# 高等数学入门——理解无穷小的阶（微积分的“科学计数法”）

## 微积分的“科学记数法”

### 抓住极限定义的重点

$\lim_{x \to a}f(x) = A$的定义是: 对任意$\varepsilon>0$存在$\delta>0$，使得当$0<|x-a|<\delta$时，成立$|f(x)-A|<\varepsilon$.

### 记号$o,O,\sim$

#### $\sim$，同阶和等价无穷小（大）量

1. 自反性：$f \sim f$
2. 对称性：$f \sim g$当且仅当$g \sim f$
3. 传递性：如果$f \sim g$$g \sim h$，那么一定有$f \sim h$

##### 为什么引入这个概念？

$\alpha \neq 0$, 求极限 $\lim_{x \to 0}\frac{(1+x)^\alpha-1}{x}$

$y=(1+x)^\alpha-1$，那么简单计算就能得到$\ln(1+y)=\alpha\ln(1+x)$。又知道$\ln(1+x)\sim x$，综合一下我们就有 \begin{aligned} \lim_{x \to 0}\frac{y(x)}{x}&=\lim_{x \to 0}\frac{y(x)}{x}\frac{\ln(1+y)}{\ln(1+y)} \\ &=\lim_{x \to 0}\frac{\ln(1+y)}{x}\frac{y}{\ln(1+y)}\\ &=\lim_{x \to 0}\frac{\alpha\ln(1+x)}{x}\lim_{x \to 0}\frac{y}{\ln(1+y)} \\ &=\alpha \cdot 1\cdot 1 \end{aligned} 这里利用了极限的一个基本性质：

#### 记号$o$和$O$又在表达什么？

$f(x)=O(g(x))(x \to a)$的定义是，存在$a$的某个去心邻域和非负常数$M$，使$\left\vert\frac{f(x)}{g(x)}\right\vert \leq M$在这个邻域上恒成立。

# The Lebesgue-Radon-Nikodym theorem and how von Neumann proved it

## An introduction

If one wants to learn the fundamental theorem of Calculus in the sense of Lebesgue integral, properties of measures have to be taken into account. In elementary calculus, one may consider something like $df(x)=f'(x)dx$ where $f$ is differentiable, say, everywhere on an interval. Now we restrict $f$ to be a differentiable and nondecreasing real function defined on $I=[a,b]$. There we got a one-to-one function defined by $g(x)=x+f(x)$

For measurable sets $E\in\mathfrak{M}$, it can be seen that if $m(E)=0$, we have $m(g(E))=0$. Moreover, $g(E) \in \mathfrak{M}$, and $g$ is one-to-one. Therefore we can define a measure like $\mu(E)=m(g(E))$ If we have a relation $\mu(E)=\int_{E}hdm$ (in fact, this is the Radon-Nikodym theorem we will prove later), the fundamental theorem of calculus for $f$ becomes somewhat clear since if $E=[a,x]$, we got $g(E)=[a+f(a),x+f(x)]$, thus we got \begin{aligned} \mu(E)=m(g(E))&=g(x)-g(a)\\ &=f(x)-f(a)+\int_a^xdt \\ &=\int_a^xh(t)dt \end{aligned} which trivially implies $f(x)-f(a)=\int_a^x[h(t)-1]dt$ the function $h$ looks like to be $g'=f'+1$.

We are not proving the fundamental theorem here. But this gives rise to a question. Is it possible to find a function such that $\mu(E)=\int_{E}hdm$ one may write as $d\mu=hdm$ or, more generally, a measure $\mu$ with respect to another measure $\lambda$? Does this $\mu$ exist with respect to $\lambda$? Does this $h$ exist? Lot of questions. Luckily the Lebesgue decomposition and Radon-Nikodym theorem make it possible.

### Notations

Let $\mu$ be a positive measure on a $\sigma$-algebra $\mathfrak{M}$, let $\lambda$ be any arbitrary measure (positive or complex) defined on $\mathfrak{M}$.

We write $\lambda \ll \mu$ if $\lambda(E)=0$ for every $E\in\mathfrak{M}$ for which $\mu(E)=0$. (You may write $\mu \ll m$ in the previous section.) We say $\lambda$ is absolutely continuous with respect to $\mu$.

Another relation between measures worth consideration is being mutually singular. If we have $\lambda(E)=\lambda(A \cap E)$ for every $E \in \mathfrak{M}$, we say $\lambda$ is concentrated on $A$.

If we now have two measures $\mu_1$ and $\mu_2$, two disjoint sets $A$ and $B$ such that $\mu_1$ is concentrated on $A$, $\mu_2$ is concentrated on $B$, we say $\mu_1$ and $\mu_2$ are mutually singular, and write $\mu_1 \perp \mu_2$

Let $\mu$ be a positive $\sigma$-finite measure on $\mathfrak{M}$, and $\lambda$ a complex measure on $\mathfrak{M}$.

• There exists a unique pair of complex measures $\lambda_{ac}$ and $\lambda_{s}$ on $\mathfrak{M}$ such that

$\lambda = \lambda_{ac}+\lambda_s \quad \lambda_{ac}\ll\mu\quad \lambda_s \perp \mu$

• There is a unique $h \in L^1(\mu)$ such that

$\lambda_{ac}(E)=\int_{E}hd\mu$

for every $E \in \mathfrak{M}$.

The unique pair $(\lambda_{ac},\lambda_s)$ is called the Lebesgue decomposition; the existence of $h$ is called the Radon-Nikodym theorem, and $h$ is called the Radon-Nikodym derivative. One also writes $d\lambda_{ac}=hd\mu$ or $\frac{d\lambda_{ac}}{d\mu}=h$ in this situation.

These are two separate theorems, but von Neumann gave the idea to prove these two at one stroke.

If we already have $\lambda \ll \mu$, then $\lambda_s=0$ and the Radon-Nikodym derivative shows up in the natural of things.

Also, one cannot ignore the fact that $m$ the Lebesgue measure is $\sigma$-finite.

## Proof explained

### Step 1 - Construct a bounded functional

We are going to employ Hilbert space technique in this proof. Precisely speaking, we are going to construct a bounded linear functional to find another function, namely $g$, which is the epicentre of this proof.

The boundedness of $\lambda$ is clear since it's complex, but $\mu$ is only assumed to be $\sigma$-finite. Therefore we need some adjustment onto $\mu$.

#### 1.1 Replacing $\mu$ with a finite measure

If $\mu$ is a positive $\sigma$-finite measure on a $\sigma$-algebra $\mathfrak{M}$ in a set $X$, then there is a function $w$ such that $w \in L^1(\mu)$ and $0<w(x)<1$ for every $x \in X$.

The $\sigma$-finiteness of $\mu$ denotes that, there exist some sets $E_n$ such that $X=\bigcup_{n=1}^{\infty}E_n$ and that $\mu(E_n)<\infty$ for all $n$.

Define w_n(x)= \begin{aligned} \begin{cases} \frac{1}{2^n(1+\mu(E_n))}\quad &x \in E_n \\ 0 \quad &x\notin E_n \end{cases} \end{aligned} (you can also say that $w_n=\frac{1}{2^n(1+\mu(E_n))}\chi_{E_n}$), then we have \begin{aligned} w &= \sum_{n=1}^{\infty}w_n \\ \end{aligned} satisfies $0<w<1$ for all $x$. With $w$, we are able to define a new measure, namely $\tilde{\mu}(E)=\int_{E}wd\mu.$ The fact that $\tilde{\mu}(E)$ is a measure can be validated by considering $\int_{E}wd\mu=\int_{X}\chi_{E}wd\mu$. It's more important that $\tilde{\mu}(E)$ is bounded and $\tilde{\mu}(E)=0$ if and only if $\mu(E)=0$. The second one comes from the strict positivity of $w$. For the first one, notice that \begin{aligned} \tilde{\mu}(X) &\leq \sum_{n=1}^{\infty}\tilde{\mu}(E_n) \\ &= \sum_{n=1}^{\infty}\frac{1}{2^n(1+\mu(E_n))} \\ &\leq \sum_{n=1}^{\infty}\frac{1}{2^n} \end{aligned}

#### 1.2 A bounded linear functional associated with $\lambda$

Since $\lambda$ is complex, without loss of generality, we are able to assume that $\lambda$ is a positive bounded measure on $\mathfrak{M}$. By 1.1, we are able to obtain a positive bounded measure by $\varphi=\lambda+\tilde{\mu}$ Following the construction of Lebesgue measure, we have $\int_{X}fd\varphi=\int_{X}fd\lambda+\int_{X}fwd\mu$ for all nonnegative measurable function $f$. Also, notice that $\lambda \leq \varphi$, we have $\left\vert \int_{X}fd\lambda \right\vert \leq \int_{X}|f|d\lambda \leq \int_{X}|f|d\varphi \leq \sqrt{\varphi(X)}\left\Vert f \right\Vert_2$ for $f \in L^2(\varphi)$ by Schwarz inequality.

Since $\varphi(X)<\infty$, we have $\Lambda{f}=\int_{X}fd\lambda$ to be a bounded linear functional on $L^2(\varphi)$.

### Step 2 - Find the associated function with respect to $\lambda$

Since $L^2(\varphi)$ is a Hilbert space, every bounded linear functional on a Hilbert space $H$ is given by an inner product with an element in $H$. That is, by the completeness of $L^2(\varphi)$, there exists a function $g$ such that $\Lambda{f}=\int_{X}fd\lambda=\int_{X}fgd\varphi=(f,g).$ The properties of $L^2$ space shows that $g$ is determined almost everywhere with respect to $\varphi$.

For $E \in \mathfrak{M}$, we got $0 \leq (\chi_{E},g)=\int_{E}gd\varphi=\int_{E}d\lambda=\lambda(E)\leq\varphi(E)$ which implies $0 \leq g \leq 1$ for almost every $x$ with respect to $\varphi$. Therefore we are able to assume that $0 \leq g \leq 1$ without ruining the identity. The proof is in the bag once we define $A$ to be the set where $0 \leq g < 1$ and $B$ the set where $g=1$.

### Step 3 - Generate $\lambda_{ac}$ and $\lambda_{s}$ and the Radon-Nikodym derivative at one stroke

We claim that $\lambda(A \cap E)$ and $\lambda(B \cap E)$ form the decomposition we are looking for, $\lambda_{ac}$ and $\lambda_s$, respectively. Namely, $\lambda_{ac}=\lambda(A \cap E)$, $\lambda_s=\lambda(B \cap E)$.

#### Proving $\lambda_s \perp \mu$

If we combine $\Lambda{f}=(f,g)$ and $\varphi=\lambda+\tilde{\mu}$ together, we have $\int_{X}(1-g)fd\lambda=\int_{X}fgwd\mu.$ Put $f=\chi_{B}$, we have $\int_{B}wd\mu=0.$ Since $w$ is strictly positive, we see that $\mu(B)=0$. Notice that $A \cap B = \varnothing$ and $A \cup B=X$. For $E \in \mathfrak{M}$, we write $E=E_A \cup E_B$, where $E_A \subset A$ and $E_B \subset B$. Therefore $\mu(E)=\mu(E_A)+\mu(E_B)=\mu(E \cap A)+\mu(E \cap B)=\mu(E \cap A).$ Therefore $\mu$ is concentrated on $A$.

For $\lambda_s$, observe that $\lambda_s(E)=\lambda(E \cap B)=\lambda((E \cap B) \cap B)=\lambda_s(E \cap B).$ Hence $\lambda_s$ is concentrated on $B$. This observation shows that $\lambda_s \perp \mu$.

#### Proving $\lambda_{ac} \ll \mu$ by the Radon-Nikodym derivative

The relation that $\lambda_{ac} \ll \mu$ will be showed by the existence of the Radon-Nikodym derivative.

If we replace $f$ by $(1+g+\cdots+g^n)\chi_E,$ where $E \in \mathfrak{M}$, we have $\int_X(1-g)fd\lambda=\int_E(1-g^{n+1})d\lambda=\int_Eg(1+g+\cdots+g^n)wd\mu.$ Notice that \begin{aligned} \int_{E}(1-g^{n+1})d\lambda &=\int\limits_{E \cap A}(1-g^{n+1})d\lambda + \int\limits_{E \cap B}(1-g^{n+1})d\lambda \\ &=\int\limits_{E \cap A}(1-g^{n+1})d\lambda \\ &\to\lambda(E \cap A) = \lambda_{ac}(E)\quad(n\to\infty) \end{aligned} Define $h_n=g(1+g+g^2+\cdots+g^n)w$, we see that on $A$, $h_n$ converges monotonically to h= \begin{aligned} \begin{cases} \frac{gw}{1-g} \quad &x\in{A}\\ 0 \quad &x\in{B} \end{cases} \end{aligned} By monotone convergence theorem, we got $\lim_{n\to\infty}\int_{E}h_nd\mu = \int_{E}hd\mu=\lambda_{ac}(E).$ for every $E\in\mathfrak{M}$.

The measurable function $h$ is the desired Radon-Nikodym derivative once we show that $h \in L^1(\mu)$. Replacing $E$ with $X$, we see that $\int_{X}|h|d\mu=\int_{X}hd\mu=\lambda_{ac}(X)\leq\lambda(X)<\infty.$ Clearly, if $\mu(E)=0$, we have $\lambda_{ac}(E)=\int_{E}hd\mu=0$ which shows that $\lambda_{ac}\ll\mu$ as desired.

### Step 3 - Generalization onto complex measures

By far we have proved this theorem for positive bounded measure. For real bounded measure, we can apply the proceeding case to the positive and negative part of it. For all complex measures, we have $\lambda=\lambda_1+i\lambda_2$ where $\lambda_1$ and $\lambda_2$ are real.

### Step 4 - Uniqueness of the decomposition

If we have two Lebesgue decompositions of the same measure, namely $(\lambda_{ac},\lambda_s)$ and $(\lambda'_{ac},\lambda'_s)$, we shall show that $\lambda_{ac}-\lambda_{ac}'=\lambda_s'-\lambda_s=0$ By the definition of the decomposition we got $\lambda_{ac}-\lambda'_{ac}=\lambda'_s-\lambda_s$ with $\lambda_{ac}-\lambda_{ac}' \ll \mu$ and $\lambda_{s}'-\lambda_{s}\perp\mu$. This implies that $\lambda'_{s}-\lambda_{s} \ll \mu$ as well.

Since $\lambda'_s-\lambda_s\perp\mu$, there exists a set with $\mu(A)=0$ on which $\lambda'_s-\lambda_s$ is concentrated; the absolute continuity shows that $\lambda'_s(E)-\lambda_s(E)=0$ for all $E \subset A$. Hence $\lambda_s'-\lambda_s$ is concentrated on $X-A$. Therefore we got $(\lambda'_s-\lambda_s)\perp(\lambda'_s-\lambda_s)$, which forces $\lambda'_s-\lambda_s=0$. The uniqueness is proved.

(Following the same process one can also show that $\lambda_{ac}\perp\lambda_s$.)

# Counterexamples of Fubini's theorem

## Hypotheses in Fubini's theorem cannot be dispensed with

In this post we proved Fubini's theorem in the sense of Lebesgue measure, which makes it easier to evaluate multi variable integral. But these two classic counterexamples in this post prevent you from using Fubini's theorem without enough consideration.

## Counterexamples

So we said $f(x,y)$ has to be integrable. What if $f$ is not? First let's see this function. $f(x,y)=\frac{y^2-x^2}{(x^2+y^2)^2}=\frac{\partial^2}{\partial x \partial y}\arctan(\frac{y}{x})$ This function is not Lebesgue integrable on $[0,1] \times [0,1]$, since we have \begin{aligned} \iint\limits_{D} |f(x,y)|dxdy & = \int_\varepsilon^1 \int_0^{\frac{\pi}{2}}\frac{r^2|\cos{2\theta}|}{r^4}rd\theta{dr} \\ &=\int_\varepsilon^1\frac{1}{r}dr \\ &\to \infty \quad (\varepsilon \to 0) \end{aligned} where $D=\{(x,y):\varepsilon^2 \leq x^2+y^2 \leq 1, x \geq 0, y \geq 0\}$. Since $D \subset [0,1] \times [0,1]$, we see the integral of $|f(x,y)|$ is not finite. Fubini's theorem fails then since \begin{aligned} \int_0^1 dx \int_0^1 f(x,y)dy &= \int_0^1 \frac{-1}{1+x^2}dx \\ &= -\frac{\pi}{4} \end{aligned} meanwhile \begin{aligned} \int_0^1 dy \int_0^1 f(x,y)dx &= \int_0^1\frac{1}{1+y^2}dy \\ &=\frac{\pi}{4} \end{aligned} See, everything messes up, and the identity disappears. This function is too 'large' for Fubini's, so is the next one.

The following function is generated by series. First consider the sequence on $[0,1]$ generated by $0= \delta_1<\delta_2<\cdots,\delta_n \to 1$

And a sequence of functions $g_n$ generated by $\int_0^1 g_ndx=1$ with supports in $(\delta_n,\delta_{n+1})$.

Define $f(x,y)$ on $[0,1] \times [0,1]$ such that $f(x,y)=\sum_{n=1}^{\infty}[g_n(x)-g_{n+1}(x)]g_n(y)$ The right hand is convergent since for each point $(x,y)$ there is at least one term in this sum that is different from $0$.

An easy computation shows that $\int_0^1\int_0^1 |f(x,y)|dxdy = \infty$ and $\int_0^1 dx \int_0^1 f(x,y)dy = 1 \neq 0 = \int_0^1 dy \int_0^1 f(x,y)dx$

# Fubini's theorem in Euclidean space (Understanding 'almost everywhere')

## General Idea

In elementary calculus, integrals of continuous functions of several variables are often calculated by iterating one-dimensional integrals. But the properties of measurability give rise to a lot of issues for Lebesgue integration on $\mathbb{R}^d$. What we are looking for is the equation $\int_{\mathbb{R}^m}\left(\int_{\mathbb{R}^n}f(x,y)dy\right)dx=\int_{\mathbb{R}^n}\left(\int_{\mathbb{R}^m}f(x,y)dx\right)dy=\int_{\mathbb{R}^d}fdm$ where $d=m+n$ and $m,n$ are positive integers. If this equation holds for $f$, the integration would be relatively easy, as the iteration can be taken in any order. In fact, this equation can be generalized to some other abstract measure space, but that's beyond what this post could cover.

### Notations

For $d=m+n$, we write $\mathbb{R}^d=\mathbb{R}^m\times\mathbb{R}^n$ A point in $\mathbb{R}^d$ therefore takes the form $(x,y)$, where $x\in\mathbb{R}^m$ and $y\in\mathbb{R}^n$. If $f$ is defined on $\mathbb{R}^d$, the slice of $f$ is respectively \begin{aligned} f^y(x)&=f(x,y)\quad y\in\mathbb{R}^n \\ f_x(y)&=f(x,y)\quad x\in\mathbb{R}^m \end{aligned} For $E \subset \mathbb{R}^m\times\mathbb{R}^n$, we defines its slices by \begin{aligned} E^y&=\{x\in\mathbb{R}^m:(x,y)\in{E}\} \\ E_x&=\{y\in\mathbb{R}^n:(x,y)\in{E}\} \end{aligned}

### But why 'almost everywhere'?

Unfortunately, even if we assume that $f$ is measurable on $\mathbb{R}^d$, it can be shown that $f^y$ is not necessarily measurable for each $y$. It's easy to construct a non-measurable set on $\mathbb{R}$ (x-axis), namely $A$. Then $A$ has Lebesgue measure $0$ in $\mathbb{R} \times \mathbb{R}$. But $E^y$ is not measurable for $y=0$. Nevertheless, the consideration of 'almost everywhere' is able to save us from this.

## Fubini's Theorem

Suppose $f(x,y)$ is integrable on $\mathbb{R}^m \times \mathbb{R}^n$. Then for almost every $y \in \mathbb{R}^n$, we have

1. $f^y$ is integrable on $\mathbb{R}^m$.

2. The function defined by $\int_{\mathbb{R}^m}f^y(x)dx$ is integrable on $\mathbb{R}^n$.

3. This equation holds ($m$ denotes the Lebesgue measure on $\mathbb{R}^d$): $\int_{\mathbb{R}^m}\left(\int_{\mathbb{R}^n}f(x,y)dx\right)dy=\int_{\mathbb{R}^n}\left(\int_{\mathbb{R}^m}f(x,y)dy\right)dx=\int_{\mathbb{R}^d}fdm$ The symmetric conclusion can be obtained for $x$.

### General and more rigorous version

The general version of Fubini's theorem is developed in abstract product space, which will not be proved here. But it's worth a peek. Of course, feel free to jump to the next section if you are not interested.

Let $(X,\mathscr{S},\mu)$ and $(Y,\mathscr{T},\lambda)$ be $\sigma$-finite measure spaces, and let $f$ be an $(\mathscr{S} \times \mathscr{T})$-measurable function defined on $X \times Y$.

1. If $f$ is an nonnegative real function, and if $\varphi(x)=\int_Y f_xd\lambda \quad \psi(y) = \int_X f^yd\mu\qquad (x\in X, y \in Y)$ then $\varphi$ is $\mathscr{S}$-measurable, and $\psi$ is $\mathscr{T}$-measurable, and $\int_Xd\mu(x)\int_Yf(x,y)d\lambda(y)=\int_{Y}d\lambda(y)\int_{X}f(x,y)d\mu(x)$

2. If $f$ is complex and if $\varphi^\ast(x)=\int_Y|f|_xd\lambda\quad\text{and}\quad\int_{X}\varphi^\ast d\mu<\infty$ then $f \in L^1(\mu\times\lambda)$.

3. If $f \in L^1(\mu \times \lambda)$, then $f_x \in L^1(\lambda)$ for almost all $x \in X$, $f^y \in L^1(\mu)$ for almost all $y \in Y$. The function therefore defined in 1 a.e. are in $L^1(\mu)$ and $L^1(\lambda)$ respectively, and the equation holds.

Clearly, if we replace $X$, $Y$ with $\mathbb{R}^m$ and $\mathbb{R}^n$, $\mathscr{S}$ and $\mathscr{T}$ with the respective Lebesgue $\sigma$-algebra, $\lambda$ and $\mu$ with Lebesgue measure, then we obtained the Euclidean version. Notice that $f$ is integrable means that $\int_X|f|d\mu < \infty$.

### Before the proof

The proof is relatively long. Instead of proving that $f$ as an integrable function satisfies the three conclusions, we shall show that, however, the family of functions satisfy the three conclusions (say, $\mathcal{F}$) contains all integrable functions. If you check the general version of Fubini's theorem, you see that integrability was explicitly discussed.

First, we shall show that $\mathcal{F}$ is not empty. This is important because we might have been discussing something that never exists. Second. Considering the fact that any integrable function can be "approximated" by simple functions, where simple functions can be generated linear combination, it encourage us to discuss limits and linear combinations in $\mathcal{F}$. Finally, we shall show that if $f$ is integrable, then $f \in \mathcal{F}$. The power of almost-everywhere will show up along the proof.

## Complete proof of Fubini's Theorem (With explanation)

### Step 1 - $\mathcal{F}$ is not empty

It's somewhat absurd to discuss the property of $\mathcal{F}$ without proving that it's not empty. But that can be done easily.

Suppose $E$ is a bounded open cube in $\mathbb{R}^d$ such that $E = Q_1 \times Q_2$, where $Q_1$ and $Q_2$ are open cubes in $\mathbb{R}^m$ and $\mathbb{R}^n$. Then $\chi_E \in \mathcal{F}$.

For each $y$, $\chi_E(x,y)$ is measurable. And the integrability of $\chi_E(x,y)$ follows with \begin{aligned} g(y)=\int_{\mathbb{R}^m}\chi_E(x,y)dx=\begin{cases}\text{vol}(Q_1)\quad &y \in Q_2 \\ 0 \quad &y\notin Q_2\end{cases} \end{aligned} It shows that $g(y)=\text{vol}(Q_1)\chi_{Q_2}$, which is measurable and integrable as well. Further, $\int_{\mathbb{R}^{n}}g(y)dy=\text{vol}(Q_1)\text{vol}(Q_2)$ Since we initially have $\int_{\mathbb{R}^d}\chi_Edm=\text{vol}(E)=\text{vol}(Q_1)\text{vol}(Q_2)$, we see that $\chi_E$ satisfies these three properties, hence $\chi_E \in \mathcal{F}$.

### Step 2 - $\mathcal{F}$ is closed under finite linear combination

We have only judged open cubes in $\mathbb{R}^d$, which are far from Lebesgue $\sigma$-algebra. To get there, we may have to check some $G_\delta$ sets, but we can't do that since we have no idea about limits in $\mathcal{F}$. We are also looking for some simple functions, which are linear combinations of character functions.

Any finite linear combination of functions in $\mathcal{F}$ also belongs to $\mathcal{F}$.

Since there are arbitrarily many bounded open cubes in $\mathbb{R}^d$, we are able to find arbitrarily many members in $\mathcal{F}$. Say, $f_1,f_2,\cdots,f_n\in\mathcal{F}$ Following the definition of $\mathcal{F}$, for each $1 \leq k \leq n$, we are able to find a set $A_k \subset \mathbb{R}^n$ such that $A_k$ has measure $0$ and whenever $y \notin A_k$, $f_k^y$ is integrable on $\mathbb{R}^m$. If we collect these sets altogether, namely $A=\cup A_k$, we see that in $\mathbb{R}^n-A$, all $f_k$'s has the desired property, so does their arbitrary finite linear combination (due to the linear property of Lebesgue integral). Since $A$ has measure zero as well, it turns out that the finite linear combinations belong to $\mathcal{F}$.

### Step 3 - Monotone convergence in $\mathcal{F}$

Limits and convergence come into play. One may think about something like complete metric space, where Cauchy sequences converges. In this step we show that the monotone limit does exist in $\mathcal{F}$.

Suppose $f_k$ is a sequence of measurable functions in $\mathcal{F}$ so that $f_{k} \leq f_{k+1}$ or $f_k \geq f_{k+1}$ holds for all $k$, and $f_k \to f$ where $f$ is integrable on $\mathbb{R}^d$, then $f \in \mathcal{F}$.

Without loss of generality, it suffices to assume that $0 \leq f_1 \leq f_2 \leq \cdots \leq f_n$ Since for other situations, we can take some $-f_k$ or $f_k-f_1$ or something like that. An application of monotone convergence theorem yields that $\lim\limits_{k \to \infty}\int_{\mathbb{R}^d}f_kdm = \int_{\mathbb{R}^d}fdm$ Also, we can find some sets with measure $0$, namely $A_k$, carrying the same meaning as is in Step 2. For $A=\bigcup_{k=1}^{\infty}A_k$, we also have $m(A)=0$ in $\mathbb{R}^n$. Also, for $y \in \mathbb{R}^n - A$, $f_k^y$ is integrable on $\mathbb{R}^m$ for all $k$. Thus by monotone convergence theorem, we see that $g_k(y)=\int_{\mathbb{R}^m}f_k^ydx \to g(y)=\int_{\mathbb{R}^m}f^ydx\quad(k\to\infty)$ Clearly we have $g_k \leq g_{k+1}$ for all $k$, and by assumption, $g_k$ is integrable. Use monotone convergence theorem again, we see that $\int_{\mathbb{R}^n}g_kdy\to\int_{\mathbb{R}^n}gdy \quad (k\to\infty)$ Combining these two limits, we see $\int_{\mathbb{R}^n}gdy =\int_{\mathbb{R}^d}fdm$

We'll show that $f \in \mathcal{F}$ by checking its properties one by one.

1. Since $f$ is integrable, we see that $\int_{\mathbb{R}^n}g = \int_{\mathbb{R}^d}f<\infty$. Thus $g$ is integrable.

2. Since $g$ is integrable, we have $g(y)<\infty$ a.e. for $y$, consequently $f^y$ is integrable a.e. for $y$.

3. By the definition of $g$, we have $\int_{\mathbb{R}^n}\left(\int_{\mathbb{R}^m}f(x,y)dx\right)dy=\int_{\mathbb{R}^d}fdm$

Thus $f \in \mathcal{F}$ as proved.

### Step 4 - Characteristic functions of measurable sets

#### 4.1 - Final destination

We are pretty close to simple functions now. To get rid of infinity, we are going to prove this:

If $E$ is any measurable subset in $\mathbb{R}^d$ with $m(E)<\infty$, then $\chi_E\in\mathcal{F}$.

Once it's done, we can construct simple functions, which approximate to any integrable functions, with ease. Fortunately, with the help of the property of Lebesgue measurable sets, we are able to break "measurable subsets" into several pieces. Recall the fact that

$E \subset \mathbb{R}^d$ is Lebesgue measurable if and only if there are sets $A$ and $B\subset\mathbb{R}^d$ such that $A \subset E \subset B$, $A$ is a $F_{\sigma}$ and $B$ is a $G_{\delta}$, and $m(B-A)=0$.

Since $B-E \subset B-A$, we also have $m(B-E)=0$. Also, since $E \cup (B-E)=B$, $E \cap (B-E) = \varnothing$, we have $\chi_B=\chi_E+\chi_{B-E}$ which is equivalent to$\chi_{E}=\chi_{B}-\chi_{B-E}$. Notice that the right hand of this equation is a finite combination of functions (Step 2 comes into play). If we prove that $\chi_{B},\chi_{B-E} \in \mathcal{F}$, then we are done.

We are going to prove that if $E$ is a $G_{\delta}$ set, or $E$ has measure $0$, then $\chi_{E}\in\mathcal{F}$. That is, we are going to generalize all Lebesgue measurable sets by proving these two key situations.

#### 4.2 - Finite measure $G_{\delta}$ sets

In Step 1 we proved $\chi_{E} \in \mathcal{F}$ if $E$ is a bounded open cube. Now we are going to generalize this to $G_\delta$, which is a countable intersection of open sets. Also, since every open sets can be a countable union of closed cubes ($\mathbb{R}^d$ is a locally compact Hausdorff space in which every open set is $\sigma$-compact). You will see how Step 2 and Step 3 play a role in this section.

##### 4.2.1 - Characteristic function of closed cubes

If $Q$ a closed cube in $\mathbb{R}^d$, then $\chi_{Q} \in \mathcal{F}$.

Since $Q = \text{int}(Q) \cup \partial{Q}$, where $\text{int}(Q)$ denotes its interior and $\partial{Q}$ denotes its boundary, we have $\chi_{Q}=\chi_{\text{int}(Q)}+\chi_{\partial{Q}}$ As proved in Step 1, $\text{int}(Q) \in \mathcal{F}$. So we have to prove that $\chi_{\partial{Q}}\in\mathcal{F}$, and the conclusion follows from Step 2.

Since $m(\partial{Q})=0$, we have $\int_{\mathbb{R}^d}\chi_{\partial{Q}}dm=0$. Also, it can be seen that for almost every $y$, we have $\partial{Q}^y$ has measure $0$ in $\mathbb{R}^m$, and therefore $g(y)=\int_{\mathbb{R}^m}\chi_{\partial{Q}}dx=0$ a.e. for $y$. Consequently, $\int_{\mathbb{R}^n}gdy=0$, therefore $\chi_{\partial{Q}} \in \mathcal{F}$.

##### 4.2.2 - Finitely many almost disjoint closed cubes

Suppose $E = \bigcup_{k=1}^{K}Q_k$, where $Q_k$ is closed cube, and $\text{int}(Q_i)\cap\text{int}(Q_j)=\varnothing$ for $i \neq j$, then $\chi_{E} \in \mathcal{F}$.

This conclusion is obvious if one notice that $\chi_{E} = \sum_{k=1}^{K}\chi_{Q_k}.$ In 4.2.1 we showed that $\chi_{Q_k} \in \mathcal{F}$. Hence $\chi_{E} \in \mathcal{F}$ according to Step 2.

##### 4.2.3 - Arbitrary open sets with finite measure

Since every open sets in $\mathbb{R}^d$ can be a countable union of almost disjoint cubes, we have $E = \bigcup_{k=1}^\infty Q_k$ If we take $E_{K}=\bigcup_{k=1}^{K}Q_k$, we have $f_K=\chi_{E_K}=\sum_{k=1}^{K}\chi_{Q_k}$. And we are going to follow Step 3 to show that $\chi_{E} \in \mathcal{F}$ if $m(E)<\infty$.

Since the Lebesgue $\sigma$-measure contains all Borel sets, and $E$ is open, we see that $E$ is measurable. If $m(E)<\infty$, then we see that $\chi_{E}$ is integrable. Also we have $f_{K+1} \geq f_{K}$ for all $K$, and $f_{K} \to \chi_{E}$; hence $f_{K}$ is what we described in Step 3. Say, $\chi_{E} \in \mathcal{F}$.

##### 4.2.4 - Arbitrary $G_\delta$ sets

If $E$ is a $G_\delta$ set of finite measure, then $\chi_{E} \in \mathcal{F}$.

By the definition of $G_\delta$ sets, we have $E=\bigcap_{k=1}^{\infty}R_k$ where $R_k$ are open sets. Since $m(E)<\infty$, $m$ is regular, we have a open set $S_0 \supset E$ such that $m(S_0)<\infty$. Let $S_k = S_0 \cap \left(\bigcap_{j=1}^{k}R_j\right)$ Then we have $E= \bigcap_{k=1}^{\infty}S_k$ For $S_k$'s, observe that $S_0 \supset S_1 \supset \cdots$, we have $f_k=\chi_{S_k}$ decreases to the limit $f=\chi_E$. Following Step 3, we see that $\chi_E \in \mathcal{F}$.

#### 4.3 - Sets with measure $0$

If $m(E)=0$, then $\chi_{E} \in \mathcal{F}$

If $E$ is a $G_{\delta}$ set, then we are done by following 4.2. If not, it comes to the issue of $m$'s being a complete measure.

Again, by the regularity of $m$, we may choose a set $G$ of $G_\delta$ such that $E \subset G$ and that $m(G)=0$. As proved, $\chi_{G} \in \mathcal{F}$. Therefore $\int_{\mathbb{R}^m}\chi_{G}dx=0\quad\text{for a.e. }y$ Thus, the slice $G^y$ has measure $0$ a.e. for $y$, since $E^y \subset G^y$, we have $E^y$ has measure $0$ a.e. for $y$. Therefore the fact that $\chi_{E}\in\mathcal{F}$ can be verified by simple calculation.

### Step 5 - All integrable functions

If $f$ is integrable, then $f \in \mathcal{F}$.

Like the construction of Lebesgue integral, $f$ has the decomposition that $f= f^+-f^-$. Thus it suffice to prove this for nonnegative $f$ (by Step 1).

There exists a sequence of integrable and nonnegative simple functions $s_k$ that monotonically converges to $f$. Since each integrable $s_k$ is a finite combination of sets with finite measure, by Step 2 and 4, $s_k \in \mathcal{F}$. By Step 3, clearly we have $f \in \mathcal{F}$.

Fubini's theorem shows us that we might be able to evaluate multidimensional integrals in the sense of measure theory with ease (at least 'almost everywhere'). However there are some counterexamples showing that Fubini's theorem will fall, which will be discussed later.

This proof is a good example of how to play with the elements of Lebesgue integral. Let's take a rewind. We want to obtain all integrable functions in $\mathcal{F}$, which however can't be done directly. So we are looking for simple functions, which are generated by characteristic functions. And luckily we obtained a wide enough range of characteristic functions. With linear combinations and limits, we finally achieved the goal to describe all integrable functions. The properties of 'almost everywhere' played a critical role.

# Topological properties of the zeros of a holomorphic function

### What's going on

If for every $z_0 \in \Omega$ where $\Omega$ is a plane open set, the limit $\lim_{z \to z_0}\frac{f(z)-f(z_0)}{z-z_0}$ exists, we say that $f$ is holomorphic (a.k.a. analytic) in $\Omega$. If $f$ is holomorphic in the whole plane, it's called entire. The class of all holomorphic functions (denoted by $H(\Omega)$) has many interesting properties. For example it does form a ring.

But what happens if we talk about the points where $f$ is equal to $0$? Is it possible to find an entire function $g$ such that $g(z)=0$ if and only if $z$ is on the unit circle? The topological property we will discuss in this post answers this question negatively.

### Zeros

Suppose $\Omega$ is a region, the set $Z(f)=\{z_0\in\Omega:f(z_0)=0\}$ is a at most countable set without limit point, as long as $f$ is not identically equal to $0$ on $\Omega$.

Trivially, if $f(\Omega)=\{0\}$, we have $Z(f)=\Omega$. The set of unit circle is not at most countable and every point is a limit point. Hence if an entire function is equal to $0$ on the unit circle, then the function equals to $0$ on the whole plane.

Note: the connectivity of $\Omega$ is important. For example, for two disjoint open sets $\Omega_0$ and $\Omega_1$, define $f(z)=0$ on $\Omega_0$ and $f(z)=1$ on $\Omega_1$, then everything fails.

### A simple application (Feat. Baire Category Theorem)

Before establishing the proof, let's see what we can do using this result.

Suppose that $f$ is an entire function, and that in every power series $f(z)=\sum c_n(z-a)^n$ has at leat one coefficient is $0$, then $f$ is a polynomial.

Clearly we have $n!c_n=f^{(n)}(a)$, thus for every $a \in \mathbb{C}$, we can find a postivie integer $n_0$ such that $f^{(n_0)}(a)=0$. Thus we establish the identity: $\bigcup_{n=0}^{\infty} Z(f^{(n)})=\mathbb{C}$ Notice the fact that $f^{(n)}$ is entire. So $Z(f^{n})$ is either an at most countable set without limit point, or simply equal to $\mathbb{C}$. If there exists a number $N$ such that $Z(f^{N})=\mathbb{C}$, then naturally $Z(f^{n})=\mathbb{C}$ holds for all $n \geq N$. Whilst we see that $f$'s power series has finitely many nonzero coefficients, thus polynomial.

So the question is, is this $N$ always exist? Being an at most countable set without limit points , $Z(f^{(n)})$ has empty interior (nowhere dense). But according to Baire Category Theorem, $\mathbb{C}$ could not be a countable union of nowhere dense sets (of the first category if you say so). This forces the existence of $N$.

### Proof

The proof will be finished using some basic topology techniques.

Let $A$ be the set of all limit points of $Z(f)$ in $\Omega$. The continuity of $f$ shows that $A \subset Z(f)$. We'll show that if $A \neq \varnothing$, then $Z(f)=\Omega$.

First we claim that if $a \in A$, then $a \in \bigcap_{n \geq 0}Z(f^{(n)})$. That is, $f^{(k)}(a) = 0$ for all $k \geq 0$. Suppose this fails, then there is a smallest positive integer $m$ such that $c_m \neq 0$ for the power series on the disc $D(a;r)$: $f(z)=\sum_{n=1}^{\infty}c_n(z-a)^{n}.$

Define

\begin{aligned} ​ g(z)=\begin{cases} ​ (z-a)^{-m}f(z)\quad&(z\in\Omega-\{a\}) \\\ ​ c_m\quad&(z=a) ​ \end{cases} \end{aligned}

It's clear that $g \in H(D(a;r))$ since we have $g(z)=\sum_{n=1}^{\infty}c_{m+n}(z-a)^{n}\quad(z\in D(a;r))$

But the continuity shows that $g(a)=0$ while $c_m \neq 0$. A contradiction.

Next fix a point $b \in \Omega$. Choose a curve (continuous mapping) defined $\gamma$ on $[0,1]$ such that $\gamma(0)=a$ and $\gamma(1)=b$. Let

$\Gamma=\{t\in[0,1]:\gamma(t)\in\bigcap_{n \geq 0}Z(f^{(n)})\}$ By hypothesis, $0 \in \Gamma$. We shall prove that $1 \in \Gamma$. Let $s = \sup\Gamma$ There exists a sequence $\{t_n\}\subset\Gamma$ such that $t_n \to s$. The continuity of $f^{(k)}$ and $\gamma$ shows that $f^{(k)}(\gamma(s))=0$

Hence $s \in \Gamma$. Choose a disc $D(\gamma(s);\delta)\subset\Omega$. On this disc, $f$ is represented by its power series but all coefficients are $0$. It follows that $f(z)=0$ for all $z \in D(\gamma(s);\delta)$. Further, $f^{(k)}(z)=0$ for all $z \subset D(\gamma(s);\delta)$ for all $k \geq 0$. Therefore by the continuity of $\gamma$, there exists $\varepsilon>0$ such that $\gamma(s-\varepsilon,s+\varepsilon)\subset D(\gamma(s);\delta)$, which implies that $(s-\varepsilon, s+\varepsilon)\cap[0,1]\subset\Gamma$. Since $s=\sup\Gamma$, we have $s=1$, therefore $1 \in \Gamma$.

So far we showed that $\Omega = \bigcap_{n \geq 0}Z(f^{(n)})$, which forces $Z(f)=\Omega$. This happens when $Z(f)$ contains limit points, which is equivalent to what we shall prove.

When $Z(f)$ contains no limit point, all points of $Z(f)$ are isolated points; hence in each compact subset of $\Omega$, there are at most finitely many points in $Z(f)$. Since $\Omega$ is $\sigma$-compact, $Z(f)$ is at most countable. $Z(f)$ is also called a discrete set in this situation.

# 准素分解与准素标准型

### 特征向量的推广

Matlab(以及很多支持矩阵运算的编程语言)里，知道了一个方阵的特征值，怎么计算特征向量? 有一个比较形象的解决方案:

null(A-c.*I)

$\text{ker}(\mathscr{A}-c\mathscr{I})=\{\alpha|(\mathscr{A}-c\mathscr{I})\alpha=0\{$

1. $\text{ker}(1)=0$$\text{ker}(m)=V$
2. $g|f$，那么$\text{ker}(g)\subset\text{ker}(f)$
3. $\text{ker}(f)\cap\text{ker}(g)=\text{ker}(\text{gcd}(f,g))$
4. $\text{ker}(f)+\text{ker}(g)=\text{ker}(\text{lcm}(f,g))$
5. $\text{gcd}(f,g)=1$，那么$\text{ker}(fg)=\text{ker}(f)\oplus\text{ker}(g)$

### 空间准素分解——基于极小多项式的广义特征子空间分解

$\mathbb{F}$上任一非常数多项式$f\in\mathbb{F}[X]$均可以唯一分解成不可约多项式乘积，即 $f=p_1p_2\cdots{}p_m$ 其中$p_i$是不可约多项式，且是唯一的(不考虑常数和次序)。对于$i\neq{j}$，一定有$\text{gcd}(p_i,p_j)=1$.

$V=W_1\oplus W_2\oplus\cdots\oplus W_s.$

### 矩阵的准素标准形

#### 准素分解与极小多项式

$\mathscr{A}_i=\mathscr{A}\vert_{W_i}$的极小多项式为$p_i(\lambda)^{r_i}$.

$q=m/p_i^{r_i}$。那么$q$$p_i^{r_i-1}$互质。假设$W_i=\text{ker}(p_i^{r_i-1})$，那么一定有 \begin{equation} \begin{aligned} \text{ker}(p_i^{r_i-1}q)&=W_i\oplus\text{ker}(q) \\ &=W_i\oplus(W_1\oplus\cdots\oplus W_{i-1}\oplus W_{i+1}\oplus\cdots\oplus W_s )\\ &=V \end{aligned} \end{equation}

#### 准素分解和特征多项式

$\mathscr{A}_i$的特征多项式是$p_i(\lambda)^{d_i}$

(注意，这个等式可以看成一个Laplace展开的递归运用。)

### 准素分解的实例，以及如何判断矩阵是否可以对角化

$A$的极小多项式为$m(\lambda)=(\lambda-1)^2(\lambda^2+1)$。那么$\mathbb{R}^4$应该分解为 $\mathbb{R}^4=\text{ker}[(\lambda-1)^2]\oplus\text{ker}(\lambda^2+1)$

null((A-I)^2)

$W_1$中取基$x_1=(0，0，0，1)^T$$x_2=(2，0，1，0)^T$，在$W_2$中取基$(1，0，0，-1)^T$$(1，1，0，-1)^T$，得到矩阵 $P=\begin{bmatrix} 0&2&1&1 \\ 0&0&0&1 \\ 0&1&0&0 \\ 1&0&-1&-1 \end{bmatrix}$

$\mathbb{F}$上的矩阵$A$可以对角化的充分必要条件是，$m(\lambda)$可以写成这种形式 $m(\lambda)=(\lambda-\lambda_1)(\lambda-\lambda_2)\cdots(\lambda-\lambda_s)$ 其中$\lambda_1，\cdots，\lambda_s\in\mathbb{F}$，且互异.

$P^{-1}AP=\text{diag}(A_1，\cdots，A_s)$

# 线性空间的循环分解

### 循环分解

$V$$\mathbb{F}$上的$n$维线性空间，$\mathscr{A}$$V$的线性变换，则 $V=\bigoplus_{i=1}^r\mathbb{F}\big[\mathscr{A}\big]\alpha_i$ 其中，$\mathbb{F} [\mathscr{A}] \alpha_i\neq\{0\}$$\alpha_i\in{V}$生成的循环子空间.

#### 循环分解的证明

$W_1=\mathbb{F}[\mathscr{A}]\alpha_1$，并为方便起见设$W_0=0$. 又可以选取$\beta_2 \in V \setminus W_1$使得 $R_2(\lambda)=P_{\text{cond }W_1,\mathscr{A}}(\lambda)=P_{\text{cond }W_1,\mathscr{A},\beta_2}(\lambda)$

$W_2=\mathbb{F}[\mathscr{A}]\alpha_1+\mathbb{F}[\mathscr{A}]\alpha_2$。现在证明子空间的和为直和。

$P(\lambda)$是一个零化多项式，因此$\alpha=0$。这说明，子空间的和为直和。因此得到了子空间的直和 $W_2=\mathbb{F}[\mathscr{A}]\alpha_1\oplus\mathbb{F}[\mathscr{A}]\alpha_2$

### 有理标准形与特征多项式

$k_i=\text{deg }m_i$，则在对于基 $\alpha_1,\mathscr{A}\alpha_1,\cdots,\mathscr{A}^{k_1-1}\alpha_1,\cdots,\alpha_r,\mathscr{A}\alpha_r,\cdots,\mathscr{A}^{k_r-1}\alpha_r$

1. $V$是循环空间当且仅当$f=m$。也就是说，极小多项式和特征多项式相等。

2. $m|f$$m$$f$有相同的不可约因子(尽管次数可能不同)。也就是说，对于$f$可以唯一分解成不可约因子$f=p_1^{d_1}\cdots p_s^{d_s}$ 那么一定有正整数$r_1,\cdots,r_s$使得 $m=p_1^{r_1}\cdots p_s^{r_s}$ 其中$0<r_i\leq d_i$

3. 循环分解也成了Cayley-Hamilton的证明。因为$m|f$，所以特征多项式$f$一定是零化多项式。

### 循环分解的实例

$f(\lambda)=(\lambda-1)^2(\lambda+1)(\lambda+5) \\ m(\lambda)=(\lambda-1)(\lambda+1)(\lambda+5)$

# 线性变换不变子空间的导子及其性质

### 问题的引入

$W\subset{V}$$V$的子空间，如果对任意$\alpha\in{W}$都有$\mathscr{A}\alpha\in{W}$，那么$W$就是$\mathscr{A}$不变子空间。设多项式$g(\lambda)=\lambda$，那么就有$g(\mathscr{A})\alpha\in{W}$。如果$g(\lambda)$复杂一些会怎样？有没有什么特殊的例子和性质？对于所有$\alpha\in{V}$又是怎样？这就是这篇博客要关注的问题。线性空间的循环分解也要用到这一个工具。

### 导子多项式(Conductor)

$V$是定义在数域$\mathbb{F}$上的有限维线性空间，定义线性变换$\mathscr{A}:V\to{V}$，设$W$$\mathscr{A}$的线性子空间。取$\textbf{v}\in{V}$。那么将$\textbf{v}$映入$W$的导子多项式$P_{\text{cond }W,\mathscr{A},\textbf{v}}(\lambda)$是指满足$P(\mathscr{A})\textbf{v}\in{W}$的次数最小的首项系数为$1$的多项式(首一多项式，monic polynomial); 多项式$P_{\text{cond }W,\mathscr{A}}(\lambda)$是指将全体$\textbf{v}\in{V}$都有$P(\mathscr{A})\in{W}$的次数最小的首项系数为$1$的多项式。

$A=\begin{pmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&0 \end{pmatrix}$

#### 两种导子多项式的存在性、唯一性

$P(\lambda)=\prod_{i=1}^{n}P_{\text{cond }W,\mathscr{A},\textbf{e}_i}(\lambda)$

### 导子的性质

$Q(\lambda)=S(\lambda)P_{\text{cond }W,\mathscr{A}}(\lambda)+R(\lambda)$

（引理1）设两多项式满足$\text{gcd}(P(\lambda),Q(\lambda))=1$，对于向量$\textbf{u},\textbf{v}\in{V}$，有$P_{\text{cond }W,\mathscr{A},\textbf{u}}(\lambda)=P(\lambda)$$P_{\text{cond }W,\mathscr{A},\textbf{v}}(\lambda)=Q(\lambda)$，那么

$P_{\text{cond }W,\mathscr{A},\textbf{u+v}}(\lambda)=P(\lambda)Q(\lambda)$

$P(\mathscr{A})Q(\mathscr{A})(\textbf{u+v})=Q(\mathscr{A})P(\mathscr{A})\textbf{u}+P(\mathscr{A})Q(\mathscr{A})\textbf{v}\in{W}$

$P_{\text{cond }W,\mathscr{A},\textbf{u+v}}(\mathscr{A})(\textbf{u}+\textbf{v})=S(\mathscr{A})Q(\mathscr{A})\textbf{v}+Q(\mathscr{A})S(\mathscr{A})\textbf{u}\in{W}$

（引理2）设$P(\lambda)$是一个在$\mathbb{F}$上不可约的多项式，设有正整数$m$使得$(P(\lambda))^m|P_{\text{cond }W,\mathscr{A}(\lambda)}$，那么存在$\textbf{v}\in{V}$使得$P_{\text{cond },W,\mathscr{A},\textbf{v}}(\lambda)=(P(\lambda))^m$

$P_{\text{cond }W,\mathscr{A}}(\lambda)=(P(\lambda))^mQ(\lambda)$

$(P(\mathscr{A}))^{m-1}Q(\mathscr{A})\textbf{v}_0\notin{W}$

$P_{\text{cond },W,\mathscr{A},\textbf{u}}(\lambda)=(P(\lambda))^m$

$f=\prod_{i=1}^{m}p_i$

$P_{\text{cond },W,\mathscr{A}}(\lambda)=(P_1(\lambda))^{r_1}(P_2(\lambda))^{r_2}\cdots(P_n(\lambda))^{r_n}$

$P_{\text{cond }W,\mathscr{A}}(\lambda)=\prod_{i=1}^{n}P_{\text{cond }W,\mathscr{A},\textbf{v}_i}(\lambda)$

$P_{\text{cond }W,\mathscr{A}}(\lambda)=P_{\text{cond }W,\mathscr{A},\textbf{v}}(\lambda)$

# 线性空间的循环子空间与简单应用

### 矩阵的有理标准形

$C(c_0+c_1t+\cdots+c_{n-1}t^{n-1}+t^n)=\begin{pmatrix} 0&0&\cdots&0&-c_0 \\ 1&0&\cdots&0&-c_1 \\ 0&1&\cdots&0&-c_2 \\ \vdots&\vdots&\ddots&\vdots&\vdots \\ 0&0&\cdots&1&-c_{n-1} \end{pmatrix}$

### 循环空间

$V$是数域$\mathbb{F}$上的线性空间，对于固定的线性变换$\mathscr{A}$，和一个指定的向量$\alpha \in V$，应该有 $\mathscr{A}\alpha, \mathscr{A}^2\alpha, \cdots\in{V}$

### 零化子

$\alpha$的次数最低的首一零化多项式称为最小零化子。显然，零化子是最小零化子的倍。注意这里的零化子是相对于$\alpha$，这和线性变换的零化子是两回事。接下来会通过分析线性相关性找出最小零化子。

#### 如何找到最小零化子

$g(\lambda)$$\alpha$的零化多项式。那么$g(\lambda)$可以写成 $g(\lambda)=m_{\alpha}(\lambda)q(\lambda)+r(\lambda)$

### 循环空间的实例

$\mathscr{A}\varepsilon_1=(1,0,1)^T,\mathscr{A}^2\varepsilon_1=(1,1,1),\mathscr{A}^3\varepsilon_1=(1,2,2)$

$B=\begin{pmatrix} 0&0&-1 \\ 1&0&0 \\ 0&1&2 \end{pmatrix}$

$P=\begin{pmatrix} 1&1&1 \\ 0&0&1 \\ 0&1&1 \end{pmatrix}$

# 抽象Lebesgue积分的构建

### 从一个”难题”入手

$\{f_n\}$是一个定义在$[0，1]$上的连续函数列，且$0\leq f_n\leq 1$. $n\to\infty$时，对任意$x\in[0，1]$$f_n(x)\to{0}$. 求证 $\lim_{n\to\infty}\int_{0}^{1}f_n(x)dx=0$

Riemann积分在讨论函数列的时候往往需要考虑是否一致收敛，这往往很麻烦。19世纪末，很多数学家都主张，高等数学课中的Riemann积分(这也是每个人都要学的)应该被新的一种更普遍、更灵活、更方便解决极限问题的积分替代。那个时期很多的数学家都进行了尝试，Lebesgue的办法可以说是集大成者。粗略地说，Riemann积分是由下面这个和式逼近的: $\sum_{i=1}^{n}f(t_i)\Delta{x_i}$

### $\pi$-系统、$\lambda$-系统、$\sigma$-代数

#### 函数的值域

$\pi$-系统保证了这个集合族在有限次交运算的封闭性。一个最简单的$\pi$-系统是$\mathbb{R}$中所有闭区间(注意把$\varnothing$也算上)构成的集合。两个闭区间的并集必定是闭区间或$\varnothing$，而$\varnothing$和闭区间的并集是$\varnothing$。这就是一个$\pi$-系统。但是不一定保证无穷次运算的封闭，也不保证并集的封闭。

1. $X\in\mathcal{L}$.
2. $A, B\in\mathcal{L}$，且$B\subset{A}$，那么$A-B\in\mathcal{L}$.
3. $A_n\in\mathcal{L}$，且$A_n\subset{A_{n+1}}$，那么有$\bigcup_{n=1}^{\infty}A_n\in\mathcal{L}$.

#### $\sigma$-代数，两个系统的结合

1. $X\in\mathfrak{M}$.
2. $A\in\mathfrak{M}$，那么$A^c\in\mathfrak{M}$(这里$A^c=X-A$).
3. 若对$n=1, 2, \cdots$$A_n\in\mathfrak{M}$，那么$\bigcup A_n\in\mathfrak{M}$.

#### 一些评注和补充

1. 不难证明，$\sigma$-代数既是$\pi$-系统又是$\lambda$-系统。也就是说，它满足这两个系统本身的性质，所以集合的差，有限个集合的交、并自然不在话下。
2. $\sigma$-代数中的元素可以有很多个，比如$\mathcal{P}(X)$，也可以有两个，比如$\{\varnothing，X\}$。实际上，$X$的任何子集族都可以生成一个最小的$\sigma$-代数。特别地，由$X$的全体开子集生成的$\sigma$-代数$\mathcal{B}$是一个有特殊地位的代数，它能和谐地处理连续函数(广义的)。$\mathcal{B}$的元素称为Borel集。
3. $\pi-\lambda$定理(两种系统的关系)：设$\mathcal{P}$$\mathcal{L}$分别是一个$\pi$-系统和一个$\lambda$-系统，而且$\mathcal{P}\subset\mathcal{L}$，设包含$\mathcal{P}$的最小$\sigma$-代数为$\sigma(\mathcal{P})$，那么有$\sigma(\mathcal{P})\subset\mathcal{L}$

### 特征函数、简单函数

1. $0\leq s_1\leq s_2\leq\cdots\leq f$.
2. 对任意$x\in{X}$$s_n(x)\to f(x)(n\to\infty)$.