I was trying to solve the exercise 3.17 from the book of real analysis by Folland and I've found a problem. The first part of the exercise is the following:
Let $(X, M, \mu) $ be a $\sigma$-finite measure space, $ N $ a sub-$\sigma $-algebra of M and $\nu$ the restriction of $\mu $ to $N $. If $ f \in L^1 (\mu) $, there exist $ g \in L^1 (\nu) $ such that $\int_E f \,d\mu = \int_E g\, d\nu $ for all $ E \in N $.
Let $ X $ the set of natural numbers, $\mu $ the counting measure, $ N $ containing just $ X $ and the empty set and $ f $ the characteristic function of the singlepoint $1 $ which is a $ L^1 $ function. Then, we have
$$ 1 = \int_X g\, d\nu ,$$
for some $ g \in L^1 (\nu) $. But if $ h $ is positive and $\nu $-measurable function, we can take an increasing sequence of simple and measurable functions which converges to $ h $. By definition of $ N $, a simple measurable function is a linear combination of the characeristic of $ X $ and the empty set. So, the integral of $ h $ must be 0 or infinity. Then $ L^1 (\nu) $ is trivial, and we have an absurd.
Is there something wrong in this argument? I can't see. And since he doesn't define sub-$\sigma $-algebra before (I'm assuming that is just a $\sigma $-algebra contained in the first), perhaps we need to assume more things. If I could conclude that $ N $ is also $ \sigma$- finite, then we can apply the Lebesgue Radon Nikodym to solve the problem, but the same example above shows that we don't have this in general.
Best Answer
Your counterexample is correct. You have in addition to assume that $\nu$ is also $\sigma$-finite (and apply Radon-Nidokym-Lebesgue, as you say correctly). Folland refers to the example of conditional expectation from probability theory, where $\mu$ is a (probability and hence) finite[!] measure. In this case of course, $\nu$ is also finite (and hence, $\sigma$-finite). Perhaps, when generalizing the statement, the additional assumption on $\nu$ (which is no longer automatic) was just forgotten.