1. ## questions about lebesgue measure

$(X,F,\mu)=([-1,1],\Lambda,\lambda)$ , where $\Lambda$ is a sigma algebra of Lebesgue measurable sets and $\lambda$ is Lebesgue measure. Define $v(A)= \int_A x d\lambda$ for all $A \in \Lambda$.

a) prove $v$ is a signed measure.
b) prove $v << \lambda$
c)Find the Hahn-Jordan decomposition of $v$ and its total variation.

i dont even know how to get start. please help me. and im not sure what << means on b).

2. 1) varify that v satisfy the definition of sigh measure.
2) the v << u notation mean v is absolutely continous with respect to u, that is, if A is a measureable set with u(A)=0, then v(A)=0.
3)make sure that you understand the Hahn-Jordan decomposition and total variation of a measure in your textbook.

3. Originally Posted by Shanks
1) varify that v satisfy the definition of sigh measure.
2) the v << u notation mean v is absolutely continous with respect to u, that is, if A is a measureable set with u(A)=0, then v(A)=0.
3)make sure that you understand the Hahn-Jordan decomposition and total variation of a measure in your textbook.
on 1), the definition says that v assumes at most one of $\infty$, $-\infty$. but how do i compute this integral? $\int_A x d\lambda \neq \int_A x dx$, correct? how do i compute the smallest value and the biggest value it assumes? could you help me on this?

4. the equality holds since they define the same integral in the sense of lebesgue integration.
for the second problem,let f(x)=x, what is the positive part of f, and the negative part? the biggesr value it assumes is the integral of the positive part of f, similarly for the smallest value.

5. Originally Posted by PRLM
on 1), the definition says that v assumes at most one of $\infty$, $-\infty$. but how do i compute this integral? $\int_A x d\lambda \neq \int_A x dx$, correct? how do i compute the smallest value and the biggest value it assumes? could you help me on this?
i am really not sure if this is correct so i would like someone to confirm what i did is correct, please.

1). to show that v is a signed measure, i verified that v checked the definition.
a)v assumes at most one of $\infty ,-\infty$.
$vA=\int_A x d\lambda = \int_A x dx$ since $x$ is bounded on a finite interval and it is reimann integrable. and since the integral is all finite on any set $A \subset [-1,1]$, v does not assume any of $\infty,-\infty$.
b) $v(\emptyset)=0$ since the integral over a set of measure zero is zero.
c) by the equation $\int \Sigma f =\Sigma \int f$, it is easy to verify.
Therefore it is a signed measure.

2).To show $|v|<<\mu$, show that for each subset $A$ such that $\mu A=0$, $|v|A=0$. again since the integral over a set of measure zero is zero, it is obvious.

i am not sure what $\frac{dv}{d\mu}$ is. but i assume that it is Randon Nikodym derivative. Since v is a signed measure, by restricting v to a positive set, we get a measure. So v is restricted to [0,1]. So $\frac{dv}{d\mu}$ = x $\in [0,1]$.

3)Hahn-Jordan decomposition is $v^+(E)=v(E \cap [0,1]),v^-(E)=-v(E \cup [-1,0))$. $|v|(X)=\int_0^1 xdx-\int_{-1}^0 xdx = 1$.

please someone correct me if im wrong.

6. the Radon Nikodym derivative is f(x)=x, x belongs to [-1,1].

7. Originally Posted by Shanks
the Radon Nikodym derivative is f(x)=x, x belongs to [-1,1].
i thought that Radon N derivative is a nonnegative function.

8. No, it is possible that the Radon Nikodym derivative assumes negative value.

9. Originally Posted by Shanks
No, it is possible that the assumes negative value.
would you explain how u get Radon Nikodym derivative ?

10. One way to do this is by the definition of derivative of a measure with respect to another measure.(it is complicated )
another easier way is using the Radon Nikodym theorem. it asserts the existence and uniqueness of Radon Nikodym derivative of a measure which is absolutely continous with respect to another measure.
for more information you can see Rudin's book Real and Complex Analysis.