outer measure on R
\[
\newcommand\R{\mathbf R}
\]
Outer Measure on \(\R\)
启发性讨论
Goal
-
To extend integration to a larger class of functions
-
mainly the domain of a function as the union of subsets
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assign a size to each of those subsets
The length $\ell(I)$ of an open interval $I$ is defined by $$
\ell(I)= \begin{cases}b-a & \text { if } I=(a, b) \text { for some } a, b \in \mathbf{R} \text { with } a
We have $A \subseteq \mathbf{R}$, then the size of $A$ shall at most the
sum of len of a sequence of open intervals whose union contains $A$.
(包含关系的集合之和不能比原来的集合的长度还要大)
The outer measure $|A|$ of a set $A \subseteq \mathbf{R}$ is defined by
$|A|=\inf \left\{\sum_{k=1}^{\infty} \ell\left(I_k\right): I_1, I_2, \ldots\right.$
are open intervals such that
$\left.A \subseteq \bigcup_{k=1}^{\infty} I_k\right\}$.
有限集合的外测度为0.
Suppose $A=\left\{a_1, \ldots, a_n\right\}$ is a finite set of real
numbers. Suppose $\varepsilon>0$. Define a sequence $I_1, I_2, \ldots$
of open intervals by $$
I_k= \begin{cases}\left(a_k-\varepsilon, a_k+\varepsilon\right) & \text { if } k \leq n, \\ \varnothing & \text { if } k>n .\end{cases}
$$
Then $I_1, I_2, \ldots$ is a sequence of open intervals whose union
contains $A$. Clearly
$\sum_{k=1}^{\infty} \ell\left(I_k\right)=2 \varepsilon n$. Hence
$|A| \leq 2 \varepsilon n$. Because $\varepsilon$ is an arbitrary
positive number, this implies that $|A|=0$.
## 外测度的好性质
可数集合的测度为0
Every countable subset of $\mathbf{R}$ has outer measure 0 .
对于 $A=\left\{a_1, a_2, \ldots\right\}$
只要选取$I_k=\left(a_k-\frac{\varepsilon}{2^k}, a_k+\frac{\varepsilon}{2^k}\right)$就好了.
外测度保序
Suppose $A$ and $B$ are subsets of $\mathbf{R}$ with $A \subseteq B$.
Then $|A| \leq|B|$.
假设$I_1, I_2, \ldots$ 是开区间, 他们的并包含了B,
那么他们的并一定包含了A.
于是$|A| \leq \sum_{k=1}^{\infty} \ell\left(I_k\right)$. 取下极限就有了.
平移变换(translation)
If $t \in \mathbf{R}$ and $A \subseteq \mathbf{R}$, then the translation
$t+A$ is defined by $$
t+A=\{t+a: a \in A\} .
$$
外侧度平移不变
Suppose $t \in \mathbf{R}$ and $A \subseteq \mathbf{R}$. Then
$|t+A|=|A|$.
这是因为 $\ell$ 两个东西减过之后就会是0.
$|t+A| \leq \sum_{k=1}^{\infty} \ell\left(t+I_k\right)=\sum_{k=1}^{\infty} \ell\left(I_k\right)$
可数可加性
Suppose $A_1, A_2, \ldots$ is a sequence of subsets of $\mathbf{R}$.
Then $$
\left|\bigcup_{k=1}^{\infty} A_k\right| \leq \sum_{k=1}^{\infty}\left|A_k\right| .
$$
- If $\left|A_k\right|=\infty$ for some $k \in \mathbf{Z}^{+}$,
apparently
- $\varepsilon>0$. For each $k \in \mathbf{Z}^{+}$, let
$I_{1, k}, I_{2, k}, \ldots$ be a sequence of open int. contains $A_k$
s.t.
$\sum_{j=1}^{\infty} \ell\left(I_{j, k}\right) \leq \frac{\varepsilon}{2^k}+\left|A_k\right|$.
- thus
$\sum_{k=1}^{\infty} \sum_{j=1}^{\infty} \ell\left(I_{j, k}\right) \leq \varepsilon+\sum_{k=1}^{\infty}\left|A_k\right|$.
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