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# 高等微積分....急急

Let I be a generalized rectangle in R^n and let the function f:I→R be integrable.Denote the interior of I by D. Show that the restriction f:D→R

is integrable and that ∫_I▒f(下限是I)=∫_D▒f (下限是D)

可以麻煩把過程寫出來嗎??

感激不敬...

### 1 Answer

- Scharze spaceLv 71 decade agoFavorite Answer
Claim: If D has Jordan content 0, f is integrable on D then ∫Df(x)dx=0

Proof of claim: Choose a compact rectangle A⊃D Define g(x)=f(x) if x∈D

g(x)=0 if x∈A\D

It suffices to show that ∫Ag(x)dx=0

f is bounded function let M=sup{|f(x)| :x∈D}=sup{|g(x)|:x∈A}=M

Suppose M>0

Since D has Jordan content 0, for ε>0 there is partition P of A ,P={P1 P2,…,PM} so that 0≤U(χD P)<ε/M

Then for any partition P’ finer than P,we have

|R(g,P’)|=|Σif(xi)v(Pi)|≤MΣv(Pi)=MU(χD,P’)≤MU(χD,P)<ε/M*M=ε

So g is integrable and has value 0

Now if I is open rectangle then I=D and ∫Df(x)dx=∫If(x)dx

If not then I=∂I∪D and ∂I∩D=∅

∂I has Jordan content 0 =>∫∂If(x)dx=0

=>∫If(x)dx=∫Df(x)dx+∫∂If(x)dx=∫Df(x)dx

Q.E.D

QED