勒貝格微分定理




數學上，勒貝格微分定理是實分析的一條定理。這條定理大致是說，一個局部可積函數在幾乎每點的值，都是函數在該點為中心的無限小的球上的平均。換言之，該函數的定義域上幾乎處處都是勒貝格點。



定理敘述


設f∈Lloc1(Rk){displaystyle fin mathrm {L_{loc}^{1}} (mathbb {R} ^{k})}{displaystyle fin mathrm {L_{loc}^{1}} (mathbb {R} ^{k})}為实值或复值的局部可積函數，m為Rk{displaystyle mathbb {R} ^{k}}{displaystyle mathbb {R} ^{k}}的勒貝格測度。那麼Rk{displaystyle mathbb {R} ^{k}}{displaystyle mathbb {R} ^{k}}中幾乎處處的x都符合


limr→01m(B(x,r))∫B(x,r)|f(y)−f(x)|dm(y)=0{displaystyle lim _{rto 0}{frac {1}{m(B(x,r))}}int _{B(x,r)}left|f(y)-f(x)right|dm(y)=0}{displaystyle lim _{rto 0}{frac {1}{m(B(x,r))}}int _{B(x,r)}left|f(y)-f(x)right|dm(y)=0}

使上式成立的点称为f{displaystyle f}f的勒贝格点。



證明


因為這定理是關於函數的局部性質，不失一般性，可假設函數f定義在有界集合中，故f為可積函數。


定義



(Trf)(x)=1m(B(x,r))∫B(x,r)|f(y)−f(x)|dm(y){displaystyle (T_{r}f)(x)={frac {1}{m(B(x,r))}}int _{B(x,r)}left|f(y)-f(x)right|dm(y)}{displaystyle (T_{r}f)(x)={frac {1}{m(B(x,r))}}int _{B(x,r)}left|f(y)-f(x)right|dm(y)}

(Tf)(x)=lim supr→0(Trf)(x){displaystyle (Tf)(x)=limsup _{rto 0}(T_{r}f)(x)}{displaystyle (Tf)(x)=limsup _{rto 0}(T_{r}f)(x)}


那麼這定理就是對幾乎處處的x有Tf = 0。只需證對任何y > 0，集合{Tf > y}的測度為零。


對連續函數，這定理顯然成立。連續函數在L1(Rk){displaystyle mathrm {L} ^{1}(mathbb {R} ^{k})}{displaystyle mathrm {L} ^{1}(mathbb {R} ^{k})}中稠密，故此對任意正整數n，有連續函數g使得‖f−g‖L1<1/n{displaystyle |f-g|_{mathrm {L} ^{1}}<1/n}{displaystyle |f-g|_{mathrm {L} ^{1}}<1/n}。


令h=f−g{displaystyle h=f-g}{displaystyle h=f-g}。由於g連續，有Tg = 0。


用三角不等式有


(Trh)(x)≤1m(B(x,r))∫B(x,r)|h|dm+|h(x)|{displaystyle (T_{r}h)(x)leq {frac {1}{m(B(x,r))}}int _{B(x,r)}left|hright|dm+|h(x)|}{displaystyle (T_{r}h)(x)leq {frac {1}{m(B(x,r))}}int _{B(x,r)}left|hright|dm+|h(x)|}

設Mh=supr>01m(B(x,r))∫B(x,r)|h|dm{displaystyle Mh=sup _{r>0}{frac {1}{m(B(x,r))}}int _{B(x,r)}left|hright|dm}{displaystyle Mh=sup _{r>0}{frac {1}{m(B(x,r))}}int _{B(x,r)}left|hright|dm}。（Mh為h的哈代－李特爾伍德極大函數。）從上式得


Th≤Mh+|h|{displaystyle Thleq Mh+|h|}{displaystyle Thleq Mh+|h|}

因為Trf≤Trg+Trh=Trh{displaystyle T_{r}fleq T_{r}g+T_{r}h=T_{r}h}{displaystyle T_{r}fleq T_{r}g+T_{r}h=T_{r}h}，所以有


Tf≤Th≤Mh+|h|{displaystyle Tfleq Thleq Mh+|h|}{displaystyle Tfleq Thleq Mh+|h|}

若Tf > y，則有Mh > y/2或者|h| > y/2。因此{Tf>y}⊂{Mh>y/2}∪{|h|>y/2}{displaystyle {Tf>y}subset {Mh>y/2}cup {|h|>y/2}}{displaystyle {Tf>y}subset {Mh>y/2}cup {|h|>y/2}}


由哈代－李特爾伍德極大不等式得


m{Mh>y/2}≤3k(2/y)‖h‖L1<3k⋅2/(ny){displaystyle m{Mh>y/2}leq 3^{k}(2/y)|h|_{mathrm {L} ^{1}}<3^{k}cdot 2/(ny)}{displaystyle m{Mh>y/2}leq 3^{k}(2/y)|h|_{mathrm {L} ^{1}}<3^{k}cdot 2/(ny)}

由積分的基本性質有


m{|h|>y/2}y/2≤‖h‖L1{displaystyle m{|h|>y/2}y/2leq |h|_{mathrm {L} ^{1}}}{displaystyle m{|h|>y/2}y/2leq |h|_{mathrm {L} ^{1}}}

故得


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

因此


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

因為上式對所有正整數n成立，從而知m{Tf > y}=0。定理得證。



參考


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



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