Conditional Expectation for $\sigma$-finite measures

One can define a reasonable notion of conditional expectation for arbitrary localizable measurable spaces, not necessarily σ-finite. This is explained in great detail in the answer to Is there an introduction to probability theory from a structuralist/categorical perspective?

The “pushforward for L_1-spaces” mentioned there is precisely the conditional expectation. Let me offer a few comments and expand this a little bit.

First, one doesn't really need a measure μ to talk about conditional expectations, only a measure class [μ], or equivalently, a σ-ideal N of negligible sets (alias sets of measure 0). Given a set X with a σ-algebra M of measurable subsets and a σ-ideal N of negligible subsets, one can define the set of (finite complex-valued) measures on (X,M,N) as the set of additive functions M→C that vanish on N. Infinite measures can be defined using [0,∞] instead of C. Furthermore, given a faithful finite measure μ on (X,M,N), one can identify the set of μ-integrable functions f with the set of finite measures ν via the isomorphism f↦fν supplied by the Radon-Nikodym theorem.

Given a morphism (X,M,N)→(X',M',N'), one can pushforward a finite measure on (X,M,N) and get a finite measure on (X',M',N'), by taking the preimage of a measurable subset of X' and computing its measure as a subset of X. This is the conditional expectation. In particular, in the notation of the original post, one uses the morphism (X,B,N)→(X,F,N), where N is the σ-ideal of sets with ν-measure zero. Although both the domain and codomain have the same underlying set X, it's best to think of them as different spaces, and the pushforward can then be thought of as the fiberwise integration. (For instance, take X=[0,1]×[0,1] with the Borel σ-algebra and the σ-algebra of “vertical” Borel sets. The resulting morphism will be isomorphic to the projection [0,1]×[0,1]→[0,1], and the pushforward will be the fiberwise integration map.)

Let's illustrate the above construction on the example by Iosif Pinelis. The product of the infinite measure ν and the function f is the finite measure fν. Its pushforward along the map (X,B,∅)→(X,F,∅) can be computed as follows. The codomain is isomorphic to the measurable space consisting of a single point (which is how one should think about it geometrically). By definition, measures on (X,F,∅) can be identified with complex numbers, and the pushforward of a finite measure on (X,B,∅) simply computes the measure of X. Thus, in this example the conditional expectation is the measure on (X,F,∅) that assigns the number 2 (the sum of 1/2^x) to X.

Of course, most analysts are more comfortable with functions rather than measures and prefer to use the Radon-Nikodym theorem with respect to the pushforward of ν to convert the pushforward of fν to an integrable function. However, this is only possible if the pushforward of ν is a faithful semifinite measure, whereas in this example it is a purely infinite, nonsemifinite measure.

However, the failure of the Radon-Nikodym theorem doesn't mean that the conditional expectation doesn't exist, but rather that it exists only as a finite measure that cannot be converted to an integrable function.

$\newcommand{\N}{\mathbb N} \newcommand{\R}{\mathbb R} \newcommand{\B}{\mathcal B} \newcommand{\F}{\mathcal F} \newcommand{\la}{\lambda} \newcommand{\si}{\sigma} \newcommand{\Si}{\Sigma} \renewcommand{\c}{\circ} \newcommand{\tr}{\operatorname{tr}}$

The definition you quoted is correct.

However, there can be no reasonable notion of the conditional expectation without the sigma-finiteness condition, even in the discrete setting. E.g., let $X=\N$, $\B=2^\N$, and let $\F$ be any sigma-algebra over $\N$ containing an infinite atom $A\subseteq\N$; for instance, one may take $\F=\{\emptyset,\N\}$, with $A=\N$. Let $\nu$ be the counting measure on $\B=2^\N$, and let $f(x)=1/2^x$ for $x\in\N$. Then $E_\nu f=1\in\R$.

However, on the atom $A$ one cannot reasonably ascribe any value to the conditional expectation $E_\nu(f|\F)$, because such a value (say $v$) could reasonably be only the $\nu$-average of $f$ on $A$. Indeed, if you take $v=0$, this would imply $\int_A f\,d\nu=0$, which is false; if you take $v\ne0$, this would imply $|\int_A f\,d\nu|=|v|\nu(A)=\infty$, which is also false.

The problem here is that, while the measure $\nu$ is sigma-finite, its restriction $\nu|_\F$ to $\F$ is not.