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By admin 数学代写 2022年2月23日

First we need to define what is meant by finite total variation.
CHAPTER 9. INTEGRATION
186
Definition 9.3.1 Let $g$ be an arbitrary function defined on $[a, b] .$ For $P_{[a, x]} \equiv$ $\left{x_{0}, \cdots, x_{n}\right}$ a partition of $[a, x]$, define $V\left(P_{[a, x]}, g\right)$ by
$$
\sum_{i=1}^{n}\left|g\left(x_{i}\right)-g\left(x_{i-1}\right)\right|
$$
Define the total variation of $g$ on $[a, x]$ by
$$
V_{[a, x]}(g) \equiv \sup \left{V\left(P_{[a, x]}, g\right): P_{[a, x]} \text { is a partition of }[a, x]\right} .
$$
Then $g$ is said to be of bounded variation on $[a, b]$ if $V_{[a, b]}(g)$ is finite. Also, for $P=$ $\left{x_{0}, x_{1}, \cdots, x_{n}\right}$ where $x_{k} \geq x_{k-1},|P| \equiv \max \left{\left|x_{k}-x_{k-1}\right|: k=1,2, \cdots, n\right}$.

Then with this definition, one has an important proposition which pertains to the case of principal interest here in which the functions are all real valued. The above definition of finite total variation works for functions which have values in some normed linear space however.

Proposition $9.3 .2$ Every real valued function $g$ of bounded variation can be written as the difference of two increasing function, one of which is the function $x \rightarrow V_{[a, x]}(g)$. Furthermore, the functions of bounded variation are exactly those functions which are the difference of two increasing functions.

Proof: Let $g$ be of bounded variation. It is obvious from the definition that $x \rightarrow$ $V_{[a, x]}(g)$ is an increasing function. Also $g(x)=V_{[a, x]}(g)-\left(V_{[a, x]}(g)-g(x)\right)$ The first part of the proposition is proved if I can show $x \rightarrow V_{[a, x]}(g)-g(x)$ is increasing. Let $x \leq y$. Is it true that $V_{[a, x]}(g)-g(x) \leq V_{[a, y]}(g)-g(y) ?$ This is true if and only if
$$
g(y)-g(x) \leq V_{[a, y]}(g)-V_{[a, x]}(g)
$$
To show this is so, first note that $V\left(P_{[a, x]}, g\right) \leq V\left(Q_{[a, x]}, g\right)$ whenever the partition $Q_{[a, x]} \supseteq P_{[a, x]}$. You demonstrate this by adding in one point at a time and using the triangle inequality. Now let $P_{y}$ and $P_{[a, x]}$ be partitions of $[a, y]$ and $[a, x]$ respectively such that
$$
V\left(P_{[a, x]}, g\right)+\varepsilon>V_{[a, x]}(g), V\left(P_{y}, g\right)+\varepsilon>V_{[a, y]}(g)
$$
Without loss of generality $P_{y}$ contains $x$ because from what was just shown you could add in the point $x$ and the approximation of $V\left(P_{y}, g\right)$ to $V_{[a, y]}(g)$ would only be better. Then from the definition,
$$
\begin{gathered}
V_{[a, y]}(g)-V_{[a, x]}(g) \geq V\left(P_{y}, g\right)-\left(V\left(P_{[a, x]}, g\right)+\varepsilon\right) \
\geq|g(y)-g(x)|-\varepsilon \geq g(y)-g(x)-\varepsilon
\end{gathered}
$$
and since $\varepsilon$ is arbitrary, this establishes $9.3$. This proves the first part of the proposition.
Now suppose $g(x)=g_{1}(x)-g_{2}(x)$ where each $g_{i}$ is an increasing function. Why is $g$ of bounded variation? Letting $x0$, there exists $\delta>0$ such that if
$$
P \equiv\left{x_{0}, x_{1}, \cdots, x_{n}\right}, a=x_{0}<\cdots<x_{n}=b
$$
is any partition hoving $|P|<\delta$, and $z_{i} \in\left[x_{i-1}, x_{i}\right]$,
$$
\left|I-\sum_{i=1}^{n} f\left(z_{i}\right)\left(g\left(x_{i}\right)-g\left(x_{i-1}\right)\right)\right|<\varepsilon .
$$
The number $\int_{a}^{b} f(x) d g(x)$ is defined as 1 . I will denote this Riemann Stieltjes sum approximating I as $\sum_{P} f\left(z_{i}\right)\left(g\left(x_{i}\right)-g\left(x_{i-1}\right)\right)$. When $f$ is Riemann Stieltjes integrable on $[a, b]$ with respect to $g$ as just described, this is denoted as $f \in R([a, b], g)$ or simply as $R[a, b]$ if the definition is clear for $g$.
A special case is the following definition.
Definition 9.3.4 The Riemann integral is a special case of the above in which $g(x)=x$. We write $\int_{a}^{b} f(x) d g(x)$ in the form $\int_{a}^{b} f(x) d x$ to signify the Riemann integral.
There is only one possible number $I$ satisfying the above definition.
Lemma 9.3.5 The integral $\int_{a}^{b} f(x) d g(x)$ is well defined in the sense that if there is such a number $I$, then there is only one.

Proof: Suppose you have two of them $I, \hat{I}$ and that $P, \hat{P}$ are corresponding partitions such that $|P|,|\hat{P}|$ are both small enough that
$$
\left|I-\sum_{P} f\left(z_{i}\right)\left(g\left(x_{i}\right)-g\left(x_{i-1}\right)\right)\right|<\varepsilon,\left|\hat{I}-\sum_{\hat{P}} f\left(\hat{z}{i}\right)\left(g\left(x{i}\right)-g\left(x_{i-1}\right)\right)\right|<\varepsilon
$$
whenever $z_{i}$ or $\hat{z}{i}$ are in $\left[x{i-1}, x_{i}\right]$. Let $Q \equiv P \cup \hat{P}$ and choose $z_{i}$ and $\hat{z}{i}$ to be the left endpoint of the sub intervals defined by the partition $Q$. Then $|Q| \leq \min (|P|,|\hat{P}|)$ and so $|I-S|<\varepsilon,|\hat{I}-S|<\varepsilon$ where $S=\sum{Q} f\left(x_{i-1}\right)\left(g\left(x_{i}\right)-g\left(x_{i-1}\right)\right)$. Then $|I-\hat{I}| \leq|I-S|+|S-\hat{I}|<2 \varepsilon$. Since $\varepsilon$ is arbitrary, $I=\hat{I}$.

Next is a fairly obvious theorem which says essentially that things which hold for sums typically hold for integrals also, provided the integrals exist.

Theorem 9.3.6 Assuming all integrals make sense, the following relation exists for $f, g$ functions and $a, b$ scalars.
$$
\int_{a}^{b}(a f+b \hat{f}) d g=a \int_{a}^{b} f d g+b \int_{a}^{b} \hat{f} d g
$$
Assuming all integrals make sense and $g$ is increasing, it follows that
$$
\int_{a}^{b}|f| d g \geq\left|\int_{a}^{b} f d g\right|
$$
Also, if $a<c<b$ and all integrals make sense for $I=[a, c],[a, b],[c, b]$, then

首先，我们需要定义有限总变异的含义。
第 9 章集成
186
定义 9.3.1 令 $g$ 是定义在 $[a, b] 上的任意函数。$ For $P_{[a, x]} \equiv$ $\left{x_{0}, \cdots, x_{ n}\right}$ $[a, x]$ 的一个分区，定义 $V\left(P_{[a, x]}, g\right)$ 为
$$
\sum_{i=1}^{n}\left|g\left(x_{i}\right)-g\left(x_{i-1}\right)\right|
$$
定义 $[a, x]$ 上 $g$ 的总变化
$$
V_{[a, x]}(g) \equiv \sup \left{V\left(P_{[a, x]}, g\right): P_{[a, x]} \text { 是}[a, x]\right} 的分区。
$$
如果 $V_{[a, b]}(g)$ 是有限的，则称 $g$ 在 $[a, b]$ 上有界变化。此外，对于 $P=$ $\left{x_{0}, x_{1}, \cdots, x_{n}\right}$ 其中 $x_{k} \geq x_{k-1},\ |P| \equiv \max \left{\left|x_{k}-x_{k-1}\right|: k=1,2, \cdots, n\right}$。

然后有了这个定义，就有了一个重要的命题，它与这里的主要利益的情况有关，在这种情况下，函数都是实值的。然而，上述有限总变分的定义适用于在某些规范线性空间中具有值的函数。

命题 $9.3 .2$ 每个有界变化的实值函数 $g$ 可以写成两个递增函数的差，其中一个是函数 $x \rightarrow V_{[a, x]}(g)$。此外，有界变分函数正是两个递增函数之差的函数。

证明：令$g$ 是有界变化的。从定义可以看出$x \rightarrow$ $V_{[a, x]}(g)$ 是一个递增函数。还有 $g(x)=V_{[a, x]}(g)-\left(V_{[a, x]}(g)-g(x)\right)$ 证明了命题的第一部分如果我可以显示 $x \rightarrow V_{[a, x]}(g)-g(x)$ 正在增加。设 $x \leq y$。 $V_{[a, x]}(g)-g(x) \leq V_{[a, y]}(g)-g(y) 是真的吗？$ 当且仅当
$$
g(y)-g(x) \leq V_{[a, y]}(g)-V_{[a, x]}(g)
$$
为了证明这一点，首先注意 $V\left(P_{[a, x]}, g\right) \leq V\left(Q_{[a, x]}, g\right)$ 每当分区$Q_{[a, x]} \supseteq P_{[a, x]}$。您可以通过一次添加一个点并使用三角不等式来证明这一点。现在让 $P_{y}$ 和 $P_{[a, x]}$ 分别是 $[a, y]$ 和 $[a, x]$ 的分区，使得
$$
V\left(P_{[a, x]}, g\right)+\varepsilon>V_{[a, x]}(g), V\left(P_{y}, g\right)+\varepsilon> V_{[a, y]}(g)
$$
不失一般性 $P_{y}$ 包含 $x$ 因为从刚刚显示的内容中，您可以将点 $x$ 和 $V\left(P_{y}, g\right)$ 的近似值添加到 $ V_{[a, y]}(g)$ 只会更好。那么根据定义，
$$
\开始{聚集}
V_{[a, y]}(g)-V_{[a, x]}(g) \geq V\left(P_{y}, g\right)-\left(V\left(P_{[a , x]}, g\right)+\varepsilon\right) \
\geq|g(y)-g(x)|-\varepsilon \geq g(y)-g(x)-\varepsilon
\结束{聚集}
$$
由于 $\varepsilon$ 是任意的，因此建立了 $9.3$。这证明了命题的第一部分。
现在假设 $g(x)=g_{1}(x)-g_{2}(x)$ 其中每个 $g_{i}$ 是一个递增函数。为什么 $g$ 的有界变化？让 $x0$，存在 $\delta 的性质>0$ 使得如果
$$
P \equiv\left{x_{0}, x_{1}, \cdots, x_{n}\right}, a=x_{0}<\cdots<x_{n}=b
$$
是任何包含 $|P|<\delta$ 和 $z_{i} \in\left[x_{i-1}, x_{i}\right]$ 的分区，
$$
\left|I-\sum_{i=1}^{n} f\left(z_{i}\right)\left(g\left(x_{i}\right)-g\left(x_{i- 1}\right)\right)\right|<\varepsilon 。
$$
数 $\int_{a}^{b} f(x) d g(x)$ 定义为 1 。我将这个 Riemann Stieltjes 和近似 I 表示为 $\sum_{P} f\left(z_{i}\right)\left(g\left(x_{i}\right)-g\left(x_{i- 1}\右）\右）$。当 $f$ 是 $[a, b]$ 上关于 $g$ 的黎曼斯蒂尔切可积时，这表示为 $f \in R([a, b], g)$ 或简单地表示为 $R [a, b]$ 如果 $g$ 的定义是明确的。
一个特例是下面的定义。
定义 9.3.4 黎曼积分是上述 $g(x)=x$ 的特例。我们将 $\int_{a}^{b} f(x) d g(x)$ 写成 $\int_{a}^{b} f(x) d x$ 的形式来表示 Rie