The following exercise is 3.17 in Folland's Real Analysis, Second Edition:
''Let $(X, \scr{M}, \mu)$ be a $\sigma$-finite measure space, $\scr{N}$ a sub-$\sigma$-algebra of $\scr{M}$, and $\nu = \mu|\scr{N}$. If $f \in L^1(\mu)$, there exists $g \in L^1(\nu)$ (thus $g$ is $\scr{N}$-measurable) such that $\int_E f d\mu = \int_E g d\nu$ for all $E \in \scr{N}$; if $g'$ is another such function then $g = g'$ $\nu$-a.e. (In probability theory, $g$ is called the conditional expectation of $f$ on $\scr{N}$.)''
My question is this: Is Folland missing a hypothesis that $\nu$ is $\sigma$-finite? This is not necessarily true even if $\mu$ is $\sigma$-finite. For example, if $\mu(X) = \infty$ and $\scr{N} = \{X, \emptyset\}$, then $\nu$ is not $\sigma$-finite.
The obvious way to go about solving this exercise is to define a measure $\lambda$ on $(X, \scr{N})$ by $\lambda(E) = \int_E f d\mu$. Then $\lambda <\!\!< \nu$. So if $\nu$ is assumed to be $\sigma$-finite, one may apply the Radon-Nikodym theorem in the form of Exercise 3.14, which I quote here, to obtain the desired result (using Exercise 3.14 eliminates the need to worry about whether $\lambda$ is $\sigma$-finite):
''If $\nu$ is an arbitrary signed measure and $\mu$ is a $\sigma$-finite measure on $(X, \scr{M})$ such that $\nu <\!\!< \mu$, there exists an extended $\mu$-integrable function $f \colon X \to [-\infty, \infty]$ such that $d\nu = f d\mu$.''
Of course, in my proof of Exercise 3.17, I am making the replacements $\nu \to \lambda$, $\mu \to \nu$, and $\scr{M} \to \scr{N}$ in the statement of Exercise 3.14.
Thanks for your help.
Best Answer
That's a good observation, and I think you are right.
In particular, if as in your example we have $\mu(X) = \infty$ and $\mathcal{N} = \{X, \emptyset\}$, then $L^1(\nu)$ contains only the zero function. So if $f \in L^1(\mu)$ is any function with $\int_X f\,d\mu \ne 0$ then there will be no $g \in L^1(\nu)$ with $\int_X f \,d\mu = \int_X g \,d\nu$. Thus without the extra hypothesis the theorem is not true.
Note that in the usual setup of "conditional probability" the measure $\mu$ is actually a probability measure. In particular it is finite, hence so is $\nu$.