覆盖引理的证明

覆盖引理   

令 $E\subset F\subset B_1$ 为可测集. 假设 $\delta\in(0,1)$, 且满足
(1)$|E|<\delta|B_1|$;
(2)对任意的开球 $B\subset B_1$, 如果 $|E\cap B|\geq\delta|B|$, 则 $B\subset F$.
那么 $|E|\leq\mu|F|$ 对某个常数 $0<\mu<1$.

证明: 不失去一般性,  我们假设 $|E|\neq0$. 则由Lebesgue微分定理可知, 存在 $|\tilde{E}|$ 使得 $|\tilde{E}|=|E|$， 且对任意的 $x\in \tilde{E}$, 总存在 $B_x\subset B_1$ 满足 $x\in B_x$ 以及 $|E\cap B_x|=\delta|B_x|$.

 

(其存在性由形变连续性和介值定理可知.M.V.Safonov， 构造如下) 

令$x\in \tilde{E}\subset B_1$, 先选择充分小的 $\theta>0$ 满足
\begin{align}
\frac{|B_\theta(x)\cap \tilde{E}|}{|B_\theta(x)|}>\delta,\ 且 \ B_\theta(x)\subset B_1.
\end{align}
对于 $\lambda\in[0,1]$, 定义 $\{B_{1-(1-\theta)\lambda}(\lambda x)\}$, 实际是线性同伦变化,  容易验证:
\begin{align}
B_{1-(1-\theta)\lambda}(\lambda x)&\subset B_1,\\
x\in B_{\theta}(x)&\subset B_{1-(1-\theta)\lambda}(\lambda x).
\end{align}
当 $\lambda=0,1$, 则 $B^\lambda=B_{1-(1-\theta)\lambda}(\lambda x)$ 变为 $B_1(0)$ 且 $B_\theta(x)$.

注意到$\frac{|E\cap B_1(0)|}{|B_1(0)|}<\epsilon$. 因此, 由连续性, 可选择最大的 $\lambda_x\in(0,1)$ 满足 $|B^\lambda\cap E|=\delta|B^\lambda|$.

对于这样的球族，  注意到他们的半径都不大于1.

由Vitali覆盖引理可知,  我们能选择出至多可数个球 $\{B_i\}$ 满足 $\cup_{x\in \tilde{E}} B_x\subset \cup_{i} 5B_i$. 于是可知
$|F\setminus E|\geq |(F\setminus E)\cap\cup_{i} B_i|\geq \sum_{i} |(F\cap B_i)\setminus (E\cap B_i)|=\sum_{i} (1-\delta)|B_i|\geq\frac{1-\delta}{5^n}|\cup_{i} 5B_i|\geq\frac{(1-\delta)}{5^n}|E|.$ 

因此  $|F|=|F\setminus E|+|E|\geq (1+\frac{(1-\delta)}{5^n})|E|.$ 

证毕!

证明: (另外一个证明)  我们现对上面的球族的并集来做估计.
$|\cup_{x\in \tilde{E}} B_x|=|\cup_{x\in \tilde{E}} B_x\cap E|+|(\cup_{x\in \tilde{E}} B_x)\setminus E|\geq |E|+|(\cup_{i} B_i)\setminus E|$

$=|E|+(1-\delta)\sum_i|B_i|\geq |E|+ \frac{1-\delta}{5^n}|\cup_i5B_i|\geq |E|+\frac{1-\delta}{5^n}|\cup_{x\in \tilde{E}} B_x|$.

我们有

$ (1-\frac{1-\delta}{5^n})|F|\geq (1-\frac{1-\delta}{5^n})|\cup_{x\in \tilde{E}} B_x|\geq|E|$.

 

若将引理变形如下或者其他的等价形式即对$|F|$加条件结论也总是正确的.

令 $E\subset F\subset B_1$ 为可测集. 假设 $\delta\in(0,1)$, 且满足
(1)$|E|<\delta|B_1|$;
(2)对任意的开球 $B\subset B_1$, 如果 $|E\cap B|\geq\delta|B|$, 则 $|F\cap B|\geq\tilde{\delta}|B|$, 其中 $0<\delta<\tilde{\delta}\leq1$.
那么 $|E|\leq\mu|F|$ 对某个常数 $0<\mu<1$.

 

Lihe Wang, 也给出了如下的覆盖引理:

令 $E\subset F\subset B_1$ 为可测集. 假设 $\epsilon\in(0,1)$, 且满足
(1)$|E|<\epsilon |B_1|$;
(2)对任意的 $x\in B_1$ 的开球 $B_r(x)$, 如果 $|E\cap B_r(x)|\geq\epsilon |B_r(x)|$, 则 $B_r(x)\cap B_1\subset F$.
那么 $|E|\leq 20^n\epsilon |F|$，    （且$|E|\leq (1-\frac{1-\epsilon}{20^n})|F|$.）

 证明:  见Wang的文章，与[CC] 书上的CZ分解证明一致. 如下   

Since $|E|<\varepsilon\left|B_{1}\right|$, we see that for almost every $x \in E$, there is an $r_{x}<2$ so that $\left|E \cap B_{r_{x}}(x)\right|=\varepsilon\left|B_{r_{x}}\right|$ and $\left|E \cap B_{r}(x)\right|<\varepsilon\left|B_{r}\right|$ for all $\infty>r>r_{x}$. By Vitali' covering lemma, there are $x_{1}, x_{2}, \cdots$, so that $B_{r_{x_{1}}}\left(x_{1}\right), B_{r_{x_{2}}}\left(x_{2}\right), \cdots$ are disjoint and $\cup_{k} B_{5 r_{x_{k}}}\left(x_{k}\right) \cap B_{1} \supset \tilde{E}$.

From the choice of $B_{r_{x_{k}}}$, we have
$$
\left|E \cap B_{5 r_{x_{k}}}\left(x_{k}\right)\right|<\epsilon\left|B_{5 r_{x_{k}}}\left(x_{k}\right)\right|=5^{n} \epsilon\left|B_{r_{x_{k}}}\left(x_{k}\right)\right|=5^{n}\left|E \cap B_{r_{x_{k}}}\left(x_{k}\right)\right| .
$$
We also notice that
$$
\left|B_{r_{x_{k}}}\left(x_{k}\right)\right| \leq 4^{n}\left|B_{r_{x_{k}}}\left(x_{k}\right) \cap B_{1}\right|
$$
since $x_{k} \in B_{1}$ and $r_{x_{k}} \leq 2$.

Putting everything together,
$$
\begin{aligned}
|E| &=\left|\cup_{k} B_{5 r_{x_{k}}}\left(x_{k}\right) \cap E\right| \\
& \leq \sum_{k}\left|B_{5 r_{x_{k}}}\left(x_{k}\right) \cap E\right| \\
& \leq 5^{n} \sum_{k} \varepsilon\left|B_{r_{x_{k}}}\left(x_{k}\right)\right| \\
& \leq 20^{n} \sum_{k} \varepsilon\left|B_{r_{x_{k}}}\left(x_{k}\right) \cap B_{1}\right| \\
&=20^{n} \varepsilon\left|\cup B_{r_{x_{k}}}\left(x_{k}\right) \cap B_{1}\right| \\
& \leq 20^{n} \varepsilon|F| .
\end{aligned}
$$
This finishes the proof.

 

In the proof of Lihe Wang's lemma, it omit some details for that its proof is elementary, so we give a direct proof.  可将上述式子中的 $20^n$ 改进为$10^n$.
Note that for $B_r(x)$ with $x\in B_1$ and $r\in(0,2)$, by elementary geometry we have
\begin{align}
B_\frac{r}{2}((1-r/2)x)\subset B_r(x)\cap B_1.
\end{align}
Hence, we have
\begin{align}
\frac{|B_r(x)\cap B_1|}{|B_r(x)|}\geq \frac{|B_\frac{r}{2}((1-r/2)x)\cap B_1|}{|B_r(x)|}=\frac{1}{2^n}.
\end{align}
In fact, we have use the fact that
\begin{align}
B_\frac{r}{2}((1-r/2)x)\subset B_1,\ \ \ B_\frac{r}{2}((1-r/2)x)\subset B_r(x),
\end{align}
because for any $y\in B_\frac{r}{2}((1-r/2)x)$, we have
\begin{align}
|y-0|\leq |y-(1-r/2)x|+|0-(1-r/2)x|<\frac{r}{2}+(1-\frac{r}{2})=1,
\end{align}
and
\begin{align}
|y-x|\leq |y-(1-r/2)x|+|(1-r/2)x-x|<\frac{r}{2}+\frac{r}{2}=r.
\end{align}

Therefore, for any $0<r<2$ and $x\in B_1$, we have
\begin{align}
\frac{|B_r(x)\cap B_1|}{|B_r(x)|}\geq\frac{1}{2^n}
\end{align}

Originally, I divide it into two cases, and if $0<r\leq 1+|x|$, there exists a $B_\frac{r}{2}(y)\subset B_r(x)\cap B_1$ , then
\begin{align}
\frac{|B_r(x)\cap B_1|}{|B_r(x)|}\geq \frac{|B_\frac{r}{2}(y)}{|B_r(x)|}= \frac{1}{2^n};
\end{align}
If $1+|x|<r<2$, we have
\begin{align}
\frac{|B_r(x)\cap B_1|}{|B_r(x)|}\geq \frac{|B_1|}{|B_2|}=\frac{1}{2^n},
\end{align}
where we have used the fact that
\begin{align}
B_1\subset B_r(x)
\end{align}
if $1+|x|<r<2$.
As in the above argument, in fact, there is no need to separate $r$ into two cases.

 

以上是不扩张球的情形，一般的如C-Z分解所诱导的测度估计引理需要扩张方体或者球，其实总是可以可以等到类似的结果. 需要注记的是，以上的原始想法都归功于Krylov和Safonov, 或者更早的(Krylov的老师)E.M.Landis, (有人曾在某本书的前言部分说Landis对于极值原理的灵活运用从未被超越）， 其他都是变体而已.