0
$\begingroup$

I'm trying to do solve a problem from Ziemer's lecture notes Modern Real Analysis (Exercise 6.38). The statement is the following:

Let $(X, M, \mu)$ be a $\sigma$-finite measure space, and let $f\in L^1(X,\mu)$. In particular $f$ is $M$-measurable. Suppose $M_0\subset M$ be a $\sigma$-algebra. Of course, $f$ may not be $M_0$-measurable. However, prove that there is a unique $M_0$-measurable function $f_0$ such that $$\int fgd\mu=\int f_0gd\mu$$ for each $M_0$-measurable $g$ for which the integrals are finite. Hint: Use the Radon-Nikodym Theorem.

I tried to apply Radon-Nikodym Theorem in various ways to see what happen but nothing works. Any hint will be really appreciated.

2 Answers 2

5

Hint:

Suppose that $f\geq 0$.

Let $\nu$ be the measure defined on $(X,M)$ by $$ \nu(E):=\int_E f\,d\mu. $$

Now, let $\mu_0=\mu\vert_{M_0}$ and $\nu_0=\nu\vert_{M_0}$ be the restriction of these $M$-measures to $M_0$. These are both measures on $M_0$; and, since $\nu\ll\mu$, we also have $\nu_0\ll\mu_0$.

  • 0
    Thanks!!! I forgot that $v(E)=\int_E f d\mu$ implies $\int g dv = \int fg d\mu$. – 2017-02-10
1

Hint: scale (I mean, prove for indicator functions by Radon nicodym, then extend to simple by linearity and then to $L^1$ by convergence theorems)

  • 0
    You are right, man... I completely forgot this property. I'de already show it in Bartle Problem 8.N. thanks for the help – 2017-02-10