勒貝格微分定理

定理敘述

${\displaystyle f\in \mathrm {L_{loc}^{1}} (\mathbb {R} ^{k})}$為实值或复值的局部可積函數，m${\displaystyle \mathbb {R} ^{k}}$勒貝格測度。那麼${\displaystyle \mathbb {R} ^{k}}$幾乎處處x都符合

${\displaystyle \lim _{r\to 0}{\frac {1}{m(B(x,r))}}\int _{B(x,r)}\left|f(y)-f(x)\right|dm(y)=0}$

證明

${\displaystyle (T_{r}f)(x)={\frac {1}{m(B(x,r))}}\int _{B(x,r)}\left|f(y)-f(x)\right|dm(y)}$
${\displaystyle (Tf)(x)=\limsup _{r\to 0}(T_{r}f)(x)}$

${\displaystyle h=f-g}$。由於g連續，有Tg = 0。

${\displaystyle (T_{r}h)(x)\leq {\frac {1}{m(B(x,r))}}\int _{B(x,r)}\left|h\right|dm+|h(x)|}$

${\displaystyle Mh=\sup _{r>0}{\frac {1}{m(B(x,r))}}\int _{B(x,r)}\left|h\right|dm}$。（Mhh哈代－李特爾伍德極大函數。）從上式得

${\displaystyle Th\leq Mh+|h|}$

${\displaystyle Tf\leq Th\leq Mh+|h|}$

Tf > y，則有Mh > y/2或者|h| > y/2。因此${\displaystyle \{Tf>y\}\subset \{Mh>y/2\}\cup \{|h|>y/2\}}$

${\displaystyle m\{Mh>y/2\}\leq 3^{k}(2/y)\|h\|_{\mathrm {L} ^{1}}<3^{k}\cdot 2/(ny)}$

${\displaystyle m\{|h|>y/2\}y/2\leq \|h\|_{\mathrm {L} ^{1}}}$

${\displaystyle m\{|h|>y/2\}\leq 2/(ny)}$

{\displaystyle {\begin{aligned}&m\{Tf>y\}\\&\leq m\{Mh>y/2\}+m\{|h|>y/2\}\\&<2(3^{k}+1)/(ny)\end{aligned}}}

參考

• Rudin, Walter (1987), Real and complex analysis, International Series in Pure and Applied Mathematics (3rd ed.), McGraw-Hill.