann integration does not handle functions with many discontinuities.

2. Riemann integration does not handle unbounded functions.

3. Riemann integration does not work well with limits.

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Definition. (Riemann sum)

令 \(f: [a, ~ b] \rightarrow \mathbb{R}\) 为一个有界实值函数，其中 \(a < b\) 为实数。在定义域 \([a, ~ b]\) 上的一个 partition，为一个有限集 \(P = \left\{ a_{0}, ~ a_{1}, ~ a_{2}, \ldots, ~ a_{n} \right\}\)，满足 \(a = a_{0} < a_{1} < a_{2} < \ldots < a_{n} = b\)。Partition \(P\) 从而定义了函数 \(f\) 的 lower Riemann sum 和 upper Riemann sum：

\[\begin{align*} &L(P, f):= \sum\limits^{n}_{i=1}m_{i}\cdot\Delta a_{i} = \sum\limits^{n}_{i=1}\inf_{\scriptscriptstyle a_{i-1}\leq x \leq a_{i}}{f(x)\cdot \left(a_{i}-a_{i-1} \right)}\\ &U(P, f):= \sum\limits^{n}_{i=1}M_{i}\cdot\Delta a_{i} = \sum\limits^{n}_{i=1}\sup_{\scriptscriptstyle a_{i-1}\leq x \leq a_{i}}{f(x)\cdot \left(a_{i}-a_{i-1} \right)} \end{align*} \]

注意其中：

\[\Delta a_{i} = a_{i} - a_{i-1}; \; \; M_{i} = \sup\limits_{a_{i-1} \leq x \leq a_{i}} f(x); \; \; m_{i} = \inf\limits_{a_{i-1} \leq x \leq a_{i}} f(x) \; \; \mbox{ for } \forall i \leq n \]

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Corollary 5.1

对于任意给定的 partition \(P\), 有：

\[L(P, ~ f) \leq U(P, ~ f) \]

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Proof. (Corollary 5.1)

对于 \(\forall i \leq n\)，有 \(\inf\limits_{a_{i-1} \leq x \leq a_{i}} f(x) \leq \sup\limits_{a_{i-1} \leq x \leq a_{i}} f(x)\) 自然成立，则显然：

\[\begin{align*} & \forall i \leq n: ~ \inf\limits_{a_{i-1} \leq x \leq a_{i}} f(x) \cdot \Delta a_{i} ~ = ~ m_{i} \cdot \Delta a_{i} ~ \leq M_{i} ~ \cdot \Delta a_{i} ~ = ~ \sup\limits_{a_{i-1} \leq x \leq a_{i}} f(x) \cdot \Delta a_{i} \\ & \Longrightarrow \\ & \sum\limits^{n}_{i=1}M_{i}\cdot\Delta a_{i} ~ = U(P, f) ~ \leq ~ L(P, f) = ~ \sum\limits^{n}_{i=1} m_{i}\cdot\Delta a_{i} \end{align*} \]

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Corollary 5.2

如果对于函数 \(f\) 的两个 partition \(P, ~ P^{'}\) 满足 \(P \subset P^{'}\)，那么

\[L(P, f) ~ \leq ~ L(P^{'}, f)\\ U(P^{'}, f) ~ \leq ~ U(P,f) \]

这其实在表述：一个更细致的分割（refinement）使 lower Riemann sum 增大（not strictly），使 upper Riemann sum 减小（not strictly）。

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Proof. (Corollary 5.2)

假设对于有界实值函数 \(f: [a, ~ b] \rightarrow \mathbb{R}\)，
令：\(P = \left\{ a_{0}, ~ a_{1}, \ldots, ~ a_{m-1}, ~ a_{m}, \ldots, ~ a_{n-1}, ~ a_{n} \right\}\)，满足：\(1 \leq m \leq n \;\) such that：

\[a = a_{0} < a_{1} < \cdots < a_{m-1} < a_{m} < \cdots a_{n-1} < a_{n} = b \]

令： \(P^{'} = \left\{ a_{0}, ~ a_{1}, \ldots, a_{m-1}, ~ a_{t}, ~ a_{m}, \ldots, ~ a_{n-1}, ~ a_{n} \right\}\) such that：

\[a = a_{0} < a_{1} < \cdots < a_{m-1} < a_{t} < a_{m} < \cdots a_{n-1} < a_{n} = b \]

• 注：这里显然有：\(P \subset P^{'}\)。因此这一步我构造了条件所需的两个 partition：\(P\) 和 \(P^{'}\)。其实我就是往 \(P\) 中某两个相邻元素 \(a_{m-1}, ~ a_{m}\) 中间插了一个新值 \(a_{t}\)。为什么我要用这么麻烦的方法表述？因为此处从 \(a_{0}\) 到 \(a_{n}\) 是由证明需要而严格按照大小排序的，但是，集合具有无序性和互异性。因此，采用一系列下标表示其中元素以及它们对应的大小关系是严格遵守集合性质的做法。为什么我只插入了一个值 \(a_{t}\)？因为根据 partition 的定义，partition 是有限集，而两个有限集 \(P \subset P^{'}\) 只能相差有限个元素，i.e.，\(| P^{'} \Delta P | < \infty\)。因此，\(P \subset P^{'}\) 的一般情况可由相差任意一个元素的情况 by induction 得到。这种 technique 后续文章中不再作解释。

那么：

\[\begin{align*} L(P, f) = & \sum\limits^{n}_{i=1} \left( a_{i} - a_{i-1} \right) \cdot \inf_{a_{i-1} \leq x \leq a_{i}} f(x) \\ L(P^{'}, f) = & \sum\limits^{m-1}_{i=1} \left( a_{i} - a_{i-1} \right) \cdot \inf_{a_{i-1} \leq x \leq a_{i}} f(x) + ~ (a_{t} - a_{m-1}) \cdot \inf_{a_{m-1} \leq x \leq a_{t}} f(x) ~ + ~ \\ & (a_{m} - a_{t}) \cdot \inf_{a_{t} \leq x \leq a_{m}} f(x) ~ + ~ \sum\limits^{n}_{i=m+1} \left( a_{i} - a_{i-1} \right) \cdot \inf_{a_{i-1} \leq x \leq a_{i}} f(x) \end{align*} \]

则：

\[L(P^{'}, f) - L(P, f) = \left( a_{t} - a_{m-1} \right) \cdot \inf_{a_{m-1} \leq x \leq a_{t}} f(x) ~ + ~ (a_{m} - a_{t}) \cdot \inf_{a_{t} \leq x \leq a_{m}} f(x) ~ - ~ (a_{m} -a_{m-1}) \cdot \inf_{a_{m-1} \leq x \leq a_{m}} f(x) \]

可以发现：\(\left( a_{m-1}, ~ a_{t} \right) \subset \left( a_{m-1}, ~ a_{m} \right); \quad \left( a_{t}, ~ a_{m} \right) \subset \left( a_{m-1}, ~ a_{m} \right)\)

所以：\(\inf_{a_{m-1} \leq x \leq a_{m}} \leq \inf_{a_{m-1} \leq x \leq a_{t}}, \quad \inf_{a_{m-1} \leq x \leq a_{m}} \leq \inf_{a_{t} \leq x \leq a_{m}}\)

因此：

\[\begin{align*} L(P^{'}, f) - L(P, f) & \; \geq \; \left( a_{t} - a_{m-1} \right) \cdot \inf_{a_{m-1} \leq x \leq a_{m}} f(x) ~ + ~ (a_{m} - a_{t}) \cdot \inf_{a_{m-1} \leq x \leq a_{m}} f(x) ~ - ~ (a_{m} -a_{m-1}) \cdot \inf_{a_{m-1} \leq x \leq a_{m}} f(x) \\ & = \; 0 \end{align*} \]

所以：$L(P^{'}, f) \geq L(P, f) $ 成立。
(相似地，可证 \(U(P^{'}, f) \leq U(P, f)\)，我懒得写了。)

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Corollary 5.3

对于函数 \(f\) 的任意两个 partition \(P, ~ Q\)， 必有：\(L(P, f) \leq U(Q, f)\)

证明很简单。 因为 \(P \cup Q\) 必定同时为 \(P\) 和 \(Q\) 的 refinement（更细致的分割，也就是 \(P \subset P \cup Q, ~ Q \subset P \cup Q\)），所以：

\[L(P, f) ~ \leq ~ L(P \cup Q, f) ~ \leq ~ U(P \cup Q, f) ~ \leq ~ U(Q, f) \]

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Definition. (Lower, upper and Riemann integral)

令 \(f: [a, ~ b] \rightarrow \mathbb{R}\) 为一个有界实值函数，其中 \(a < b\) 为实数。定义 \(f\) 在 \([a, ~ b]\) 上的 lower integral：

\[\sup\left\{L(P, f):\mbox{ $P$ is a partition on }[a, ~ b]\right\} \]

相似地，定义 \(f\) 在 \([a, ~ b]\) 上的 upper integral：

\[\inf\left\{U(P, f):\mbox{ $P$ is a partition on }[a, ~ b]\right\} \]

当且仅当函数 \(f\) 在 \([a, ~ b]\) 上的 lower integral 等于其 upper integral 时，即：

\[\sup\left\{L(P, f):\mbox{ $P$ is a partition on }[a, ~ b]\right\} = \inf\left\{U(P, f):\mbox{ $P$ is a partition on }[a, ~ b]\right\} = L \]

称 \(f\) 在 \([a, ~ b]\) 上的黎曼积分（Riemann integral）存在，且积分为 \(L\)。

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现在我将举例说明黎曼积分面对上述三种情况时的缺陷。

Example 5.1（Problem with many discontinuities）

Define \(f:[0, ~ 1] \rightarrow \mathbb{R}\) by:

\[f(x) = \begin{cases} 1 & \mbox{if }\; x \in [0, ~ 1]\cap\mathbb{Q}\\ 0 & \mbox{if }\; x \in [0, ~ 1]\setminus\mathbb{Q} \end{cases} \]

The Riemann integral exists if and only if the lower integral equals to the upper integral of \(f\) on \([a, ~ b]\), i.e.,

\[\sup\left\{L(P, f):\mbox{ $P$ is a partition on }[a, ~ b]\right\} = \inf\left\{U(P, f):\mbox{ $P$ is a partition on }[a, ~ b]\right\} \]

For any partition \(P\) and for any interval \([a_{i-1}, ~ a_{i}]\) it gives, since both rational numbers and irrational numbers exist in any interval, thus we have:

\[\begin{align*} &\inf_{\scriptscriptstyle a_{i-1}\leq x \leq a_{i}}{f(x)\cdot(a_{i}-a_{i-1})} = 0\\ &\sup_{\scriptscriptstyle a_{i-1}\leq x \leq a_{i}}{f(x)\cdot(a_{i}-a_{i-1})} = a_{i}-a_{i-1} \end{align*} \]

Then: \(L(P, f) = 0\) and \(U(P,f) = \sum\limits^{n}_{i=1}(a_{i}-a_{i-1}) = 1\) for any partition \(P\), and hence the upper integral does not equal to the lower integral, since:

\[\begin{align*} &\sup\left\{L(P, f):\mbox{ $P$ is a partition on }[a, ~ b]\right\} = 0\\ &\inf\left\{U(P, f):\mbox{ $P$ is a partition on }[a, ~ b]\right\} = 1 \end{align*} \]

Therefore we can conclude that \(f\) is not Riemann integrable, but since there are far more irrational number than rationals numbers, \(f\) should, in some sense, have integral 0. However, the Riemann integral of \(f\) is not defined.

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Example 5.2（Problem with unbounded functions}

Define \(f: [0, 1] \rightarrow \mathbb{R}\) by:

\[f(x)= \begin{cases} \frac{1}{\sqrt{x}} & \text{if }\; 0 < x \leq 1\\ 0 & \text{if }\; x = 0 \end{cases} \]

For any partition \(P:= \left\{a_{0}, a_{1}, ..., a_{n}\right\}\) on [0, 1], for the first interval \([a_{0}, a_{1}] = [0, a_{1}]\), we always have:

\[\sup_{\scriptscriptstyle a_{0}= 0 \leq x \leq a_{1}}f(x) \cdot a_{1} = \infty \]

and then for any partition \(P\) of \([0, 1]\):

\[U(P, f):= \sum\limits^{n}_{i=1}\sup_{\scriptscriptstyle a_{i-1}\leq x \leq a_{i}}{f(x)\cdot(a_{i}-a_{i-1})} = \infty \]

However, we should consider the area under the graph of \(f\) to be \(2\), not \(\infty\), because \(\lim\limits_{a\rightarrow0^{+}}\int^{1}_{a}f(x) ~ dx = \lim\limits_{a\rightarrow 0^{+}}(2-2\sqrt{a}) = 2\). Calculus courses deal with this example by defining \(\int^{1}_{0}\frac{1}{\sqrt{x}} ~ dx = \lim\limits_{a\rightarrow 0^{+}}\int^{1}_{a}\frac{1}{\sqrt{x}} ~ dx\).That is, we can use the approach in Calculus to calculating Riemann integrals for these functions.

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Example 5.3（Problem with unbounded functions)

Using the same approach above, and

\[f(x) = \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{1-x}} \]

Then we would define \(\int^{1}_{0} f(x) ~ dx\) to be:

\[\int^{1}_{0} f(x) ~ dx = \lim\limits_{a\rightarrow 0^{+}}\int^{\frac{1}{2}}_{a} f(x) ~ dx + \lim\limits_{b\rightarrow 1^{-}}\int^{b}_{\frac{1}{2}} f(x) ~ dx \]

However, this technique (i.e. taking Riemann integrals over subdomains and then taking limits) can fail with more complicated functions, as shown in the following example.

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Example 5.4（Problem with unbounded functions)

Let \(\left\{r_{k}\right\}_{k \in \mathbb{Z}^{+}}\), be a sequence which includes each rational number in \((0, ~ 1)\) exactly once. For \(k \in \mathbb{Z}^{+}\), define \(f_{k}: [0, ~ 1] \rightarrow \mathbb{R}\) by:

\[f_{k}(x) = \begin{cases} \frac{1}{\sqrt{x-r_{k}}} & \mbox{ if } \; x > r_{k}\\ 0 & \mbox{ if } x \; \leq r_{k} \end{cases} \]

Define \(f: [0, ~ 1] \rightarrow [0, ~ \infty]\) by:

\[f(x) = \sum\limits^{\infty}_{k=1}\frac{f_{k}(x)}{2^{k}} \]

For every non-empty open sub-interval \((a, ~ b) \subset [0, ~ 1]\), it must contains a rational number \(r_{h} \in (a, ~ b)\), then \(a < r_{h} < b\). Note that \(r_{h}\) is also belongs to the sequence \(\left\{ r_{k}\right\}_{k \in \mathbb{Z}^{+} }\) because \(r_{h}\) is a rational number in \([0, ~ 1]\). Hence for that \(r_{h}\), we have:

\[f_{h}(x) = \begin{cases} \frac{1}{\sqrt{x-r_{h}}} & \mbox{ if } \; x > r_{h}\\ 0 & \mbox{ if } \; x \leq r_{h} \end{cases} \]

For \(x\rightarrow r^{+}_{h}, ~ f_{h}(x)\rightarrow \infty\), meaning \(f_{h}(x)\) is unbounded. Thus, function \(f\) is unbounded on every its non-empty open sub-interval. Therefore, the Riemann integral of \(f\) is undefined on every its non-empty open sub-interval.

However, the area under the graph of each \(f_{k}\) is less than 2:

\[\int^{1}_{0} f_{k}(x) ~ dx =\lim\limits_{\scriptscriptstyle t \rightarrow 1^{-}, s \rightarrow r^{+}_{k}}\int^{t}_{s} \frac{1}{\sqrt{x-r_{k}}} ~ dx \;=\; 2\sqrt{1-r_{k}} \;< \; 2 \]

Then the formula defining \(f: f(x) = \sum\limits^{\infty}_{k=1}\frac{f_{k}(x)}{2^{k}}\) shows that we should expect the area under the graph of \(f\) to be less than 2, instead of being undefined.

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Example 5.5（Problem with point-wise limits)

Let \(\left\{r_{k}\right\}_{k \in \mathbb{Z}^{+}}\), be a sequence which includes each rational number in \([0, ~ 1]\) exactly once. For \(k \in \mathbb{Z}^{+}\), define \(f_{k}: [0, ~ 1] \rightarrow \mathbb{R}\) by:

\[f_{k}(x) = \begin{cases} 1 & \mbox{ if } \; x \in \left\{r_{1},..., r_{k}\right\}\\ 0 & \mbox{ otherwise} \end{cases} \]

Remark: Note that the difference between each function \(f_{k}\) in this example and function \(f\) mentioned in Example 5.1 is that, \(f_{k}\) is a function which takes \(1\) on finite points \(r_{1},...,r_{k}\), while \(f\) is a function which takes \(1\) on all rational numbers on \([0, ~ 1]\).

For any \(k \in \mathbb{Z}^{+}\), function \(f_{k}\) is Riemann integrable: For lower integral, \(L(P, f_{k}) = 0\) for any partition \(P\) on \([0, ~ 1]\), since \(\inf\limits_{\scriptscriptstyle a_{i-1} \leq x \leq a_{i}}f_{k}(x) = 0\) for all \(a_{i-1}, ~ a_{i} \in P\). Hence the lower integral of \(f_{k}\) is \(0\).

For upper integral and the pair \(a_{m-1}, a_{m}\) such that \(r_{k} \in [a_{m-1}, ~ a_{m}]\), we have:

\[\sup\limits_{\scriptscriptstyle a_{m-1} \leq x \leq a_{m}}f_{k}(x)\cdot(a_{m}-a_{m-1}) = a_{m}-a_{m-1}, \]

and \(f_{k} = 0\) on other intervals which do not include \(r_{k}\), thus \(\inf(a_{m}-a_{m-1}) = 0\) by refining partition \(P\). Hence the upper integral of \(f_{k}\) is also \(0\). Because both lower and upper integral are 0, \(f_{k}\) is Riemann integrable with \(\int^{1}_{0}f_{k}(x) ~ dx = 0\).

Define \(f: [0, ~ 1] \rightarrow \mathbb{R}\) by:

\[f(x) = \begin{cases} 1 & \; \mbox{ if x is rational}\\ 0 & \; \mbox{ if x is irrational} \end{cases} \]

Clearly \(\lim\limits_{k \rightarrow \infty}f_{k}(x) = f(x)\) for each \(x \in [0, ~ 1]\).

However, \(f\) is not Riemann integrable (Example 5.1) even though \(f\) is the point-wise limit of a sequence of integrable functions bounded by 1.