Radon-Nikodym with a smaller $\sigma$-algebra

Question

Let $(X,M, \mu)$ be a $\sigma$-finite measure. Let $M' \subset M$ be a smaller $\sigma$-algebra. Let $\mu' = \mu|_M$.

Show that there exists a unique $g \in L^1(\mu')$ such that $\int_E f d\mu = \int_E g d\mu'$ for all $E\in M'$ with $\|g\|_1 \leq \|f\|_1$.

My attempt:

I want to use Radon-Nikodym theorem here. But I'm not sure how to deal with the $\mu'$, since the version I learned requires both measures to be $\sigma$-finite.

Thank you for your help.

Answer 1

The question is:

Let $(X,M, \mu)$ be a $\sigma$-finite measure space. Let $M' \subset M$ be a smaller $\sigma$-algebra. Let $\mu' = \mu|_{M'}$.

Show that for all $f \in L^1(\mu)$, there exists a unique $g \in L^1(\mu')$ such that $\int_E f d\mu = \int_E g d\mu'$ for all $E\in M'$ with $\|g\|_1 \leq \|f\|_1$.

This result, as stated, is false. Here is a counter-example.

Consider $(X,M, \mu)$ where $X=\Bbb R$, $M$ is the Borel $\sigma$-algebra and $\mu$ is the Lebesgue measure. We know that $(X,M, \mu)$ is a $\sigma$-finite measure space.

Now, let $M'=\{\emptyset , \Bbb R\}$. Then $M'$ is a $\sigma$-algebra and $M' \subset M$. Let $\mu' = \mu|_{M'}$.

Let $f$ be any function in $L^1(\mu)$ such that $0< \int_{\Bbb R} f d\mu < +\infty$. Note that $g \in L^1(\mu')$ if and only if $g=0$.

So there is no $g \in L^1(\mu')$ such that $\int_{\Bbb R} f d\mu = \int_{\Bbb R} g d\mu'$

Remark: The key issue is that, even $\mu$ being $\sigma$-finite, $\mu'$ may not be $\sigma$-finite.

If we add to the question the additional hypothesis that $\mu'$ is $\sigma$-finite, then the result holds.

Just define on $M'$, the measure $\nu$ defined by, for all $E \in M'$, $\nu(E)= \int_E f d\mu$. Clearly, $\nu \ll \mu'$ and since now $\mu'$ is $\sigma$-finite, we can apply Radon-Nikodym theorem and conclude that there exists a unique $g \in L^1(\mu')$ such that, for all $E\in M'$, $$\int_E f d\mu = \nu(E)= \int_E g d\mu'$$

Then, let $\textrm{sign}(g) = \chi_{[g>0]} - \chi_{[g<0]}$. Since $g$ is $M'$-measurable function, so is $\textrm{sign}(g)$. So we have \begin{align*} \int_X |g| d\mu' & = \int_X \textrm{sign}(g) g d\mu'= \int_X \textrm{sign}(g) d\nu = \int_X \textrm{sign}(g) f d\mu \leqslant \int_X |f| d\mu \end{align*} So $\|g\|_1 \leq \|f\|_1$.

Answer 2

I suppose $f$ is a given element of $L^{1}(\mu)$.

Consider $(X,M)$ with the measures $\mu'$ and $v$ wher $\nu (E)=\int_E f d\mu$. $\nu$ is real (or comlex) measure and $\nu << \mu'$. Hence. there is a unique $g \in L^{1}(\mu')$ such that $\nu(E)=\int_E gd\mu'$ for al $E \in M'$.

Using a simple function approximation you can see that $\int ghd\mu' =\int fh d\mu$ for any bounded measurable function $h$. Taking $h=\frac {|g|}g$ when $g \neq 0$ and $1$ when $g=0$ we se that $\int |g|d\mu' =\int fhd\mu\leq \int|f|d\mu$ since $|h|=1$